LABORATORIO QCA.UDEA

VER LA SIGUIENTE DIRECCION DEL WIKIBLOG PARA ESTUDIAR LOS TEMAS REFERENTES A SEGURIDAD Y REACCIONES DE METATESIS, VER AL LADO SUPERIOR IZQUIERDO LOS TEMAS QUE APARECEN COMO INDICE Y DAR CLICK AHI:

http://hernanquiroz.wetpaint.com

Escrito por hdqtitm el 15/04/2007 19:20 | Comentarios (0)

COMO GRAFICAR SEMILLEROS

LEE CADA TEMA Y AL FINAL HAZ CLICK PARA LA VISUALIZACION INTERACTIVA.

http://www.xtec.es/~mgarc127/

http://www.xtec.es/~mgarc127/manualillo.html

Escrito por hdqtitm el 13/03/2007 15:27 | Comentarios (0)

LECTURAS Y TALLERES DE CALCULO Y QUIMICA (HERNAN QUIROZ )

LABORATORIO QCA.UDEA

VER LA SIGUIENTE DIRECCION DEL WIKIBLOG PARA ESTUDIAR LOS TEMAS REFERENTES A SEGURIDAD Y REACCIONES DE METATESIS, VER AL LADO SUPERIOR IZQUIERDO LOS TEMAS QUE APARECEN COMO INDICE Y DAR CLICK AHI:

http://hernanquiroz.wetpaint.com

Escrito por hdqtitm el 15/04/2007 19:20 | Comentarios (0)

COMO GRAFICAR SEMILLEROS

LEE CADA TEMA Y AL FINAL HAZ CLICK PARA LA VISUALIZACION INTERACTIVA.

http://www.xtec.es/~mgarc127/

http://www.xtec.es/~mgarc127/manualillo.html

Escrito por hdqtitm el 13/03/2007 15:27 | Comentarios (0)

DOMINIO DE UNA FUNCION

El dominio de una función está formado por aquellos valores de x (números reales) para los que se puede calcular la imagen f(x).

Ejemplos:

Los primeros puntos de la gráfica que se pueden hallar, son los puntos de la función que pertenecen a los ejes coordenados.

Para hallar el punto donde la función corta al eje de ordenadas (eje Y) se resuelve el sistema:

Para hallar los puntos donde la función corta al eje de abscisas (eje X) se resuelve el sistema:

Ejemplo:

Punto de corte con el eje OY :

Puntos de corte con el eje OX :

Por tanto los puntos de corte con los ejes de coordenadas son:

TABLA DE VALORES
X	Y
0	2
1	0
2	0
-1/2	0

FUNCIÓN PAR

Una función f es PAR cuando:

Las funciones pares son simétricas respecto del eje de ordenadas (eje OY).

Ejemplo:

FUNCIÓN IMPAR

Una función f es IMPAR cuando:

Las funciones impares son simétricas respecto del origen de coordenadas.

Ejemplo:

FUNCIÓN PERIÓDICA

Una función f es PERIÓDICA cuando existe un número

tal que:

(los valores de la función se repiten de p en p).
El número p se llama periodo.

Ejemplo:

Escrito por hdqtitm el 09/03/2007 03:32 | Comentarios (0)

RIEMANN ITM

HAZ CLICK PARA VER EL TEMA:

http://www.mat.uson.mx/eduardo/calculo2/area/practica3/practica3.htm

Escrito por hdqtitm el 28/02/2007 23:58 | Comentarios (0)

TALLER INTEGRALES ITM

SI QUIERES VER LA SOLUCION HAZ CLICK SOBRE CADA EJERCICIO.

1.-

2.-

3.-

4.-

5.-

6.-

7.-

8.-

9.-

10.-

11.-

12.-

13.-

14.-

15.-

16.-

17.- Una función f es PAR cuando:

Las funciones pares son simétricas respecto del eje de ordenadas (eje OY).

Ejemplo:

FUNCIÓN IMPAR

Una función f es IMPAR cuando:

Las funciones impares son simétricas respecto del origen de coordenadas.

Ejemplo:

FUNCIÓN PERIÓDICA

Una función f es PERIÓDICA cuando existe un número

tal que:

(los valores de la función se repiten de p en p).
El número p se llama periodo.

Ejemplo:

Escrito por hdqtitm el 09/03/2007 03:32 | Comentarios (0)

RIEMANN ITM

HAZ CLICK PARA VER EL TEMA:

http://www.mat.uson.mx/eduardo/calculo2/area/practica3/practica3.htm

Escrito por hdqtitm el 28/02/2007 23:58 | Comentarios (0)

TALLER INTEGRALES ITM

SI QUIERES VER LA SOLUCION HAZ CLICK SOBRE CADA EJERCICIO.

1.-

2.-

3.-

4.-

5.-

6.-

7.-

8.-

9.-

10.-

11.-

12.-

13.-

14.-

15.-

16.-

17.-

18.-

19.-

20.-

Escrito por hdqtitm el 28/02/2007 23:48 | Comentarios (0)

TRIGONOMETRIA SEMILLERO

ECUACIONES TRIGONOMÉTRICAS

Se suponía que este tema era el último del año y que con él terminábamos el programa de la asignatura. Para variar otra vez le erramos.

Entre cuecas, empanadas y un brindis por nuestro Chile, nos vamos con la primera "patita" de ecuaciones trigonométricas.

La ecuación trigonométrica es una igualdad que se cumple para ciertos valores del argumento.

Resolver una de estas ecuaciones, significa encontrar el valor del ángulo que satisface dicha ecuación. (A veces es más de un valor).

Ejemplo:

Resolvamos la ecuación trigonométrica para 0º < x < 90º

Aquí determinamos, sin problema, el ángulo x, acordándonos de los valores anteriormente aprendidos. En otra situaciones tendremos que recurrir a la calculadora.

Resolvamos ahora la ecuación

Ahora a resolver la guía de ejercicios, que a la larga, no fue tan traumática como la de identidades, ¿o será que estamos mejorando?

SIGNOS DE LAS FUNCIONES TRIGONOMÉTRICAS

El profe nos comentó lo fácil que era obtener los signos de las funciones trigonométricas, ya que sólo bastaba determinar las de seno y coseno, y a partir de ellos los restantes. como ya no creemos en la palabra fácil optamos por "ver para creer".

Con lo anterior, y aplicando las identidades trigonométricas fundamentales, considerando sólo su signo, obtenemos que:

III

seno

coseno

tangente

cotangente

18.-

19.-

20.-

Escrito por hdqtitm el 28/02/2007 23:48 | Comentarios (0)

TRIGONOMETRIA SEMILLERO

ECUACIONES TRIGONOMÉTRICAS

Se suponía que este tema era el último del año y que con él terminábamos el programa de la asignatura. Para variar otra vez le erramos.

Entre cuecas, empanadas y un brindis por nuestro Chile, nos vamos con la primera "patita" de ecuaciones trigonométricas.

La ecuación trigonométrica es una igualdad que se cumple para ciertos valores del argumento.

Resolver una de estas ecuaciones, significa encontrar el valor del ángulo que satisface dicha ecuación. (A veces es más de un valor).

Ejemplo:

Resolvamos la ecuación trigonométrica para 0º < x < 90º

Aquí determinamos, sin problema, el ángulo x, acordándonos de los valores anteriormente aprendidos. En otra situaciones tendremos que recurrir a la calculadora.

Resolvamos ahora la ecuación

Ahora a resolver la guía de ejercicios, que a la larga, no fue tan traumática como la de identidades, ¿o será que estamos mejorando?

SIGNOS DE LAS FUNCIONES TRIGONOMÉTRICAS

Con lo anterior, y aplicando las identidades trigonométricas fundamentales, considerando sólo su signo, obtenemos que:

	I	II	III	IV
seno	+	+	-	-
coseno	+	-	-	+
tangente	+	-	+	-
cotangente	+	-	+	-
secante	+	-	-	+
cosecante	+	+	-	-

GUÍA DE EJERCICIOS Nº 4

Sabemos, porque el profe lo explicó, que estas ecuaciones trigonométricas son un primer apronte ya que vendrán otra de mayor nivel de dificultad y sin limitaciones, pero por ahora a "luchar" con esta nueva "guiamanía"

Resuelve las siguientes ecuaciones trigonométricas, pero considerando en su solución que

IDENTIDADES TRIGONOMÉTRICAS

Llegó el momento más esperado (por el profe). Aquí nos prometió que no veríamos una en este tipo de ejercicios, pero sí muchos unos. Con ese mensaje y conociéndolo, decidimos poner mucha atención y prepararnos a conciencia para evitar la muerte del electivo por desangre.

Lo primero fue entender que una identidad trigonométrica es una igualdad que contiene razones trigonométricas y que es verdadera, cualesquiera sean los valores que se asignen a los ángulos para los cuales están definidas estas razones.

Para verificar este tipo de ejercicios, Danny nos puso como condición el trabajar con un solo miembro de la identidad, transformándolo hasta lograr la identidad con el otro miembro. Luego nos dio algunos ejemplos que nos hicieron pensar que tal vez no era tan terrible la cosa como la pintaban (un nuevo error).

RAZONES TRIGONOMÉTRICAS DE 0º, 90º, 180º, 270º Y 360º

Te preguntamos: ¿qué ocurre con las funciones trigonométricas de seno y coseno para 0º?. A nosotros también nos preguntaron lo mismo y mientras nos mirábamos, pensábamos: si el ángulo de 0º ya no es ángulo, ¿para qué nos complicamos?. El profe en la pizarra hacía unos dibujos, seguramente sonriendo, por habernos dejado con la mansa interrogante.

Al mirar las figuras hechas, nos dimos cuenta que el valor de seno va disminuyendo a medida que el ángulo disminuye, llegando a ser 0 para 0º. Para el coseno pasa lo contrario, a medida que disminuye el ángulo su valor aumenta hasta ser 1, que es la medida del radio del círculo goniométrico.

Y aquí vino lo que siempre esperamos en este glorioso electivo: el "palo". Tarea: hagan el mismo proceso anterior para todas las funciones trigonométricas y obtengan los valores mencionados en el título de esta unidad. (Se ve fea la cosa)

Al final todos llegamos con nuestros valores, que resumimos aquí:

	0º	90º	180º	270º	360º
seno	0	1	0	-1	-
secante	+	-	-	+
cosecante	+	+	-	-

GUÍA DE EJERCICIOS Nº 4

Resuelve las siguientes ecuaciones trigonométricas, pero considerando en su solución que

IDENTIDADES TRIGONOMÉTRICAS

RAZONES TRIGONOMÉTRICAS DE 0º, 90º, 180º, 270º Y 360º

Al final todos llegamos con nuestros valores, que resumimos aquí:

	0º	90º	180º	270º	360º
seno	0	1	0	-1
coseno	1	0	-1	0	1
tangente	0		0		0
cotangente		0		0
secante	1		-1		1
cosecante		1		-1

GUÍA DE EJERCICIOS Nº 4

Resuelve las siguientes ecuaciones trigonométricas, pero considerando en su solución que

Escrito por hdqtitm el 28/02/2007 23:41 | Comentarios (0)

LIBRO TEORIA CALCULO II.ITM

AQUI ENCONTRARAS UN LIBRO DE CALCULO INTEGRAL PARA QUE LEAS LA TEORIA Y PRINCIPIOS BASICOS DE LOS TEMAS QUE SE HAN VISTO:

http://ladyada.usach.cl/~calculo/capit4.pdf

Escrito por hdqtitm el 22/02/2007 17:51 | Comentarios (0)

SUMAS DE RIEMANN.ITM

coseno

-1

tangente

cotangente

secante

-1

cosecante

-1

GUÍA DE EJERCICIOS Nº 4

Resuelve las siguientes ecuaciones trigonométricas, pero considerando en su solución que

Escrito por hdqtitm el 28/02/2007 23:41 | Comentarios (0)

LIBRO TEORIA CALCULO II.ITM

AQUI ENCONTRARAS UN LIBRO DE CALCULO INTEGRAL PARA QUE LEAS LA TEORIA Y PRINCIPIOS BASICOS DE LOS TEMAS QUE SE HAN VISTO:

http://ladyada.usach.cl/~calculo/capit4.pdf

Escrito por hdqtitm el 22/02/2007 17:51 | Comentarios (0)

SUMAS DE RIEMANN.ITM

HACER CLICK EN ESTA DIRECCION E INTENTA HACER LOS EJERCICIOS QUE APARECEN AQUI:

http://matematicas.uis.edu.co/calculo2/sumas.pdf

Escrito por hdqtitm el 22/02/2007 17:46 | Comentarios (0)

TABLA DE PRIMITIVAS.ITM

Cálculo de primitivas. - Ejercicios 21 al 33. Enlaces a las explicaciones en video

HACER CLICK EN ESTA DIRECCION E INTENTA HACER LOS EJERCICIOS QUE APARECEN AQUI:

http://matematicas.uis.edu.co/calculo2/sumas.pdf

Escrito por hdqtitm el 22/02/2007 17:46 | Comentarios (0)

TABLA DE PRIMITIVAS.ITM

Cálculo de primitivas

Tabla de primitivas inmediatas

Primitiva inmediata	Ejemplos
Primitiva de una suma

Primitiva de una constante por una función

Funciones potenciales



Funciones exponenciales



Funciones logarítmicas


Funciones trigonométricas

Escrito por hdqtitm el 20/02/2007 15:40 | Comentarios (0)

DERIVADAS.ITM

TABLAS USADAS SOLO PARA DERIVADAS:

DERIVADAS

Tabla de primitivas inmediatas

Primitiva inmediata	Ejemplos
Primitiva de una suma

Primitiva de una constante por una función

Funciones potenciales



Funciones exponenciales



Funciones logarítmicas


Funciones trigonométricas

Escrito por hdqtitm el 20/02/2007 15:40 | Comentarios (0)

DERIVADAS.ITM

TABLAS USADAS SOLO PARA DERIVADAS:

DERIVADAS

Problemas resueltos de Derivadas

Fecha de primera versión: 30-08-98

Fecha de última actualización: 30-08-98

Los problemas de derivadas son muy fáciles. Casi se puede garantizar que se resuelven todos. No ocurre lo mismo con las integrales.

Problema 1:

Problema 2.

Problema 3.

Problema 4.

Problema 5.

Problema 6.

Problema 7.

Problema 8.

Problema 9.

Problema 10.

Problema 11.

Volver a

BUSCAR LA DERIVADA PARA ESTOS EJERCICIOS:

a) f(x) = 5x- 2

b) g(x) = 3x² + 2x - 7

c) h(x) =

d) p(t) =

e) q(z) =

f) r(x) = (x + 3)(x - 2)

g) s(x) = e^2x+1

Escrito por hdqtitm el 14/02/2007 23:48 | Comentarios (0)

TALLER DE SERIES

http://webpersonal.uma.es/~ipcabrera/sernumericas.pdf

Escrito por hdqtitm el 06/02/2007 17:10 | Comentarios (0)

GRAFICADOR.AREAS

HAZ CLICK EN ESTA DIRECCION PARA HALLAR EL AREA ENTRE CURVAS, LEE BIEN LAS INSTRUCCIONES DE INGRESO DE LAS RESPECTIVAS FUNCIONES.

http://cs.jsu.edu/mcis/faculty/leathrum/Mathlets/twocurves.html#applettop

Escrito por hdqtitm el 05/12/2006 15:53 | Comentarios (0)

DIBUJO

primitivas

Escrito por hdqtitm el 04/12/2006 21:59 | Comentarios (0)

ANIMACIONES

Animations

Escrito por hdqtitm el 04/12/2006 17:32 | Comentarios (0)

AREAS Y VOLUMENES

Problemas resueltos de Derivadas

Fecha de primera versión: 30-08-98

Fecha de última actualización: 30-08-98

Los problemas de derivadas son muy fáciles. Casi se puede garantizar que se resuelven todos. No ocurre lo mismo con las integrales.

Problema 1:

Problema 2.

Problema 3.

Problema 4.

Problema 5.

Problema 6.

Problema 7.

Problema 8.

Problema 9.

Problema 10.

Problema 11.

Volver a

BUSCAR LA DERIVADA PARA ESTOS EJERCICIOS:

a) f(x) = 5x- 2

b) g(x) = 3x² + 2x - 7

c) h(x) =

d) p(t) =

e) q(z) =

f) r(x) = (x + 3)(x - 2)

g) s(x) = e^2x+1

Escrito por hdqtitm el 14/02/2007 23:48 | Comentarios (0)

TALLER DE SERIES

http://webpersonal.uma.es/~ipcabrera/sernumericas.pdf

Escrito por hdqtitm el 06/02/2007 17:10 | Comentarios (0)

GRAFICADOR.AREAS

HAZ CLICK EN ESTA DIRECCION PARA HALLAR EL AREA ENTRE CURVAS, LEE BIEN LAS INSTRUCCIONES DE INGRESO DE LAS RESPECTIVAS FUNCIONES.

http://cs.jsu.edu/mcis/faculty/leathrum/Mathlets/twocurves.html#applettop

Escrito por hdqtitm el 05/12/2006 15:53 | Comentarios (0)

DIBUJO

primitivas

Escrito por hdqtitm el 04/12/2006 21:59 | Comentarios (0)

ANIMACIONES

Animations

Escrito por hdqtitm el 04/12/2006 17:32 | Comentarios (0)

AREAS Y VOLUMENES

Escrito por hdqtitm el 03/12/2006 17:40 | Comentarios (0)

TABLAS DE INTEGRALES

LOUIS LEITHOLD

(1925-2005)

5)1925-2005)

Louis Leithold

(1925-2005)

Tablas de integrales

Escrito por hdqtitm el 03/12/2006 17:40 | Comentarios (0)

TABLAS DE INTEGRALES

LOUIS LEITHOLD

(1925-2005)

5)1925-2005)

Louis Leithold

(1925-2005)

Tablas de integrales

Escrito por hdqtitm el 19/10/2006 21:32 | Comentarios (0)

CALCULO II

NOTACION SIGMA

The following problems involve the algebra (manipulation) of summation notation. Summation notation is used to define the definite integral of a continuous function of one variable on a closed interval. Let's first briefly define summation notation. If f(i) represents some expression (function) involving i, then $\displaystyle{ \sum_{i=1}^{n} f(i) }$ has the following meaning :

$\displaystyle{ \sum_{i=1}^{n} f(i) }= f(1) + f(2) + f(3) + ... + f(n)$ .

Escrito por hdqtitm el 19/10/2006 21:32 | Comentarios (0)

CALCULO II

NOTACION SIGMA

$\displaystyle{ \sum_{i=1}^{n} f(i) }= f(1) + f(2) + f(3) + ... + f(n)$ .

FONT>
The "i=" part underneath the summation sign tells you which number to first plug into the given expression. The number on top of the summation sign tells you the last number to plug into the given expression. You always increase by one at each successive step. For example,
$\displaystyle{ \sum_{i=1}^{4} (2+i^2) } = (2+1^2) + (2+2^2) + (2+3^2) + (2+4^2)$
= 3 + 6 + 11 + 18
= 38 .
We will need the following well-known summation rules.
$\displaystyle{ \sum_{i=1}^{n} c = c + c + c + ... + c }$ (n times) = cn, where c is a constant.
$\displaystyle{ \sum_{i=1}^{n} i = 1 + 2 + 3 + ... + n = { n(n+1) \over 2 } }$ .
$\displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = { n(n+1)(2n+1) \over 6 } }$ .
$\displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = { n^2(n+1)^2 \over 4 } }$ .

Most of the following problems are average. A few are somewhat challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by using the formulas given above in exactly the form that they are given. For instance, make sure that a summation begins with i=1 before using the above formulas.
PROBLEM 1 : Evaluate $\displaystyle{ \sum_{i=0}^{3} (5 +\sqrt{ 4^i }) }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Evaluate $\displaystyle{ \sum_{i=1}^{100} (4 + 3i) }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Evaluate $\displaystyle{ \sum_{i=1}^{200} (i-3)^2 }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Evaluate $\displaystyle{ \sum_{i=15}^{150} (4i+1) }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Evaluate $\displaystyle{ \sum_{i=1}^{50} \big[ \ln(i+3) - \ln(i+2) \big] }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Evaluate $\displaystyle{ \sum_{i=25}^{150} \Big\{ { 1 \over i+4 }- { 1 \over i+5 } \Big\} }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 7 : Evaluate $\displaystyle{ \sum_{i=10}^{80} (i^3 + i^2) }$ .
Click HERE to see a detailed solution to problem
The "i=" part underneath the summation sign tells you which number to first plug into the given expression. The number on top of the summation sign tells you the last number to plug into the given expression. You always increase by one at each successive step. For example,
$\displaystyle{ \sum_{i=1}^{4} (2+i^2) } = (2+1^2) + (2+2^2) + (2+3^2) + (2+4^2)$
= 3 + 6 + 11 + 18
= 38 .
We will need the following well-known summation rules.
$\displaystyle{ \sum_{i=1}^{n} c = c + c + c + ... + c }$ (n times) = cn, where c is a constant.
$\displaystyle{ \sum_{i=1}^{n} i = 1 + 2 + 3 + ... + n = { n(n+1) \over 2 } }$ .
$\displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = { n(n+1)(2n+1) \over 6 } }$ .
$\displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = { n^2(n+1)^2 \over 4 } }$ .

Most of the following problems are average. A few are somewhat challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by using the formulas given above in exactly the form that they are given. For instance, make sure that a summation begins with i=1 before using the above formulas.
PROBLEM 1 : Evaluate $\displaystyle{ \sum_{i=0}^{3} (5 +\sqrt{ 4^i }) }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Evaluate $\displaystyle{ \sum_{i=1}^{100} (4 + 3i) }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Evaluate $\displaystyle{ \sum_{i=1}^{200} (i-3)^2 }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Evaluate $\displaystyle{ \sum_{i=15}^{150} (4i+1) }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Evaluate $\displaystyle{ \sum_{i=1}^{50} \big[ \ln(i+3) - \ln(i+2) \big] }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Evaluate $\displaystyle{ \sum_{i=25}^{150} \Big\{ { 1 \over i+4 }- { 1 \over i+5 } \Big\} }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 7 : Evaluate $\displaystyle{ \sum_{i=10}^{80} (i^3 + i^2) }$ .
Click HERE to see a detailed solution to probl 7.

PROBLEM 8 : Evaluate $\displaystyle{ \sum_{i=1}^{20} \sin(i\pi/2) }$ .
Click HERE to see a detailed solution to problem 8.

PROBLEM 9 : Evaluate $\displaystyle{ \sum_{i=7}^{32} \cos(i\pi) }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Prove that $\displaystyle{ \sum_{i=1}^{n} i = { n(n+1) \over 2 } }$ .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Prove that $\displaystyle{ \sum_{i=1}^{n} i^2 = { n(n+1)(2n+1) \over 6 } }$ .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Evaluate $\displaystyle{ \lim_{n \to \infty }\sum_{i=1}^{n} { 1 \over i(i+1) } }$ .
Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Evaluate $\displaystyle{ \lim_{n \to \infty }\sum_{i=1}^{n} { 3 \Big(1 + {1 \over n }\Big)^2 { 1 \over n } } }$ .
Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Compute the sum of the first 120 numbers in the following list : 3, 7, 11, 15, 19, 23, 27, ... .
Click HERE to see a detailed solution to problem 14.
UTILIZANDO LA DEFINICION DE LIMITE

$\displaystyle{ \int^{b}_{a} f(x) \, dx} = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} }$ .
We will need the following well-known summation rules.
$\displaystyle{ \sum_{i=1}^{n} c = c + c + c + \cdots + c }$ (n times) $= nc$ , where $c$ is a constant
$\displaystyle{ \sum_{i=1}^{n} i = 1 + 2 + 3 + \cdots + n = { n(n+1) \over 2 } }$
$\displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = { n(n+1)(2n+1) \over 6 } }$
$\displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = { n^2(n+1)^2 \over 4 } }$
$\displaystyle{ \sum_{i=1}^{n} k f(i) } = \displaystyle{ k \sum_{i=1}^{n} f(i) }$ , where $k$ PROBLEM 8 : Evaluate $\displaystyle{ \sum_{i=1}^{20} \sin(i\pi/2) }$ .
Click HERE to see a detailed solution to problem 8.

PROBLEM 9 : Evaluate $\displaystyle{ \sum_{i=7}^{32} \cos(i\pi) }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Prove that $\displaystyle{ \sum_{i=1}^{n} i = { n(n+1) \over 2 } }$ .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Prove that $\displaystyle{ \sum_{i=1}^{n} i^2 = { n(n+1)(2n+1) \over 6 } }$ .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Evaluate $\displaystyle{ \lim_{n \to \infty }\sum_{i=1}^{n} { 1 \over i(i+1) } }$ .
Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Evaluate $\displaystyle{ \lim_{n \to \infty }\sum_{i=1}^{n} { 3 \Big(1 + {1 \over n }\Big)^2 { 1 \over n } } }$ .
Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Compute the sum of the first 120 numbers in the following list : 3, 7, 11, 15, 19, 23, 27, ... .
Click HERE to see a detailed solution to problem 14.
UTILIZANDO LA DEFINICION DE LIMITE

$\displaystyle{ \int^{b}_{a} f(x) \, dx} = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} }$ .
We will need the following well-known summation rules.
$\displaystyle{ \sum_{i=1}^{n} c = c + c + c + \cdots + c }$ (n times) $= nc$ , where $c$ is a constant
$\displaystyle{ \sum_{i=1}^{n} i = 1 + 2 + 3 + \cdots + n = { n(n+1) \over 2 } }$
$\displaystyle{ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = { n(n+1)(2n+1) \over 6 } }$
$\displaystyle{ \sum_{i=1}^{n} i^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = { n^2(n+1)^2 \over 4 } }$
$\displaystyle{ \sum_{i=1}^{n} k f(i) } = \displaystyle{ k \sum_{i=1}^{n} f(i) }$ , where $k$ is a constant
$\displaystyle{ \sum_{i=1}^{n} (f(i) \pm g(i)) } = \displaystyle{ \sum_{i=1}^{n} f(i) \pm \sum_{i=1}^{n} g(i) }$
Most of the following problems are average. A few are somewhat challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by using the formulas given above in exactly the form that they are given. Solutions to the first eight problems will use equal-sized subintervals and right-hand endpoints as sampling points as shown in equations (*) and (**) above.

PROBLEM 1 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{4}_{0} 5 \, dx }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{0} (2x+3) \, dx }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{0}_{-4} (x-2) \, dx }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{3}_{0} (x^2-1) \, dx }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{0} (x^2-x+3) \, dx }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{-2} (3x^2+2) \, dx }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 7 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{4}_{0} x^3 \, dx }$ .
Click HERE to see a detailed solution to problem 7.

PROBLEM 8 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{2}_{0} e^x \, dx }$ .
Click HERE to see a detailed solution to problem 8.

PROBLEM 9 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( 2 \Big({i \over n}\Big)^2 + {i \over n} \Big) \Big({1 \over n}\Big) }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : is a constant
$\displaystyle{ \sum_{i=1}^{n} (f(i) \pm g(i)) } = \displaystyle{ \sum_{i=1}^{n} f(i) \pm \sum_{i=1}^{n} g(i) }$
Most of the following problems are average. A few are somewhat challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by using the formulas given above in exactly the form that they are given. Solutions to the first eight problems will use equal-sized subintervals and right-hand endpoints as sampling points as shown in equations (*) and (**) above.

PROBLEM 1 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{4}_{0} 5 \, dx }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{0} (2x+3) \, dx }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{0}_{-4} (x-2) \, dx }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{3}_{0} (x^2-1) \, dx }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{0} (x^2-x+3) \, dx }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{1}_{-2} (3x^2+2) \, dx }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 7 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{4}_{0} x^3 \, dx }$ .
Click HERE to see a detailed solution to problem 7.

PROBLEM 8 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{2}_{0} e^x \, dx }$ .
Click HERE to see a detailed solution to problem 8.

PROBLEM 9 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( 2 \Big({i \over n}\Big)^2 + {i \over n} \Big) \Big({1 \over n}\Big) }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( 5 + {3i \over n} \Big)^4 \Big({2 \over n}\Big) }$ .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 1-i+2n \over 1-i+n } \Big({1 \over n}\Big) }$ .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( {12 \over n} + {8i \over n^2} \Big) }$ .
Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 9i+3n \over 3in+2n^2} }$ .
Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{b}_{a} K \, dx }$ , where $K$ is a constant. Use an arbitrary partition $a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b$ and arbitrary sampling numbers $c_{i}$ for $i = 1, 2, 3, ..., n$ .
Click HERE to see a detailed solution to problem 14.

PROBLEM 15 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{b}_{a} {1 \over x^2} \, dx }$ . Use an arbitrary partition $a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b$ and the sampling number $c_{i} = \sqrt{ x_{i-1}x_{i} }$ for $i = 1, 2, 3, ..., n$ . Begin by showing that $x_{i-1} < c_{i} < x_{i}$ for $i = 1, 2, 3, ..., n$ . Assume that $0 < a < b$ .
Click HERE to see a detailed solution to problem 15.

Escrito por hdqtitm el 19/10/2006 00:10 | Comentarios (0)

DERIVADAS

DERIVATIVES USING THE LIMIT DEFINITION

The following problems require the use of the limit definition of a derivative, which is given by

$tex2html_wrap_inline338$ .

They range in difficulty from easy to somewhat challenging. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by making proper use of functional notation and careful use of basic algebra. Keep in mind that the goal (in most cases) of these types of problems is to be able to divide out the $tex2html_wrap_inline29$ .

Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 1-i+2n \over 1-i+n } \Big({1 \over n}\Big) }$ .

Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( {12 \over n} + {8i \over n^2} \Big) }$ .

Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Write the following limit as a definite integral : $\ \ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 9i+3n \over 3in+2n^2} }$ .

Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{b}_{a} K \, dx }$ , where $K$ is a constant. Use an arbitrary partition $a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b$ and arbitrary sampling numbers $c_{i}$ for $i = 1, 2, 3, ..., n$ .

Click HERE to see a detailed solution to problem 14.

PROBLEM 15 : Use the limit definition of definite integral to evaluate $\displaystyle{ \int^{b}_{a} {1 \over x^2} \, dx }$ . Use an arbitrary partition $a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b$ and the sampling number $c_{i} = \sqrt{ x_{i-1}x_{i} }$ for $i = 1, 2, 3, ..., n$ . Begin by showing that $x_{i-1} < c_{i} < x_{i}$ for $i = 1, 2, 3, ..., n$ . Assume that $0 < a < b$ .

Click HERE to see a detailed solution to problem 15.

Escrito por hdqtitm el 19/10/2006 00:10 | Comentarios (0)

DERIVADAS

DERIVATIVES USING THE LIMIT DEFINITION

The following problems require the use of the limit definition of a derivative, which is given by

$tex2html_wrap_inline338$ .

PROBLEM 1 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline35$ .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline39$ .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline43$ .
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline47$ .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline51$ .
This problem may be more difficult than it first appears.
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Use the limit definition to compute the derivative, f'(x), for
$tex2html_wrap_inline55$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Use the limit definition to compute the derivative, f'(x), for
$f(x) = \displaystyle { x - 1 \over x^2 + 3x }$ .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Use the limit definition to compute the derivative, f'(x), for
$f(x) = \sqrt{ x^3 - x }$ .
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : Assume that
$f(x) = \cases{ 2 + \sqrt{ x }, & if $\space x \ge 1 $\space \cr \displaystyle{ 1 \over 2 } x + \displaystyle{ 5 \over 2 } , & if $ x < 1 $\space . \cr }$
Show that f is differentiable at x=1, i.e., use the limit definition of the derivative to compute f'(1) .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Assume that
$f(x) = \cases{ x^2 \sin \Big( \displaystyle{ 1 \over x } \Big), & if $\space x \ne 0 $\space \cr \ \ \ \ \ 0 \ \ \ \ \ , & if $ x = 0 $\space . \cr }$
Show that f is differentiable at x=0, i.e., use the limit definition of the derivative to compute f'(0) .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Use the limit definition to compute the derivative, f'(x), for
f(x) = | x² - 3x | .
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Assume that
$$ f(x) = \cases{ \displaystyle{ 1\over 4 }x^3 - \displaystyle{1 \over 2 } x^2, &... ...$\space \cr \displaystyle{ -6xucdavis.edu/~kouba/CalcOneDIRECTORY/defderdirectory/img1.gif$ term so that the indeterminant form $tex2html_wrap_inline31$ of the expression can be circumvented and the limit can be calculated.
The following seven problems require more than one application of the chain rule.
The following seven problems require more than one application of the chain rule.
The following three problems require a more formal use of the chain rule.
Some of the following problems require use of the chain rule.
Some of the following problems require use of the chain rule.
PROBLEMAS DE MAXIMOS Y MINIMOS
- PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.
  Click HERE to see a detailed solution to problem 17.
- PROBLEM 18 : Consider the function $tex2html_wrap_inline105$ . Solve f'(x) = 0 for x . Solve f''(x) = 0 for x .
  Click HERE to see a detailed solution to problem 18.
- PROBLEM 19 : Find all points (x, y) on the graph of $tex2html_wrap_inline117$ where tangent lines are perpendicular to the line 8x+2y = 1 .
  Click HERE to see a detailed solution to problem 19.
  DIBUJANDO GRAFICAS CON LA PRIMERA Y SEGUNDA DERIVADAS

PROBLEM 1 : Do detailed graphing for f(x) = x³ - 3x² .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Do detailed graphing for f(x) = x⁴ - 4x³ .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Do detailed graphing for f(x) = x³ (x-2)² .
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Do detailed graphing for $f(x) = \displaystyle{ 4x \over x^2 + 1 }$ .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Do detailed graphing for $f(x) = \displaystyle{ 2x^2-3x \over x-2 }$ .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Do detailed graphing for $f(x) = \displaystyle{ (x-4)^2 \over x^2 - 4 }$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Do detailed graphing for f(x) = x - 3x^1/3 .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Do detailed graphing for $f(x) = x^{2/3} \Big( \displaystyle{ 5 \over 2 } - x \Big)$ .
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : Do detailed graphing for $f(x) = \sin x - \sqrt{ 3 } \cos x$ for x in $[0, 2\pi ]$ .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Do detailed graphing for $f(x) = x \sqrt{ 4 - x^2 }$ .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Consider the cubic polynomial y = A x³ + 6x² - Bx , where A and B are unknown constants. If possible, determine the values of A and B so that the graph of y has a maximum value at x= -1 and an inflection point at x=1 .
Click HERE to see a detailed solution to problem 11.

PROBLEMAS DE MAXIMOS Y MINIMOS

PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : An open rectangular box with square base is to be made from 48 ft.² of material. What dimensions will result in a box with the largest possible volume ?
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : A container in the shape of a right circular cylinder with no top has surface area 3 $\pi$ ft.² What height h and base radius r will maximize the volume of the cylinder ?
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Consider all triangles formed by lines passing through the point (8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse.
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Find the point (x, y) on the graph of $y=\sqrt{x}$ nearest the point (4, 0).
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : A cylindrical can is to hold 20 $\pi$ m.³ The material for the top and bottom costs $10/m.² and material for the side costs $8/m.² Find the radius r and height h of the most economical can.
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time ?
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Construct a window in the shape of a semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what dimensions will result in the rectangle having largest possible area ?
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the output per tree drops by 10 apples. How many trees should be added to the existing orchard in order to maximize the total output of trees ?
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Find the dimensions of the rectangle of largest area which can be inscribed in the closed region bounded by the x-axis, y-axis, and graph of y=8-x³ . (See diagram.)
Click HERE to see a detailed solution to problem 12.
PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume ?
Click HERE to see a detailed solution to problem 13.
PROBLEM 14 : A movie screen on a wall is 20 feet high and 10 feet above the floor. At what distance x from the front of the room should you position yourself so that the viewing angle $\theta$ of the movie screen is as large as possible ? (See diagram.)

Click HERE to see a detailed solution to problem 14.
PROBLEM 15 : Find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2.
Click HERE to see a detailed solution to problem 15.
PROBLEM 16 : What angle $\theta$ HERE to see a detailed solution to problem 1.
PROBLEM 2 : Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : An open rectangular box with square base is to be made from 48 ft.² of material. What dimensions will result in a box with the largest possible volume ?
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : A container in the shape of a right circular cylinder with no top has surface area 3 $\pi$ ft.² What height h and base radius r will maximize the volume of the cylinder ?
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Consider all triangles formed by lines passing through the point (8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse.
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Find the point (x, y) on the graph of $y=\sqrt{x}$ nearest the point (4, 0).
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : A cylindrical can is to hold 20 $\pi$ m.³ The material for the top and bottom costs $10/m.² and material for the side costs $8/m.² Find the radius r and height h of the most economical can.
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time ?
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Construct a window in the shape of a semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what dimensions will result in the rectangle having largest possible area ?
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the output per tree drops by 10 apples. How many trees should be added to the existing orchard in order to maximize the total output of trees ?
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Find the dimensions of the rectangle of largest area which can be inscribed in the closed region bounded by the x-axis, y-axis, and graph of y=8-x³ . (See diagram.)
Click HERE to see a detailed solution to problem 12.
PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume ?
Click HERE to see a detailed solution to problem 13.
PROBLEM 14 : A movie screen on a wall is 20 feet high and 10 feet above the floor. At what distance x from the front of the room should you position yourself so that the viewing angle $\theta$ of the movie screen is as large as possible ? (See diagram.)

Click HERE to see a detailed solution to problem 14.
PROBLEM 15 : Find the dimensions (radius r and height h) of the cone of maximum volume which can be inscribed in a sphere of radius 2.
Click HERE to see a detailed solution to problem 15.
PROBLEM 16 : What angle $\theta$ between two edges of length 3 will result in an isosceles triangle with the largest area ? (See diagram.)

Click HERE to see a detailed solution to problem 16.
PROBLEM 17 : Of all lines tangent to the graph of $y= \displaystyle{ 6 \over x^2+3 }$ , find the tangent lines of mimimum slope and maximum slope.
Click HERE to see a detailed solution to problem 17.
PROBLEM 18 : Find the length of the shortest ladder that will reach over an 8-ft. high fence to a large wall which is 3 ft. behind the fence. (See diagram.)

Click HERE to see a detailed solution to problem 18.
PROBLEM 19 : Find the point P = (x, 0) on the x-axis which minimizes the sum of the squares of the distances from P to (0, 0) and from P to (3, 2).
Click HERE to see a detailed solution to problem 19.
PROBLEM 20 : Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same moment car A begins moving north at 60 mph. What will be the minimum distance between the cars and at what time t does the minimum distance occur ?
Click HERE to see a detailed solution to problem 20.
PROBLEM 21 : A rectangular piece of paper is 12 inches high and six inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper (See diagram.).

Find the minimum length of the resulting crease.
Click HERE to see a detailed solution to problem 21.

PROBLEMAS DE DERIVADA IMPLICITA

PROBLEM 1 : Assume that y is a function of x . Find y' = dy/dx for x³ + y³ = 4 .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Assume that y is a function of x . Find y' = dy/dx for (x-y)² = x + y - 1 .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Assume that y is a function of x . Find y' = dy/dx for $y = \sin(3x + 4y)$ .
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Assume that y is a function of x . Find y' = dy/dx for y = x² y³ + x³ y² .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Assume that y is a function of x . Find y' = dy/dx for e^xy = e^4x - e^5y .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Assume that y is a function of x . Find y' = dy/dx for $\cos^2 x + \cos^2 y = \cos( 2x + 2y )$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Assume that y is a function of x . Find y' = dy/dx for $x = \sqrt{ x^2 + y^2 }$ .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Assume that y is a function of x . Find y' = dy/dx for $\displaystyle{ x - y^3 \over y + x^2 } = x + 2$ .
Click
Click HERE to see a detailed solution to problem 16.
PROBLEM 17 : Of all lines tangent to the graph of $y= \displaystyle{ 6 \over x^2+3 }$ , find the tangent lines of mimimum slope and maximum slope.
Click HERE to see a detailed solution to problem 17.
PROBLEM 18 : Find the length of the shortest ladder that will reach over an 8-ft. high fence to a large wall which is 3 ft. behind the fence. (See diagram.)

Click HERE to see a detailed solution to problem 18.
PROBLEM 19 : Find the point P = (x, 0) on the x-axis which minimizes the sum of the squares of the distances from P to (0, 0) and from P to (3, 2).
Click HERE to see a detailed solution to problem 19.
PROBLEM 20 : Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same moment car A begins moving north at 60 mph. What will be the minimum distance between the cars and at what time t does the minimum distance occur ?
Click HERE to see a detailed solution to problem 20.
PROBLEM 21 : A rectangular piece of paper is 12 inches high and six inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper (See diagram.).

Find the minimum length of the resulting crease.
Click HERE to see a detailed solution to problem 21.

PROBLEMAS DE DERIVADA IMPLICITA

PROBLEM 1 : Assume that y is a function of x . Find y' = dy/dx for x³ + y³ = 4 .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Assume that y is a function of x . Find y' = dy/dx for (x-y)² = x + y - 1 .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Assume that y is a function of x . Find y' = dy/dx for $y = \sin(3x + 4y)$ .
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Assume that y is a function of x . Find y' = dy/dx for y = x² y³ + x³ y² .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Assume that y is a function of x . Find y' = dy/dx for e^xy = e^4x - e^5y .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Assume that y is a function of x . Find y' = dy/dx for $\cos^2 x + \cos^2 y = \cos( 2x + 2y )$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Assume that y is a function of x . Find y' = dy/dx for $x = \sqrt{ x^2 + y^2 }$ .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Assume that y is a function of x . Find y' = dy/dx for $\displaystyle{ x - y^3 \over y + x^2 } = x + 2$ .
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : Assume that y is a function of x . Find y' = dy/dx for $\displaystyle{ { y \over x^3 } + { x \over y^3 } } = x^2y^4$ .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Find an equation of the line tangent to the graph of (x²+y²)³ = 8x²y² at the point (-1, 1) .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Find an equation of the line tangent to the graph of x² + (y-x)³ = 9 at x=1 .
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Find the slope and concavity of the graph of x²y + y⁴ = 4 + 2x at the point (-1, 1) .
Click HERE to see a detailed solution to problem 12.
PROBLEM 13 : Consider the equation x² + xy + y² = 1 . Find equations for y' and y'' in terms of x and y only.
Click HERE to see a detailed solution to problem 13.
PROBLEM 14 : Find all points (x, y) on the graph of x^2/3 + y^2/3 = 8 (See diagram.) where lines tangent to the graph at (x, y) have slope -1 .

Click HERE to see a detailed solution to problem 14.
PROBLEM 15 : The graph of x² - xy + y² = 3 is a "tilted" ellipse (See diagram.). Among all points (x, y) on this graph, find the largest and smallest values of y . Among all points (x, y) on this graph, find the largest and smallest values of x .

Click HERE to see a detailed solution to problem 15.
PROBLEM 16 : Find all points (x, y) on the graph of (x²+y²)² = 2x²-2y² (See diagram.) where y' = 0.

Click HERE to see a detailed solution to problem 16.

DERIVADAS DE LOGARITMOS

PROBLEM 1 : Differentiate y = x^x .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Differentiate y = x^{(e^x)} .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Differentiate y = (3x²+5)^1/x
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Differentiate $y = (\sin x)^{x^3}$ .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Differentiate $y = 7x (\cos x)^{x/2}$ .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Differentiate $y = \sqrt{x}^{ \sqrt{x} } e^{ x^2 }$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Di/www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffsoldirectory/ImplicitDiffSol.html#SOLUTION 8">HERE to see a detailed solution to problem 8.
PROBLEM 9 : Assume that y is a function of x . Find y' = dy/dx for $\displaystyle{ { y \over x^3 } + { x \over y^3 } } = x^2y^4$ .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Find an equation of the line tangent to the graph of (x²+y²)³ = 8x²y² at the point (-1, 1) .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Find an equation of the line tangent to the graph of x² + (y-x)³ = 9 at x=1 .
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Find the slope and concavity of the graph of x²y + y⁴ = 4 + 2x at the point (-1, 1) .
Click HERE to see a detailed solution to problem 12.
PROBLEM 13 : Consider the equation x² + xy + y² = 1 . Find equations for y' and y'' in terms of x and y only.
Click HERE to see a detailed solution to problem 13.
PROBLEM 14 : Find all points (x, y) on the graph of x^2/3 + y^2/3 = 8 (See diagram.) where lines tangent to the graph at (x, y) have slope -1 .

Click HERE to see a detailed solution to problem 14.
PROBLEM 15 : The graph of x² - xy + y² = 3 is a "tilted" ellipse (See diagram.). Among all points (x, y) on this graph, find the largest and smallest values of y . Among all points (x, y) on this graph, find the largest and smallest values of x .

Click HERE to see a detailed solution to problem 15.
PROBLEM 16 : Find all points (x, y) on the graph of (x²+y²)² = 2x²-2y² (See diagram.) where y' = 0.

Click HERE to see a detailed solution to problem 16.

DERIVADAS DE LOGARITMOS

PROBLEM 1 : Differentiate y = x^x .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Differentiate y = x^{(e^x)} .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Differentiate y = (3x²+5)^1/x
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Differentiate $y = (\sin x)^{x^3}$ .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Differentiate $y = 7x (\cos x)^{x/2}$ .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Differentiate $y = \sqrt{x}^{ \sqrt{x} } e^{ x^2 }$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : fferentiate $y = x^{ \ln x } (\sec x)^{3x}$ .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Differentiate $y = \displaystyle{ ( \ln x )^x \over 2^{ ^{3x+1} } }$ .
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : Differentiate $y = \displaystyle{ x^{2x} (x-1)^3 \over (3+5x)^4 }$ .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Consider the function $f(x) = \displaystyle{ x^{5} e^x (4x+3) \over 5^{ \ln x } (3-x)^{2} }$ . Find an equation of the line tangent to the graph of f at x=1 .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Consider the function $f(x) = \displaystyle{ \pi^2 + 2^x + x^{2 } + x^{1/x} }$ . Determine the slope of the line perpendicular to the graph of f at x=1 .
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Differentiate $y = \displaystyle{ x^{(x^{(x^4)})}}$
Click HERE to see a detailed solution to problem 12.

Escrito por hdqtitm el 18/10/2006 23:57 | Comentarios (0)

FRACCIONES PARCIALES

Integración de funciones racionales, por fracciones parciales, cuando el denominador sólo tiene factores lineales

Ejercicios resueltos En los siguientes ejercicios, obtenga la integral indefinida:

S o l u c i o n e s

$y = x^{ \ln x } (\sec x)^{3x}$ .
Click HERE to see a detailed solution to problem 7.

PROBLEM 8 : Differentiate $y = \displaystyle{ ( \ln x )^x \over 2^{ ^{3x+1} } }$ .
Click HERE to see a detailed solution to problem 8.

PROBLEM 9 : Differentiate $y = \displaystyle{ x^{2x} (x-1)^3 \over (3+5x)^4 }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Consider the function $f(x) = \displaystyle{ x^{5} e^x (4x+3) \over 5^{ \ln x } (3-x)^{2} }$ . Find an equation of the line tangent to the graph of f at x=1 .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Consider the function $f(x) = \displaystyle{ \pi^2 + 2^x + x^{2 } + x^{1/x} }$ . Determine the slope of the line perpendicular to the graph of f at x=1 .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Differentiate $y = \displaystyle{ x^{(x^{(x^4)})}}$
Click HERE to see a detailed solution to problem 12.
Escrito por hdqtitm el 18/10/2006 23:57 | Comentarios (0)

FRACCIONES PARCIALES

Integración de funciones racionales, por fracciones parciales, cuando el denominador sólo tiene factores lineales

Ejercicios resueltos En los siguientes ejercicios, obtenga la integral indefinida:

S o l u c i o n e s

Integración de funciones racionales, por fracciones parciales, cuando el denominador contiene factores cuadráticos
Ejercicios resueltos

S o l u c i o n e s

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Integración de funciones racionales, por fracciones parciales, cuando el denominador contiene factores cuadráticos
Ejercicios resueltos

S o l u c i o n e s

&np;

Integrales en las que aparecen expresiones cuadráticas
De la descomposición de fracciones parciales a veces resultan integrandos con expresiones cuadráticas ireductibles. De la integración de este tipo de funciones nos ocuparemos en los siguientes ejercicios resueltos.

Ejercicios resueltos
En los siguientes ejercicios evalúe la integral indefinida

S o l u c i o n e s

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displayGoogleAd87a0f3ff(); $\displaystyle{ \int { 1 \over x^2-4 } \,dx }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Integrate $\displaystyle{ \int { 2x+3 \over x^2-9 } \,dx }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Integrate $\displaystyle{ \int { 2-x \over x^2+5x } \,dx }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Integrate $\displaystyle{ \int { x^2-1 \over x^2-16 } \,dx }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Integrate $\displaystyle{ \int { x^4+x^3+x^2+1 \over x^2+x-2 } \,dx }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Integrate $\displaystyle{ \int { x^2+x-1 \over x(x^2-1) } \,dx }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 9 : Integrate $\displaystyle{ \int { x^2-x+1 \over (x+1)^3 } \,dx }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Integrate $\displaystyle{ \int { x^3+4 \over (x^2-1)(x^2+3x+2) } \,dx }$ .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Integrate $\displaystyle{ \int { x^3+2x-1 \over (x^2-x-2)^2 } \,dx }$ .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Integrate $\displaystyle{ \int { \sec^2 x \over \tan^3 x - \tan^2 x } \,dx }$ .
Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Integrate $\displaystyle{ \int{ x^3+8 \over (x^2-1)(x-2) } \,dx }$ .
Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Integrate $\displaystyle{ \int { e^x \over (e^x-1)(e^x+3) } \,dx }$ .
Click HERE to see a detailed solution to problem 14.

PROBLEM 15 : Integrate $\displaystyle{ \int { 1 \over e^x+1 } \,dx }$
PROBLEM 1 : Integrate $\displaystyle{ \int { 1 \over x^2-4 } \,dx }$ .
Click HERE to see a detailed solution to problem 1.

PROBLEM 2 : Integrate $\displaystyle{ \int { 2x+3 \over x^2-9 } \,dx }$ .
Click HERE to see a detailed solution to problem 2.

PROBLEM 3 : Integrate $\displaystyle{ \int { 2-x \over x^2+5x } \,dx }$ .
Click HERE to see a detailed solution to problem 3.

PROBLEM 4 : Integrate $\displaystyle{ \int { x^2-1 \over x^2-16 } \,dx }$ .
Click HERE to see a detailed solution to problem 4.

PROBLEM 5 : Integrate $\displaystyle{ \int { x^4+x^3+x^2+1 \over x^2+x-2 } \,dx }$ .
Click HERE to see a detailed solution to problem 5.

PROBLEM 6 : Integrate $\displaystyle{ \int { x^2+x-1 \over x(x^2-1) } \,dx }$ .
Click HERE to see a detailed solution to problem 6.

PROBLEM 9 : Integrate $\displaystyle{ \int { x^2-x+1 \over (x+1)^3 } \,dx }$ .
Click HERE to see a detailed solution to problem 9.

PROBLEM 10 : Integrate $\displaystyle{ \int { x^3+4 \over (x^2-1)(x^2+3x+2) } \,dx }$ .
Click HERE to see a detailed solution to problem 10.

PROBLEM 11 : Integrate $\displaystyle{ \int { x^3+2x-1 \over (x^2-x-2)^2 } \,dx }$ .
Click HERE to see a detailed solution to problem 11.

PROBLEM 12 : Integrate $\displaystyle{ \int { \sec^2 x \over \tan^3 x - \tan^2 x } \,dx }$ .
Click HERE to see a detailed solution to problem 12.

PROBLEM 13 : Integrate $\displaystyle{ \int{ x^3+8 \over (x^2-1)(x-2) } \,dx }$ .
Click HERE to see a detailed solution to problem 13.

PROBLEM 14 : Integrate $\displaystyle{ \int { e^x \over (e^x-1)(e^x+3) } \,dx }$ .
Click HERE to see a detailed solution to problem 14.

PROBLEM 15 : Integrate $\displaystyle{ \int { 1 \over e^x+1 } \,dx }$ .
Click HERE to see a detailed solution to problem 15.

PROBLEM 16 : Integrate $\displaystyle{ \int { 3-x \over x(x^2+1) } \,dx }$ .
Click HERE to see a detailed solution to problem 16.

PROBLEM 17 : Integrate $\displaystyle{ \int { 3x+1 \over x^2(x^2+25) } \,dx }$ .
Click HERE to see a detailed solution to problem 17.

PROBLEM 18 : Integrate $\displaystyle{ \int { 1 \over x^4-16 } \,dx }$ .
Click HERE to see a detailed solution to problem 18.

PROBLEM 19 : Integrate $\displaystyle{ \int { \cos x \over \sin^3 x + \sin x } \,dx }$ .
Click HERE to see a detailed solution to problem 19.

PROBLEM 20 :Integrate $\displaystyle{ \int{ 1 \over x^4+4 } \,dx }$ .
Click HERE to see a detailed solution to problem 20.
Escrito por hdqtitm el 18/10/2006 23:49 | Comentarios (0)
MAS DE INTEGRALES
TABLE OF CONTENTS
Definite Integrals
Fundamental Theorem of Calculus
Indefinite Integrals
The Total Change Theorem
The Substitution Rule
Integrals of Symmetric Functions
Integration by Parts
Trigonometric Integrals
Trigonometric Substitution
Integration of Rational Functions By Partial Fractions
Improper Integrals
Definite Integrals
The formal definition of a definite integral is stated in terms of the limit of a Riemann sum. Riemann sums are covered in the calculus lectures and in the textbook. For simplicity's sake, we will use a more informal definiton for a definite integral. We will introduce the definite integral defined in terms of area.
Let f(x) be a continuous function on the interval [a,b]. Consider the area bounded by the curve, the x-axis and the lines x=a and x=b. The area of the region that lies above the x-axis should be treated as a positive (+) value, while the area of the region that lies below the x-axis should be treated as a negative (-) value.
The image below illustrates this concept. The positive area, above the x-axis, is shaded green and labelled "+", while the negative area, below the x-axis, is shaded red and labelled "-".
The integral of the function f(x) from a to b is equal to the sum of the individual areas bounded by the function, the x-axis and the lines x=a and x=b. This integral is denoted by
where f(x) is called the integrand, a is the lower limit and b is the upper limit. This type of integral is called a definite integral. When evaluated, a definite integral results in a real number. It is independent of the choice of sample points (x, f(x)).

Properties of Definite Integrals
The following properties are helpful when calculating definite integrals.
Examples
1 | Evaluate the integral by finding the area beneath the curve

The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus defines the relationship between the processes of differentiation and integration. That relationship is that differentiation and integration are inverse processes.
The Fundamental Theorem of Cight=51 alt="$ \displaystyle{ \int { 1 \over e^x+1 } \,dx } $" src="http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/img90.gif" width=89 align=middle border=0> .
Click HERE to see a detailed solution to problem 15.

PROBLEM 16 : Integrate $\displaystyle{ \int { 3-x \over x(x^2+1) } \,dx }$ .
Click HERE to see a detailed solution to problem 16.

PROBLEM 17 : Integrate $\displaystyle{ \int { 3x+1 \over x^2(x^2+25) } \,dx }$ .
Click HERE to see a detailed solution to problem 17.

PROBLEM 18 : Integrate $\displaystyle{ \int { 1 \over x^4-16 } \,dx }$ .
Click HERE to see a detailed solution to problem 18.

PROBLEM 19 : Integrate $\displaystyle{ \int { \cos x \over \sin^3 x + \sin x } \,dx }$ .
Click HERE to see a detailed solution to problem 19.

PROBLEM 20 :Integrate $\displaystyle{ \int{ 1 \over x^4+4 } \,dx }$ .
Click HERE to see a detailed solution to problem 20.
Escrito por hdqtitm el 18/10/2006 23:49 | Comentarios (0)
MAS DE INTEGRALES
TABLE OF CONTENTS
Definite Integrals
Fundamental Theorem of Calculus
Indefinite Integrals
The Total Change Theorem
The Substitution Rule
Integrals of Symmetric Functions
Integration by Parts
Trigonometric Integrals
Trigonometric Substitution
Integration of Rational Functions By Partial Fractions
Improper Integrals
Definite Integrals
The formal definition of a definite integral is stated in terms of the limit of a Riemann sum. Riemann sums are covered in the calculus lectures and in the textbook. For simplicity's sake, we will use a more informal definiton for a definite integral. We will introduce the definite integral defined in terms of area.
Let f(x) be a continuous function on the interval [a,b]. Consider the area bounded by the curve, the x-axis and the lines x=a and x=b. The area of the region that lies above the x-axis should be treated as a positive (+) value, while the area of the region that lies below the x-axis should be treated as a negative (-) value.
The image below illustrates this concept. The positive area, above the x-axis, is shaded green and labelled "+", while the negative area, below the x-axis, is shaded red and labelled "-".
The integral of the function f(x) from a to b is equal to the sum of the individual areas bounded by the function, the x-axis and the lines x=a and x=b. This integral is denoted by
where f(x) is called the integrand, a is the lower limit and b is the upper limit. This type of integral is called a definite integral. When evaluated, a definite integral results in a real number. It is independent of the choice of sample points (x, f(x)).

Properties of Definite Integrals
The following properties are helpful when calculating definite integrals.
Examples
1 | Evaluate the integral by finding the area beneath the curve

The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus defines the relationship between the processes of differentiation and integration. That relationship is that differentiation and integration are inverse processes.
The Fundamental Theorem ofalculus : Part 1
If f is a continuous function on [a,b], then the function denoted by
is continuous on [a,b], differentiable on (a,b) and g'(x) = f(x).

If f(t) is continuous on [a,b], the function g(x) that's equal to the the area bounded by the u-axis and the function f(u) and the lines u=a and u=x will be continuous on [a,b] and differentiable on (a,b). Most importantly, when we differentiate the function g(x), we will find that it is equal to f(x). The graph to the right illustrates the function f(u) and the area g(x).

The Fundamental Theorem of Calculus : Part 2
If f is a continuous function on [a,b], then
where F is any antiderivative of f.

If f is continuous on [a,b], the definite integral with integrand f(x) and limits a and b is simply equal to the value of the antiderivative F(x) at b minus the value of F at a. This property allows us to easily solve definite integrals, if we can find the antiderivative function of the integrand.

Parts one and two of the Fundamental Theorem of Calculus can be combined and simplified into one theorem.
The Fundamental Theorem of Calculus
Let f be a continuous function on [a,b].
Indefinite Integrals
An indefinite integral has the form
When evaluated, an indefinite integral results in a function (or family of functions). An indefinite integral of a function f(x) is also known as the antiderivative of f. A function F is an antiderivative of f on an interval I, if F'(x) = f(x) for all x in I. This is a strong indication that that the processes of integration and differentiation are interconnected.

Table of Indefinite Integrals
The following tables list the formulas for antidifferentiation. These formulas allow us to determine the function that results from an indefinite integral. Since the formulas are for the most general indefinite integral, we add a constant C to each one. With these formulas and the Fundamental Theorem of Calculus, we can evaluate simple definite integrals.
The next table lists indefinite integrals involving trigonometric functions.
Note: After finding an indefinite integral, you can always check to see if your answer is correct. Since integration and differentiation are inverse processes, you can simply differentiate the function that results from integration, and see if it is equal to the integrand.

Examples
2 | Find the general indefinite integrals
3 | Evaluate the definite integral
4 | Evaluate the definite integral of the absolute value of a function

The Total Change Theorem
The total change theorem is an adaptation of the second part of the Fundamental Theorem of Calculus. The Total Change Theorem states: the integral of a rate of change is equal to the total change.
If we know that the function f(x) is the derivative of some function F(x), then the definite integral of f(x) from a to b is equal to the change in the function F(x) from a to b.

Examples
5 | Given the velocity function, find the displacement during a period of time

The Substitution Rule
Suppose that we have an integral such as
With our current knowledge of integration, we can't find the general equation of this indefinite integral. There are no antidifferentiation formulas for this type of integral. However, from our knowledge of differentiation, specifically the chain rule, we know that 4x³ is the derivative of the function within the square root, x⁴ + 7. We must also account for the chain rule when we are performing integration. To do this, we use the substitution rule.
The Substitution Rule states: if u = g(x) is a differentiable function and f is continuous on the range of g, then
Note: Recall that if u = g(x), then du = g'(x)dx. If we substitute u into the left side of the equation for g(x) and du for g'(x)dx, then we get the integral on the right side of the equation.
From our previous example, if we let u = (x⁴+7), then du = 4x³dx. If we substitutite these values into the integral, we get an integral that can be solved using the antidifferentiation formulas.
However, this answer is still in terms of u. We must substitute u = (x⁴+7) into the resulting function, so that it is a function of x, rather than u.

The substitution rule also applies to definite integrals. The Substitution Rule for Definite Integrals states: If f is continuous on the range of u = g(x) and g'(x) is continuous on [a,b], then

Examples
6 | Find the general indefinite integrals using the substitution rule
7 | Evaluate the definite integral using the substitution rule

Integrals of Symmetric Functions
If f(x) is continuous on [-a, a] and f is an even function, then
If f(x) is Calculus : Part 1
If f is a continuous function on [a,b], then the function denoted by
is continuous on [a,b], differentiable on (a,b) and g'(x) = f(x).

If f(t) is continuous on [a,b], the function g(x) that's equal to the the area bounded by the u-axis and the function f(u) and the lines u=a and u=x will be continuous on [a,b] and differentiable on (a,b). Most importantly, when we differentiate the function g(x), we will find that it is equal to f(x). The graph to the right illustrates the function f(u) and the area g(x).

The Fundamental Theorem of Calculus : Part 2
If f is a continuous function on [a,b], then
where F is any antiderivative of f.

If f is continuous on [a,b], the definite integral with integrand f(x) and limits a and b is simply equal to the value of the antiderivative F(x) at b minus the value of F at a. This property allows us to easily solve definite integrals, if we can find the antiderivative function of the integrand.

Parts one and two of the Fundamental Theorem of Calculus can be combined and simplified into one theorem.
The Fundamental Theorem of Calculus
Let f be a continuous function on [a,b].
Indefinite Integrals
An indefinite integral has the form
When evaluated, an indefinite integral results in a function (or family of functions). An indefinite integral of a function f(x) is also known as the antiderivative of f. A function F is an antiderivative of f on an interval I, if F'(x) = f(x) for all x in I. This is a strong indication that that the processes of integration and differentiation are interconnected.

Table of Indefinite Integrals
The following tables list the formulas for antidifferentiation. These formulas allow us to determine the function that results from an indefinite integral. Since the formulas are for the most general indefinite integral, we add a constant C to each one. With these formulas and the Fundamental Theorem of Calculus, we can evaluate simple definite integrals.
The next table lists indefinite integrals involving trigonometric functions.
Note: After finding an indefinite integral, you can always check to see if your answer is correct. Since integration and differentiation are inverse processes, you can simply differentiate the function that results from integration, and see if it is equal to the integrand.

Examples
2 | Find the general indefinite integrals
3 | Evaluate the definite integral
4 | Evaluate the definite integral of the absolute value of a function

The Total Change Theorem
The total change theorem is an adaptation of the second part of the Fundamental Theorem of Calculus. The Total Change Theorem states: the integral of a rate of change is equal to the total change.
If we know that the function f(x) is the derivative of some function F(x), then the definite integral of f(x) from a to b is equal to the change in the function F(x) from a to b.

Examples
5 | Given the velocity function, find the displacement during a period of time

The Substitution Rule
Suppose that we have an integral such as
With our current knowledge of integration, we can't find the general equation of this indefinite integral. There are no antidifferentiation formulas for this type of integral. However, from our knowledge of differentiation, specifically the chain rule, we know that 4x³ is the derivative of the function within the square root, x⁴ + 7. We must also account for the chain rule when we are performing integration. To do this, we use the substitution rule.
The Substitution Rule states: if u = g(x) is a differentiable function and f is continuous on the range of g, then
Note: Recall that if u = g(x), then du = g'(x)dx. If we substitute u into the left side of the equation for g(x) and du for g'(x)dx, then we get the integral on the right side of the equation.
From our previous example, if we let u = (x⁴+7), then du = 4x³dx. If we substitutite these values into the integral, we get an integral that can be solved using the antidifferentiation formulas.
However, this answer is still in terms of u. We must substitute u = (x⁴+7) into the resulting function, so that it is a function of x, rather than u.

The substitution rule also applies to definite integrals. The Substitution Rule for Definite Integrals states: If f is continuous on the range of u = g(x) and g'(x) is continuous on [a,b], then

Examples
6 | Find the general indefinite integrals using the substitution rule
7 | Evaluate the definite integral using the substitution rule

Integrals of Symmetric Functions
If f(x) is continuous on [-a, a] and f is an even function, then
If f(x) continuous on [-a, a] and f is an odd function, then
These properties of integrals of symmetric functions are very helpful when solving integration problems. Some of the more challenging problems can be solved quite simply by using this property.

Examples
8 | Evaluate the definite integral of the symmetric function

Integration By Parts
Suppose that we have an integral such as
Similar to integrals solved using the substitution method, there are no general equations for this indefinite integral. However there do not appear to be any clear substitutions that could be made to simplify this integral. This brings us to an integration technique known as integration by parts, which will call upon our knowledge of the Product Rule for differentiation.
The Product Rule states: If f and g are differentiable functions, then
By taking the indefinite integral of both sides of the equation we have:
and we can rearrange this equation as
To make it easier to remember it is commonly written in the following notation. Let u=f(x) and v=g(x). Then the differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes:
From our previous example, if we let u=x and dv=cosx, then du=dx and v=sinx. If we substitute these values into the formula we have:
Note: By choosing u=x we obtain a simpler integral than we started with. Had we chose u=cosx and dv=x then du=-sinx and v=(1/2)x² so integration by parts gives:
This equation is correct, but the integral is more difficult than the one we started with.
When choosing u and dv always try to choose u=f(x) to be a function that becomes simpler when differentiated (or at least not more complicated) and to choose dv=g'(x) to be a function that can be easily integrated to give v.
Examples
9 | Find the general indefinite integral by integration by parts
10 | Solve the definite integral by integration by parts
Trigonometric Integrals
Suppose we have an integral such as
The easy mistake is to simply make the substitution u=sinx, but then du=cosxdx. So in order to integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers of cosine we need an extra sinx factor. Thus for this example knowing we need an extra sinx factor to integrate powers of cosine we can separate one sine factor and convert the remaining sin⁴x to an expression involving cosine using the identity sin²x + cos²x = 1.
Now by using our knowledge of substitution we can evaluate the integral by letting u=cosx, then du=-sinxdx and

Now consider the integral
If we were to use the method from the previous example and separate one cosine factor we would be left with a factor of cosine of odd degree which isn't easily converted to sine. We must now consider the half angle formulas
Using the half angle formula for cos²x, we have:

Strategy for Evaluating
(a)
If the power of sine is odd (m=2k+1), save one sine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of cosine.
then substitute u=cosx.
(b)
If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of sine.
then substitute u=sinx.
(c)
If the powers of both sine and cosine are even then use the half angle identities.
In some cases it may be helpful to use the identity

11 | Solve the indefinite trigonometric integral
12 | Using the half angle formulas solve the indefinite trigonometric integral
13 | Solve the definite trigonometric integral
Now that we have learned strategies for solving integrals with factors of sine and cosine we can use similar techniques to solve integrals with factors of tangent and secant. Using the identity sec²x = 1 + tan²x we are able to convert even powers of secant to tangent and vice versa. Now we will consider two examples to illustrate two common strategies used to solve integrals of the form
Suppose we have an integral such as
Observing that (d/dx)tanx=sec²x we can separate a factor of sec²x and still be left with an even power of secant. Using the identity sec²x = 1 + tan²x we can converis continuous on [-a, a] and f is an odd function, then
These properties of integrals of symmetric functions are very helpful when solving integration problems. Some of the more challenging problems can be solved quite simply by using this property.

Examples
8 | Evaluate the definite integral of the symmetric function

Integration By Parts
Suppose that we have an integral such as
Similar to integrals solved using the substitution method, there are no general equations for this indefinite integral. However there do not appear to be any clear substitutions that could be made to simplify this integral. This brings us to an integration technique known as integration by parts, which will call upon our knowledge of the Product Rule for differentiation.
The Product Rule states: If f and g are differentiable functions, then
By taking the indefinite integral of both sides of the equation we have:
and we can rearrange this equation as
To make it easier to remember it is commonly written in the following notation. Let u=f(x) and v=g(x). Then the differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes:
From our previous example, if we let u=x and dv=cosx, then du=dx and v=sinx. If we substitute these values into the formula we have:
Note: By choosing u=x we obtain a simpler integral than we started with. Had we chose u=cosx and dv=x then du=-sinx and v=(1/2)x² so integration by parts gives:
This equation is correct, but the integral is more difficult than the one we started with.
When choosing u and dv always try to choose u=f(x) to be a function that becomes simpler when differentiated (or at least not more complicated) and to choose dv=g'(x) to be a function that can be easily integrated to give v.
Examples
9 | Find the general indefinite integral by integration by parts
10 | Solve the definite integral by integration by parts
Trigonometric Integrals
Suppose we have an integral such as
The easy mistake is to simply make the substitution u=sinx, but then du=cosxdx. So in order to integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers of cosine we need an extra sinx factor. Thus for this example knowing we need an extra sinx factor to integrate powers of cosine we can separate one sine factor and convert the remaining sin⁴x to an expression involving cosine using the identity sin²x + cos²x = 1.
Now by using our knowledge of substitution we can evaluate the integral by letting u=cosx, then du=-sinxdx and

Now consider the integral
If we were to use the method from the previous example and separate one cosine factor we would be left with a factor of cosine of odd degree which isn't easily converted to sine. We must now consider the half angle formulas
Using the half angle formula for cos²x, we have:

Strategy for Evaluating
(a)
If the power of sine is odd (m=2k+1), save one sine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of cosine.
then substitute u=cosx.
(b)
If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of sine.
then substitute u=sinx.
(c)
If the powers of both sine and cosine are even then use the half angle identities.
In some cases it may be helpful to use the identity

11 | Solve the indefinite trigonometric integral
12 | Using the half angle formulas solve the indefinite trigonometric integral
13 | Solve the definite trigonometric integral
Now that we have learned strategies for solving integrals with factors of sine and cosine we can use similar techniques to solve integrals with factors of tangent and secant. Using the identity sec²x = 1 + tan²x we are able to convert even powers of secant to tangent and vice versa. Now we will consider two examples to illustrate two common strategies used to solve integrals of the form
Suppose we have an integral such as
Observing that (d/dx)tanx=sec²x we can separate a factor of sec²x and still be left with an even power of secant. Using the identity sec²x = 1 + tan²x we can convt the remaining sec²x to an expression involving tangent. Thus we have:
Then substitute u=tanx to obtain:

Note: Suppose we tried to use the substitution u=secx, then du=secxtanxdx. When we separate out a factor of secxtanx we are left with an odd power of tangent which is not easily converted to secant.

Consider the integral
Since (d/dx)secx=secxtanx we can separate a factor of secxtanx and still be left with an even power of tangent which we can easily convert to an expression involving secant using the identity sec²x = 1 + tan²x. Thus we have:
Then substitute u=secx to obtain:
Note: Suppose we tried to use the substitution u=tanx, then du=sec²xdx. When we separate out a factor of sec²x we are left with an odd power of secant which is not easily converted to tangent.
Strategy for Evaluating
(a)
If the power of secant is even (n=2k, k>2) save a factor of sec²x and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of tanx.
then substitute u=tanx.
(b)
If the power of tangent is odd (m=2k+1), save a factor of secxtanx and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of secx.
then substitute u=secx.
Note: If the power of secant is even and the power of tangent is odd then either method will suffice, although there may be less work involved to use method (a) if the power of secant is smaller, and method (b) if the power of tangent is smaller.
14 | Solve the indefinite trigonometric integral
15 | Solve the definite trigonometric integral
Integrals of cotangent and cosecant are very similar to those with tangent and secant.

it is easy to see that integrals of the form can be solved by nearly identical methods as are integrals of the form .

16 | Solve the indefinite trigonometric integral
Unlike integrals with factors of both tangent and secant, integrals that have factors of only tangent, or only secant do not have a general strategy for solving. Use of trig identities, substitution and integration by parts are all commonly used to solve such integrals. For example,
If we make the substitution u=secx, then du=secxtanxdx, and we are left with the simple integral
Similarily we can use the same technique to solve

17 | Solve the definite trigonometric integral
18 | Solve the definite trigonometric integral
19 | Solve the indefinite trigonometric integral
Another problem that may be encountered when solving trigonometric integrals are integrals of the form
Using the product formulas which are deduced from the addition/subtraction rules we have the corresponding identities
20 | Solve the indefinite trigonometric integral using the product formulas
Trigonometric Substitution
Sometimes trigonometric substitutions are very effective even when at first it may not be so clear why such a substitution be made. For example, when finding the area of a circle or an ellipse you may have to solve an integral of the form where a>0.
It is difficult to make a substitution where the new variable is a function of the old one, (for example, had we made the substitution u = a² - x², then du= -2xdx, and we are unable to cancel out the -2x.) So we must consider a change in variables where the old variable is a function of the new one. This is where trigonometric identities are put to use. Suppose we change the variable from x to by making the substitution . Then using the trig identity we can simplify the integral by eliminating the root sign.
By changing x to a function with a different variable we are essentially using the The Substitution Rule in reverse. If x=g(t) then by restricting the boundaries on g we can assure that g has an inverse function; that is, g is one-to-one. In the example above we would require to assure has an inverse function.
If we look at the Substitution Rule and replace u with x and x with t, we obtain
Then substitute u=tanx to obtain:

Note: Suppose we tried to use the substitution u=secx, then du=secxtanxdx. When we separate out a factor of secxtanx we are left with an odd power of tangent which is not easily converted to secant.

Consider the integral
Since (d/dx)secx=secxtanx we can separate a factor of secxtanx and still be left with an even power of tangent which we can easily convert to an expression involving secant using the identity sec²x = 1 + tan²x. Thus we have:
Then substitute u=secx to obtain:
Note: Suppose we tried to use the substitution u=tanx, then du=sec²xdx. When we separate out a factor of sec²x we are left with an odd power of secant which is not easily converted to tangent.
Strategy for Evaluating
(a)
If the power of secant is even (n=2k, k>2) save a factor of sec²x and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of tanx.
then substitute u=tanx.
(b)
If the power of tangent is odd (m=2k+1), save a factor of secxtanx and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of secx.
then substitute u=secx.
Note: If the power of secant is even and the power of tangent is odd then either method will suffice, although there may be less work involved to use method (a) if the power of secant is smaller, and method (b) if the power of tangent is smaller.
14 | Solve the indefinite trigonometric integral
15 | Solve the definite trigonometric integral
Integrals of cotangent and cosecant are very similar to those with tangent and secant.

it is easy to see that integrals of the form can be solved by nearly identical methods as are integrals of the form .

16 | Solve the indefinite trigonometric integral
Unlike integrals with factors of both tangent and secant, integrals that have factors of only tangent, or only secant do not have a general strategy for solving. Use of trig identities, substitution and integration by parts are all commonly used to solve such integrals. For example,
If we make the substitution u=secx, then du=secxtanxdx, and we are left with the simple integral
Similarily we can use the same technique to solve

17 | Solve the definite trigonometric integral
18 | Solve the definite trigonometric integral
19 | Solve the indefinite trigonometric integral
Another problem that may be encountered when solving trigonometric integrals are integrals of the form
Using the product formulas which are deduced from the addition/subtraction rules we have the corresponding identities
20 | Solve the indefinite trigonometric integral using the product formulas
Trigonometric Substitution
Sometimes trigonometric substitutions are very effective even when at first it may not be so clear why such a substitution be made. For example, when finding the area of a circle or an ellipse you may have to solve an integral of the form where a>0.
It is difficult to make a substitution where the new variable is a function of the old one, (for example, had we made the substitution u = a² - x², then du= -2xdx, and we are unable to cancel out the -2x.) So we must consider a change in variables where the old variable is a function of the new one. This is where trigonometric identities are put to use. Suppose we change the variable from x to by making the substitution . Then using the trig identity we can simplify the integral by eliminating the root sign.
By changing x to a function with a different variable we are essentially using the The Substitution Rule in reverse. If x=g(t) then by restricting the boundaries on g we can assure that g has an inverse function; that is, g is one-to-one. In the example above we would require to assure has an inverse function.
If we look at the Substitution Rule and replace u with x and x with t, we obtain
This is known as the "inverse substitution".
Integration of Rational Functions By Partial Fractions
Integration of rational functions by partial fractions is a fairly simple integrating technique used to simplify one rational function into two or more rational functions which are more easily integrated.
Think back to the steps taken when adding or subtracting fractions that do not have the same denominator. First you find the lowest common multiple of the two denominators and then cross multiply with the numerators accordingly. eg.
Well the same process applies when dealing with polynomial fractions. eg.
Now by reversing this process we can simplify a function such as into two fractions which are more easily integrated.
This process is possible when the function is proper; that is the degree of the numerator is less than the degree of the denominator. If the function is improper; that is the degree of the numerator is greater than or equal to the degree of the denominator, then we must first use long division to divide the denominator into the numerator until we obtain a remainder, such that it's degree is less than the denominator. Then if possible the above process is used to simplify the proper function.

To complete some of the problems in this section it will be useful to know the table integral
In general there are 4 cases to consider to express a rational function as the sum of two or more partial fractions.
Case 1
The denominator is a product of distinct linear factors (no factor is repeated or a constant mulptiple of another).
For example,
Since the degree of the numerator is less than the degree of the denominator we don't need to divide. The denominator can be factored as follows:
Since the denominator has distinct linear factors we can write the rational fraction as the sum of two or more partial fractions as follows:
By multiplying both sides by we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 2
The denominator is a product of linear functions, some of which are repeated.
For example,
Since the degree of the numerator is greater than the degree of the denominator we must factorize by long division.
So we can now factor the denominator to obtain:
Since the linear factor (x-2) occurs twice, the partial fraction decomposition is:
When we multiply both sides by the least common denominator we get:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 3
The denominator contains irreducible quadratic factors, none of which are repeated.
When reducing such functions to partial fractions if there is a term in the denominator of the form ax² + bx + c, where b² - 4ac < 0, then the numerator for that partial fraction will be of the form Ax + B.
For example,
Since the degree of the numerator is less than the degree of the denominator we do not have to divide first.
Since x³ + 4x = x(x² + 4) can't be factored any further we have:
multiplying both sides by x(x² + 4), we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 4
The denominator contains a repeated irreducible quadtratic factor.
Functions of this form are the same as those in case 3 only there is a term in the denominator that is repeated or is a constant multiple of another.
For example,
If we were to expand the denominator we would see that its degree is greater than the the degree of the numerator so we do not have to divide first.
Since the function cannot be factored any further we have:
multiplying both sides by (x + 1)(x² + 4)², we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Improper Integrals
In all of the previous tutorials we have dealt with integrals with a continous function f on a finite interval [a,b]. In this section we will consider two types of integrals known as improper integrals. The first type of improper integral are those defined on an infinite interval, and the second are those where the function f has an infinite discontinuity in [a,b].

Type 1: Infinite Intervals
Type 2: Discontinous Integrands
For more practice with the concepts covered in this tutorial, visit the Integral Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.

To test your knowledge of integration problems, try taking the general integrals test on the iLrn website or the advanced integrals test at the link below.
This is known as the "inverse substitution".
Integration of Rational Functions By Partial Fractions
Integration of rational functions by partial fractions is a fairly simple integrating technique used to simplify one rational function into two or more rational functions which are more easily integrated.
Think back to the steps taken when adding or subtracting fractions that do not have the same denominator. First you find the lowest common multiple of the two denominators and then cross multiply with the numerators accordingly. eg.
Well the same process applies when dealing with polynomial fractions. eg.
Now by reversing this process we can simplify a function such as into two fractions which are more easily integrated.
This process is possible when the function is proper; that is the degree of the numerator is less than the degree of the denominator. If the function is improper; that is the degree of the numerator is greater than or equal to the degree of the denominator, then we must first use long division to divide the denominator into the numerator until we obtain a remainder, such that it's degree is less than the denominator. Then if possible the above process is used to simplify the proper function.

To complete some of the problems in this section it will be useful to know the table integral
In general there are 4 cases to consider to express a rational function as the sum of two or more partial fractions.
Case 1
The denominator is a product of distinct linear factors (no factor is repeated or a constant mulptiple of another).
For example,
Since the degree of the numerator is less than the degree of the denominator we don't need to divide. The denominator can be factored as follows:
Since the denominator has distinct linear factors we can write the rational fraction as the sum of two or more partial fractions as follows:
By multiplying both sides by we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 2
The denominator is a product of linear functions, some of which are repeated.
For example,
Since the degree of the numerator is greater than the degree of the denominator we must factorize by long division.
So we can now factor the denominator to obtain:
Since the linear factor (x-2) occurs twice, the partial fraction decomposition is:
When we multiply both sides by the least common denominator we get:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 3
The denominator contains irreducible quadratic factors, none of which are repeated.
When reducing such functions to partial fractions if there is a term in the denominator of the form ax² + bx + c, where b² - 4ac < 0, then the numerator for that partial fraction will be of the form Ax + B.
For example,
Since the degree of the numerator is less than the degree of the denominator we do not have to divide first.
Since x³ + 4x = x(x² + 4) can't be factored any further we have:
multiplying both sides by x(x² + 4), we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Case 4
The denominator contains a repeated irreducible quadtratic factor.
Functions of this form are the same as those in case 3 only there is a term in the denominator that is repeated or is a constant multiple of another.
For example,
If we were to expand the denominator we would see that its degree is greater than the the degree of the numerator so we do not have to divide first.
Since the function cannot be factored any further we have:
multiplying both sides by (x + 1)(x² + 4)², we have:
From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations:

Improper Integrals
In all of the previous tutorials we have dealt with integrals with a continous function f on a finite interval [a,b]. In this section we will consider two types of integrals known as improper integrals. The first type of improper integral are those defined on an infinite interval, and the second are those where the function f has an infinite discontinuity in [a,b].

Type 1: Infinite Intervals
Type 2: Discontinous Integrands
For more practice with the concepts covered in this tutorial, visit the Integral Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.

To test your knowledge of integration problems, try taking the general integrals test on the iLrn website or the advanced integrals test at the link below.
Escrito por hdqtitm el 12/10/2006 22:54 | Comentarios (0)
MOLECULAS DIBUJO
Software Químico y Biológico

AQUI PODRAS ENCONTRAR GRAFICADORES PARA MOLECULAS QUIMICAS
http://www.galeon.com/scienceducation/software.html
Escrito por hdqtitm el 05/10/2006 21:21 | Comentarios (0)
INTEGRADOR ELECTRONICO
ESTE LUGAR TE AYUDARÀ A INTEGRAR Y TE MOSTRARÀ LOS PASOS QUE SE USARON,ES DECIR PASO POR PASO EN DETALLE, ASI COMO TAMBIEN LA TECNICA USADA DE INTEGRACIÒN.

http://www.calc101.com/webMathematica/integracion.jsp

Escrito por hdqtitm el 28/09/2006 16:00 | Comentarios (0)
GRAFICADOR LAB. U.DE.A
http://www.math.uri.edu/~bkaskosz/flashmo/famplot/
Escrito por hdqtitm el 28/09/2006 15:47 | Comentarios (0)
GRAFICADOR DE PUNTOS
http://www.webmath.com/gpoints.html
Escrito por hdqtitm el 21/09/2006 16:45 | Comentarios (0)
CALCULO II

If each f(x_i) > 0 then the area of the i^th rectangle is

and the sum of the areas of the rectangles is then:
More generally, we do not require that f(x_i) > 0 as we define
A Riemann Sum of f over [a, b] is the sum

Note that the Riemann sum when each x_i is the right-hand endpoint of the subinterval [a_i-1, a_i] is
when each x_i is the left-hand endpoint of the subinterval [a_i-1, a_i] is
and when each x_i is the left-hand midpoint of the subinterval [a_i-1, a_i] is
.
Escrito por hdqtitm el 19/09/2006 22:41 | Comentarios (0)
CALCULO II
EL MUNDO DEL CALCULO II, CALCULO INTEGRAL

FUNCIÓN PRIMITIVA DE UNA FUNCIÓN
Dada una función cualquiera f(x) definida en un intervalo cerrado [a,b], se llama función primitiva de f(x) a otra función F(x) cuya derivada sea f(x) en dicho intervalo. Es decir, F'(x) = f(x) para todo x de [a,b].
Así:
La función sen x es una primitiva de cos x puesto que (sen x)' = cos x.
PROP. DE LAS PRIM. DE UNA FUNC.
Primera propiedad
Si F(x) es una primitiva de f(x) y Escrito por hdqtitm el 12/10/2006 22:54 | Comentarios (0)
MOLECULAS DIBUJO
Software Químico y Biológico

AQUI PODRAS ENCONTRAR GRAFICADORES PARA MOLECULAS QUIMICAS
http://www.galeon.com/scienceducation/software.html
Escrito por hdqtitm el 05/10/2006 21:21 | Comentarios (0)
INTEGRADOR ELECTRONICO
ESTE LUGAR TE AYUDARÀ A INTEGRAR Y TE MOSTRARÀ LOS PASOS QUE SE USARON,ES DECIR PASO POR PASO EN DETALLE, ASI COMO TAMBIEN LA TECNICA USADA DE INTEGRACIÒN.

http://www.calc101.com/webMathematica/integracion.jsp

Escrito por hdqtitm el 28/09/2006 16:00 | Comentarios (0)
GRAFICADOR LAB. U.DE.A
http://www.math.uri.edu/~bkaskosz/flashmo/famplot/
Escrito por hdqtitm el 28/09/2006 15:47 | Comentarios (0)
GRAFICADOR DE PUNTOS
http://www.webmath.com/gpoints.html
Escrito por hdqtitm el 21/09/2006 16:45 | Comentarios (0)
CALCULO II

If each f(x_i) > 0 then the area of the i^th rectangle is

and the sum of the areas of the rectangles is then:
More generally, we do not require that f(x_i) > 0 as we define
A Riemann Sum of f over [a, b] is the sum

Note that the Riemann sum when each x_i is the right-hand endpoint of the subinterval [a_i-1, a_i] is
when each x_i is the left-hand endpoint of the subinterval [a_i-1, a_i] is
and when each x_i is the left-hand midpoint of the subinterval [a_i-1, a_i] is
.
Escrito por hdqtitm el 19/09/2006 22:41 | Comentarios (0)
CALCULO II
EL MUNDO DEL CALCULO II, CALCULO INTEGRAL

FUNCIÓN PRIMITIVA DE UNA FUNCIÓN
Dada una función cualquiera f(x) definida en un intervalo cerrado [a,b], se llama función primitiva de f(x) a otra función F(x) cuya derivada sea f(x) en dicho intervalo. Es decir, F'(x) = f(x) para todo x de [a,b].
Así:
La función sen x es una primitiva de cos x puesto que (sen x)' = cos x.
PROP. DE LAS PRIM. DE UNA FUNC.
Primera propiedad
Si F(x) es una primitiva de f(x) y C una constante cualquiera (un número), la función
F(x) + C es otra primitiva de f(x).
Demostración:
Basta recordar que la derivada de una suma de funciones es igual a la suma de las derivadas de las funciones, y que la derivada de una constante es siempre cero.
(F(x) + C)' = F'(x) + C' = f(x) + 0 = f(x)
Segunda propiedad
Si una función tiene una primitiva, entonces tiene infinitas primitivas.
Demostración:
Si F(x) es una primitiva de f(x), para cualquier constante C, F(x) + C es otra primitiva según la anterior propiedad. Así, hay tantas primitivas como valores se le quieran dar
a C.
Tercera propiedad
Dos primitivas de una misma función se diferencian en una constante. Esto es, si F(x) y G(x) son primitivas de la función f(x), entonces F(x) - G(x) = C = cte.
Demostración:
Hay que recordar que si una función f(x) definida en un intervalo cualquiera tiene derivada cero en todos los puntos, entonces la función f(x) es constante. Es decir, si f'(x) = 0, entonces f(x) = C.
Pues bien, si F(x) es una primitiva de f(x), F'(x) = f(x);
si G(x) es otra primitiva de f(x), G'(x) = f(x).
Restando miembro a miembro, F'(x) - G'(x) = (F(x) - G(x))' = f(x) - f(x) = 0, de donde se deduce que F(x) - G(x) = C.

Una función F se llama antiderivada general o integral general de otra función f, en un intervalo I, si F’(x)=f(x) para todo valor de x en I. A esta función F también se le llama primitiva.

Ejemplo.
Si F se define como F(x)=4x³+x²+5, entonces F’(x)= 12x²+2x.

Así, si f es la función definida por f(x)=12x²+2x, decimos que f es la derivada de F y que F es una antiderivada de f.

Si G es la función definida por G(x)=4x³+x²-17, entonces G también es una antiderivada de f, ya que G’(x)= 12x²+2x.

En realidad, cualquier función cuyo valor esté dado por 4x³+x²+C, donde C es cualquier constante, es una antiderivada de f.

Por consiguiente, si F(x) es una integral de una derivada dada, entonces F(x)+C también lo es. Recíprocamente. Si dos funciones son integrales de una misma derivada, sólo serán diferentes en una constante.

Integral indefinida. Primitiva e integral indefinida. Cálculo de primitivas: métodos de integración. Integración por cambio de variable e integración por partes. Integración de funciones racionales e irracionales.
Definición 1 Se dice que una función es una primitiva de otra función sobre un intervalo si para todo de se tiene que .
Teorem=2>C una constante cualquiera (un número), la función
F(x) + C es otra primitiva de f(x).
Demostración:
Basta recordar que la derivada de una suma de funciones es igual a la suma de las derivadas de las funciones, y que la derivada de una constante es siempre cero.
(F(x) + C)' = F'(x) + C' = f(x) + 0 = f(x)
Segunda propiedad
Si una función tiene una primitiva, entonces tiene infinitas primitivas.
Demostración:
Si F(x) es una primitiva de f(x), para cualquier constante C, F(x) + C es otra primitiva según la anterior propiedad. Así, hay tantas primitivas como valores se le quieran dar
a C.
Tercera propiedad
Dos primitivas de una misma función se diferencian en una constante. Esto es, si F(x) y G(x) son primitivas de la función f(x), entonces F(x) - G(x) = C = cte.
Demostración:
Hay que recordar que si una función f(x) definida en un intervalo cualquiera tiene derivada cero en todos los puntos, entonces la función f(x) es constante. Es decir, si f'(x) = 0, entonces f(x) = C.
Pues bien, si F(x) es una primitiva de f(x), F'(x) = f(x);
si G(x) es otra primitiva de f(x), G'(x) = f(x).
Restando miembro a miembro, F'(x) - G'(x) = (F(x) - G(x))' = f(x) - f(x) = 0, de donde se deduce que F(x) - G(x) = C.

Una función F se llama antiderivada general o integral general de otra función f, en un intervalo I, si F’(x)=f(x) para todo valor de x en I. A esta función F también se le llama primitiva.

Ejemplo.
Si F se define como F(x)=4x³+x²+5, entonces F’(x)= 12x²+2x.

Así, si f es la función definida por f(x)=12x²+2x, decimos que f es la derivada de F y que F es una antiderivada de f.

Si G es la función definida por G(x)=4x³+x²-17, entonces G también es una antiderivada de f, ya que G’(x)= 12x²+2x.

En realidad, cualquier función cuyo valor esté dado por 4x³+x²+C, donde C es cualquier constante, es una antiderivada de f.

Por consiguiente, si F(x) es una integral de una derivada dada, entonces F(x)+C también lo es. Recíprocamente. Si dos funciones son integrales de una misma derivada, sólo serán diferentes en una constante.

Integral indefinida. Primitiva e integral indefinida. Cálculo de primitivas: métodos de integración. Integración por cambio de variable e integración por partes. Integración de funciones racionales e irracionales.
Definición 1 Se dice que una función es una primitiva de otra función sobre un intervalo si para todo de se tiene que .
Teora 1 Sean y dos primitivas de la función en . Entonces, para todo de , . Es decir dada una función sus primitivas difieren en una constante (en adelante denotaremos por a una constante cualquiera).
Definición 2 El conjunto de todas las primitivas de una función definida en se denomina integral indefinida de y se denota por $\displaystyle \int f(x) dx$ . De manera que, si es una primitiva de ,
$\begin{displaymath} \int f(x) dx = F(x) + C \end{displaymath}$ (2)

Teorema 2 (Propiedades de la integral indefinida.)
$\displaystyle\frac{d}{d x}\left[\displaystyle \int f(x) dx \right]=f(x)$
$\displaystyle \int dF(x) = F(x) + C$
$\forall\alpha,\beta\in\mbox{${\mathbb{R}}$}$ , $\displaystyle \int [\alpha f(x) + \beta g(x)] dx = \displaystyle\alpha \int f(x) dx +\beta \displaystyle \int g(x) dx$
Tabla de Integrales
$\displaystyle \int 0 dx = C$
$\displaystyle \int 1 dx = x+C$
$\displaystyle \int x^\alpha dx = \frac{x^{\alpha+1}}{\alpha+1}+C, \quad \forall \alpha\in \mbox{${\mathbb{R}}$} ,\quad \alpha\neq -1$
$\displaystyle \int \frac{1}{x} dx =\ln \vert x\vert+C$
$\displaystyle \int a^x dx = \frac{a^x}{ \ln a}+C ,\quad\quad a>0, a\neq 1$
$\displaystyle \int \mbox{sen }x dx =-\cos x+C$
$\displaystyle \int \cos x dx = \mbox{sen }x+C$
$\displaystyle \int \displaystyle\frac{1}{\cos^2 x} dx =\tan x +C$
$\displaystyle \int \displaystyle\frac{1}{\mbox{sen }^2 x} dx =-\mbox{ ctg }x +C$
$\displaystyle \int \displaystyle\frac{1}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{l} \mbox{ arcsen }x + C \\ -\arccos x +C \end{array} \right.$ Sean y dos primitivas de la función en . Entonces, para todo de , . Es decir dada una función sus primitivas difieren en una constante (en adelante denotaremos por a una constante cualquiera).
Definición 2 El conjunto de todas las primitivas de una función definida en se denomina integral indefinida de y se denota por $\displaystyle \int f(x) dx$ . De manera que, si es una primitiva de ,
$\begin{displaymath} \int f(x) dx = F(x) + C \end{displaymath}$ (2)

Teorema 2 (Propiedades de la integral indefinida.)
$\displaystyle\frac{d}{d x}\left[\displaystyle \int f(x) dx \right]=f(x)$
$\displaystyle \int dF(x) = F(x) + C$
$\forall\alpha,\beta\in\mbox{${\mathbb{R}}$}$ , $\displaystyle \int [\alpha f(x) + \beta g(x)] dx = \displaystyle\alpha \int f(x) dx +\beta \displaystyle \int g(x) dx$
Tabla de Integrales
$\displaystyle \int 0 dx = C$
$\displaystyle \int 1 dx = x+C$
$\displaystyle \int x^\alpha dx = \frac{x^{\alpha+1}}{\alpha+1}+C, \quad \forall \alpha\in \mbox{${\mathbb{R}}$} ,\quad \alpha\neq -1$
$\displaystyle \int \frac{1}{x} dx =\ln \vert x\vert+C$
$\displaystyle \int a^x dx = \frac{a^x}{ \ln a}+C ,\quad\quad a>0, a\neq 1$
$\displaystyle \int \mbox{sen }x dx =-\cos x+C$
$\displaystyle \int \cos x dx = \mbox{sen }x+C$
$\displaystyle \int \displaystyle\frac{1}{\cos^2 x} dx =\tan x +C$
$\displaystyle \int \displaystyle\frac{1}{\mbox{sen }^2 x} dx =-\mbox{ ctg }x +C$
$\displaystyle \int \displaystyle\frac{1}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{l} \mbox{ arcsen }x + C \\ -\arccos x +C \end{array} \right.$
$\displaystyle \int \displaystyle\frac{1}{ {1+ x^2}} dx = \left\{ \begin{array}{l} \arctan x + C \\ -\mbox{ arcctg }x +C \end{array} \right.$
$\displaystyle \int \displaystyle\frac{1}{\sqrt{x^2\pm 1}} dx = \ln\left\vert x+\sqrt{x^2\pm 1}\right\vert+C$
$\displaystyle \int \displaystyle\frac{1}{{x^2-1}} dx =\frac{1}{2} \ln\left\vert\frac{ x-1}{x+1}\right\vert+C$
$\displaystyle \int \mbox{sh } x dx =\mbox{ch } x+C$
$\displaystyle \int \mbox{ch } x dx =\mbox{sh } x+C$
$\displaystyle \int \frac{1}{\mbox{ch }^2 x} dx =\tanh x +C$
$\displaystyle \int \frac{1}{\mbox{sh }^2 x} dx = \mbox{cth } x + C$
Métodos de integración.
Integración por cambio de variable.
Teorema 3 Sea $t=\phi(x)$ una función derivable en y sean el dominio y $T=\phi[(a,b)]$ la imagen de $\phi(x)$ . Supongamos que sobre el conjunto existe la primitiva de la función , o sea,
$\begin{displaymath} \displaystyle \int g(t)dt =G(t) + C. \end{displaymath}$

Entonces sobre todo el conjunto la función $g[\phi(x)]\phi'(x)$ tiene una primitiva y además
$\begin{displaymath} \displaystyle \int g[\phi(x)]\phi'(x) dx = G[\phi(x)] + C. \end{displaymath}$

Ejemplos:
a) Calcular $\displaystyle \int \cos(2x) dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Para ello hacemos:
$\begin{displaymath} \displaystyle \int \cos(2x) dx =\left\{ \begin{array}{l} ... ...\frac{1}{2} \mbox{sen }y +C= \frac{1}{2} \mbox{sen }(2x) + C \end{displaymath}$

b) Calcular $\displaystyle \int e^{\cos x }\mbox{sen }x dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla:
$\begin{displaymath} \displaystyle \int e^{\cos x}\mbox{sen }x dx =\left\{ \b... ...ht\} = -\displaystyle \int e^y dy = - e^y +C= - e^{\cos x} + C \end{displaymath}$

Integración por partes. Supongamos que las funciones y
$\displaystyle \int \displaystyle\frac{1}{ {1+ x^2}} dx = \left\{ \begin{array}{l} \arctan x + C \\ -\mbox{ arcctg }x +C \end{array} \right.$
$\displaystyle \int \displaystyle\frac{1}{\sqrt{x^2\pm 1}} dx = \ln\left\vert x+\sqrt{x^2\pm 1}\right\vert+C$
$\displaystyle \int \displaystyle\frac{1}{{x^2-1}} dx =\frac{1}{2} \ln\left\vert\frac{ x-1}{x+1}\right\vert+C$
$\displaystyle \int \mbox{sh } x dx =\mbox{ch } x+C$
$\displaystyle \int \mbox{ch } x dx =\mbox{sh } x+C$
$\displaystyle \int \frac{1}{\mbox{ch }^2 x} dx =\tanh x +C$
$\displaystyle \int \frac{1}{\mbox{sh }^2 x} dx = \mbox{cth } x + C$
Métodos de integración.
Integración por cambio de variable.
Teorema 3 Sea $t=\phi(x)$ una función derivable en y sean el dominio y $T=\phi[(a,b)]$ la imagen de $\phi(x)$ . Supongamos que sobre el conjunto existe la primitiva de la función , o sea,
$\begin{displaymath} \displaystyle \int g(t)dt =G(t) + C. \end{displaymath}$

Entonces sobre todo el conjunto la función $g[\phi(x)]\phi'(x)$ tiene una primitiva y además
$\begin{displaymath} \displaystyle \int g[\phi(x)]\phi'(x) dx = G[\phi(x)] + C. \end{displaymath}$

Ejemplos:
a) Calcular $\displaystyle \int \cos(2x) dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Para ello hacemos:
$\begin{displaymath} \displaystyle \int \cos(2x) dx =\left\{ \begin{array}{l} ... ...\frac{1}{2} \mbox{sen }y +C= \frac{1}{2} \mbox{sen }(2x) + C \end{displaymath}$

b) Calcular $\displaystyle \int e^{\cos x }\mbox{sen }x dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla:
$\begin{displaymath} \displaystyle \int e^{\cos x}\mbox{sen }x dx =\left\{ \b... ...ht\} = -\displaystyle \int e^y dy = - e^y +C= - e^{\cos x} + C \end{displaymath}$

Integración por partes. Supongamos que las funciones y son derivables en un intervalo y existe la primitiva de la función en . Entonces, sobre existe la primitiva de y se cumple que
$\begin{displaymath} \displaystyle \int u(x) v'(x) dx = u(x)v(x) - \displaystyle \int v(x)u'(x) dx , \end{displaymath}$ (3)

o en forma diferencial
$\begin{displaymath} \displaystyle \int u(x) dv(x)= u(x)v(x) - \displaystyle \int v(x)du(x). \end{displaymath}$ (4)

Ejemplos:
a) Calcular $\displaystyle \int x^n \ln x dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Utilicemos la integración por partes:
$\begin{displaymath} \begin{array}{l} \displaystyle \int x^n \ln x dx = \left\... ...{n+1}}{n+1}\left( \ln x - \frac{1}{n+1} \right) + C \end{array}\end{displaymath}$

b) Calcular $I= \displaystyle \int e^{ax} \cos bx dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Utilizemos la integración por partes:
$\begin{displaymath} \begin{array}{l} I = \displaystyle \int e^{ax}\cos bx dx ... ...{b} \displaystyle \int e^{ax}\mbox{sen }bx dx . \end{array}\end{displaymath}$

La integral $\displaystyle \int e^{ax}\mbox{sen }bx dx$ es de la misma forma que la original así que volveremos a aplicar integración por partes:
$\begin{displaymath} \begin{array}{l} \displaystyle \int e^{ax}\mbox{sen }bx ... ...frac{a}{b} \displaystyle \int e^{ax}\cos bx dx . \end{array}\end{displaymath}$

Juntando las dos fórmulas anteriores concluimos que
$\begin{displaymath} I= \frac{e^{ax}\mbox{sen }bx }{b} + \frac{a}{b^2} {e^{ax}\cos bx } - \frac{a^2}{b^2} I, \end{displaymath}$

de donde, resolviendo la ecuación respecto a obtenemos:
$\begin{displaymath} I = \displaystyle \int e^{ax}\cos bx dx = \frac{b \mbox{sen }bx + a \cos bx }{a^2+b^2}e^{ax}+C. \end{displaymath}$

Algunas de las integrales que pueden ser calculadas $ height=31 alt=$v(x)$ src="http://euler.us.es/~renato/clases/programa/img138.png" width=34 align=middle border=0> son derivables en un intervalo y existe la primitiva de la función en . Entonces, sobre existe la primitiva de y se cumple que
$\begin{displaymath} \displaystyle \int u(x) v'(x) dx = u(x)v(x) - \displaystyle \int v(x)u'(x) dx , \end{displaymath}$ (3)

o en forma diferencial
$\begin{displaymath} \displaystyle \int u(x) dv(x)= u(x)v(x) - \displaystyle \int v(x)du(x). \end{displaymath}$ (4)

Ejemplos:
a) Calcular $\displaystyle \int x^n \ln x dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Utilicemos la integración por partes:
$\begin{displaymath} \begin{array}{l} \displaystyle \int x^n \ln x dx = \left\... ...{n+1}}{n+1}\left( \ln x - \frac{1}{n+1} \right) + C \end{array}\end{displaymath}$

b) Calcular $I= \displaystyle \int e^{ax} \cos bx dx$ . Como la integral no es de la tabla es necesario convertirla en una de la tabla. Utilizemos la integración por partes:
$\begin{displaymath} \begin{array}{l} I = \displaystyle \int e^{ax}\cos bx dx ... ...{b} \displaystyle \int e^{ax}\mbox{sen }bx dx . \end{array}\end{displaymath}$

La integral $\displaystyle \int e^{ax}\mbox{sen }bx dx$ es de la misma forma que la original así que volveremos a aplicar integración por partes:
$\begin{displaymath} \begin{array}{l} \displaystyle \int e^{ax}\mbox{sen }bx ... ...frac{a}{b} \displaystyle \int e^{ax}\cos bx dx . \end{array}\end{displaymath}$

Juntando las dos fórmulas anteriores concluimos que
$\begin{displaymath} I= \frac{e^{ax}\mbox{sen }bx }{b} + \frac{a}{b^2} {e^{ax}\cos bx } - \frac{a^2}{b^2} I, \end{displaymath}$

de donde, resolviendo la ecuación respecto a obtenemos:
$\begin{displaymath} I = \displaystyle \int e^{ax}\cos bx dx = \frac{b \mbox{sen }bx + a \cos bx }{a^2+b^2}e^{ax}+C. \end{displaymath}$

Algunas de las integrales que pueden ser calculadautilizando la integración por partes son:
Las integrales donde aparezcan las funciones $\ln x$ , $\mbox{ arcsen }x$ , $\arccos x$ , $\ln \phi(x)$ , potencias enteras de las funciones anteriores, entre otras donde tendremos que escoger como función a alguna de las funciones anteriores (ver ejemplo a).
Las integrales $\displaystyle \int (ax+b)^n \mbox{sen }cx dx$ , $\displaystyle \int (ax+b)^n \cos cx dx$ y $\displaystyle \int (ax+b)^n e^{cx} dx$ . Donde para encontrar las primitivas hay que utilizar la fórmula de integración por partes veces tomando cada vez , $u(x)=(ax+b)^{n-1}$ , ...., respectivamente.
Las integrales de la forma $\displaystyle \int e^{ax}\mbox{sen }bx dx$ , $\displaystyle \int e^{ax}\cos bx dx$ , $\displaystyle \int \mbox{sen }(\ln x) dx$ y $\displaystyle \int \cos(\ln x) dx$ . Para encontrar las primitivas hay que denotar por a cualquiera de las integrales anteriores, aplicar dos veces integración por partes y resolver la ecuación resultante respecto a (ver ejemplo b).
Integración de funciones racionales.
$\begin{dff} Diremos que una funci\'on racional $f(x)=\displaystyle\frac{P_n(x)}{... ...mio $P_n(x)$ es menor que el del polinomio $Q_m(x)$, o sea, si $n<m$. \end{dff}$
Si entonces podemos dividir los polinomios y de tal forma que
$\begin{displaymath} \displaystyle\frac{P_n(x)}{Q_m(x)}= p_{n-m}(x)+\displaystyle\frac{R_k(x)}{Q_m(x)}, \quad \mbox{donde}\quad k<m. \end{displaymath}$

Teorema 4 Supongamos que $\displaystyle\frac{P_n(x)}{Q_m(x)}$ es una fracción simple, y que el polinomio denominador se puede factorizar de la siguiente forma
$\begin{displaymath} Q_n(x)= (x-x_1)^{n_1}\cdots (x-x_p)^{n_p} (x^2+p_1 x +q_1)^{m_1}\cdots (x^2+p_k x +q_k)^{m_k}, \end{displaymath}$ (5)

donde son las raíces reales de , y los factores no tienen raíces reales. Entonces, la fracción simple $\displaystyle\frac{P_n(x)}{Q_m(x)}$ se puede descomponer en las siguientes fracciones elementales simples:
$\mbox{ arcsen }x$ , $\arccos x$ , $\ln \phi(x)$ , potencias enteras de las funciones anteriores, entre otras donde tendremos que escoger como función a alguna de las funciones anteriores (ver ejemplo a).
Las integrales $\displaystyle \int (ax+b)^n \mbox{sen }cx dx$ , $\displaystyle \int (ax+b)^n \cos cx dx$ y $\displaystyle \int (ax+b)^n e^{cx} dx$ . Donde para encontrar las primitivas hay que utilizar la fórmula de integración por partes veces tomando cada vez , $u(x)=(ax+b)^{n-1}$ , ...., respectivamente.
Las integrales de la forma $\displaystyle \int e^{ax}\mbox{sen }bx dx$ , $\displaystyle \int e^{ax}\cos bx dx$ , $\displaystyle \int \mbox{sen }(\ln x) dx$ y $\displaystyle \int \cos(\ln x) dx$ . Para encontrar las primitivas hay que denotar por a cualquiera de las integrales anteriores, aplicar dos veces integración por partes y resolver la ecuación resultante respecto a (ver ejemplo b).
Integración de funciones racionales.
$\begin{dff} Diremos que una funci\'on racional $f(x)=\displaystyle\frac{P_n(x)}{... ...mio $P_n(x)$ es menor que el del polinomio $Q_m(x)$, o sea, si $n<m$. \end{dff}$
Si entonces podemos dividir los polinomios y de tal forma que
$\begin{displaymath} \displaystyle\frac{P_n(x)}{Q_m(x)}= p_{n-m}(x)+\displaystyle\frac{R_k(x)}{Q_m(x)}, \quad \mbox{donde}\quad k<m. \end{displaymath}$

Teorema 4 Supongamos que $\displaystyle\frac{P_n(x)}{Q_m(x)}$ es una fracción simple, y que el polinomio denominador se puede factorizar de la siguiente forma
$\begin{displaymath} Q_n(x)= (x-x_1)^{n_1}\cdots (x-x_p)^{n_p} (x^2+p_1 x +q_1)^{m_1}\cdots (x^2+p_k x +q_k)^{m_k}, \end{displaymath}$ (5)

donde son las raíces reales de , y los factores no tienen raíces reales. Entonces, la fracción simple $\displaystyle\frac{P_n(x)}{Q_m(x)}$ se puede descomponer en las siguientes fracciones elementales simples:
$\begin{displaymath}\begin{array}{rl} \displaystyle\frac{P_n(x)}{Q_m(x)}= & \disp... ...k)^{m_k}}+\cdots+\frac{L_1x+K_1}{(x^2+p_k x +q_k)}, \end{array}\end{displaymath}$ (6)

donde , , , , y son ciertas constantes reales.
Para determinar dichas constantes sumamos los términos de la derecha. Nótese que el denominador común coincide con (5) y el numerador es un polinomio de grado a lo sumo . Luego comparamos el polinomio numerador que se obtiene al sumar las fracciones más simples en (6) con . Igualando los coeficientes de ambos obtendremos un sistema de ecuaciones con incógnitas que podemos resolver para encontar los coeficientes indeterminados , , , , y . No obstante es posible encontrar el coeficiente $A_{n_i}$ de los sumandos correspondientes a uno de los ceros reales , o sea, el $A_{n_i}$ de
$\begin{displaymath} \frac{A_{n_i}}{(x-x_i)^{n_i}}+\frac{A_{n_i-1}}{(x-x_i)^{n_i-1}}+\cdots+\frac{A_1}{(x-x_i)}, \end{displaymath}$

utilizando la propiedad que
$\begin{displaymath} \lim_{x\to x_i} \displaystyle\frac{P_n(x)(x-x_i)^{n_i}}{Q_m(x)}= A_{n_i}. \end{displaymath}$ (7)

Como consecuencia de lo anterior, si tiene ceros reales y simples, o sea, si su factorización es de la forma
$\begin{displaymath} Q_n(x)= (x-x_1)(x-x_2)\cdots (x-x_{m-1})(x-x_m), \end{displaymath}$ (8)

entonces, $\displaystyle\frac{P_n(x)}{Q_m(x)}$ se puede descomponer en las fracciones elementales simples:

$\begin{displaymath}\begin{array}{rl} \displaystyle\frac{P_n(x)}{Q_m(x)}= & \disp... ...s+\frac{A_{m-1}}{(x-x_{m-1})}+ \frac{A_m}{(x-x_m)}, \end{array}\end{displaymath}$ (9)

donde ,...,
$\begin{displaymath}\begin{array}{rl} \displaystyle\frac{P_n(x)}{Q_m(x)}= & \disp... ...k)^{m_k}}+\cdots+\frac{L_1x+K_1}{(x^2+p_k x +q_k)}, \end{array}\end{displaymath}$ (6)

donde , , , , y son ciertas constantes reales.
Para determinar dichas constantes sumamos los términos de la derecha. Nótese que el denominador común coincide con (5) y el numerador es un polinomio de grado a lo sumo . Luego comparamos el polinomio numerador que se obtiene al sumar las fracciones más simples en (6) con . Igualando los coeficientes de ambos obtendremos un sistema de ecuaciones con incógnitas que podemos resolver para encontar los coeficientes indeterminados , , , , y . No obstante es posible encontrar el coeficiente $A_{n_i}$ de los sumandos correspondientes a uno de los ceros reales , o sea, el $A_{n_i}$ de
$\begin{displaymath} \frac{A_{n_i}}{(x-x_i)^{n_i}}+\frac{A_{n_i-1}}{(x-x_i)^{n_i-1}}+\cdots+\frac{A_1}{(x-x_i)}, \end{displaymath}$

utilizando la propiedad que
$\begin{displaymath} \lim_{x\to x_i} \displaystyle\frac{P_n(x)(x-x_i)^{n_i}}{Q_m(x)}= A_{n_i}. \end{displaymath}$ (7)

Como consecuencia de lo anterior, si tiene ceros reales y simples, o sea, si su factorización es de la forma
$\begin{displaymath} Q_n(x)= (x-x_1)(x-x_2)\cdots (x-x_{m-1})(x-x_m), \end{displaymath}$ (8)

entonces, $\displaystyle\frac{P_n(x)}{Q_m(x)}$ se puede descomponer en las fracciones elementales simples:

$\begin{displaymath}\begin{array}{rl} \displaystyle\frac{P_n(x)}{Q_m(x)}= & \disp... ...s+\frac{A_{m-1}}{(x-x_{m-1})}+ \frac{A_m}{(x-x_m)}, \end{array}\end{displaymath}$ (9)

donde ,..., se calculan por la fórmula
$\begin{displaymath} A_k=\lim_{x\to x_k} \displaystyle\frac{P_n(x)(x-x_k)}{Q_m(x)},\quad k=1,2,...,m. \end{displaymath}$ (10)

Teorema 5 (Primitivas de las fracciones simples más elementales) ¹

$\begin{displaymath} \begin{array}{l} 1)\quad \displaystyle\int \frac{A}{x-a} d... ...qrt{4q-p^2}}\arctan \frac{2x+p}{\sqrt{4q-p^2}} +C . \end{array}\end{displaymath}$ (11)

Ejemplos:
a) Calcular $\displaystyle\int \frac{x}{x^2-3x+2} dx$ . Primero encontraremos las fracciones simples mas elementales:
$\begin{displaymath} \frac{x}{x^2-3x+2}= \frac{x}{(x-1)(x-2)}= \frac{A}{x-1}+\frac{B}{x-2}. \end{displaymath}$

Luego, utilizando (10) obtenemos
$\begin{displaymath} A=\lim_{x\to 1} \frac{x(x-1)}{(x-1)(x-2)}=-1,\quad B=\lim_{x\to 2} \frac{x(x-2)}{(x-1)(x-2)}=2. \end{displaymath}$

Finalmente, utilizando (11) obtenemos
$\begin{displaymath} \displaystyle\int \frac{x}{x^2-3x+2} dx = \displaystyle\in... ...2}{x-2}\right) dx = -\ln\vert x-1\vert+2\ln\vert x-2\vert+C. \end{displaymath}$

a) Calcular $\displaystyle\int \frac{x}{(x-1)^2 (x^2+1)} dx$ . Primero encontraremos las fracciones simples mas elementales:
$\begin{displaymath} \frac{x}{(x-1)^2(x^2+1)}= \frac{A}{(x-1)^2}+ \frac{B}{x-1}+\... ...= \frac{A(x^2+1)+B(x^2+1)(x-1)+(Cx+D)(x-1)^2}{(x-1)^2(x^2+1)}. \end{displaymath}$

Para encontrar los coeficientes igualamos los polinomios de los numeradores:
$\begin{displaymath} x=A(x^2+1)+B(x^2+1)(x-1)+(Cx+D)(x-1)^2. \end{displaymath}$

Dos polinomios de grado 3 son iguales si los coeficientes de las potencias , , y son iguales, por lo que igualando dichos coeficientes obtenemos el sistema de ecuaciones:
$\begin{displaymath}\begin{array}{cc} \begin{array}{lr} x^3: & B+C=0 x^2: & A-... ...\frac{1}2 B= 0 C=0 D=-\frac{1}2 \end{array}\end{array}\end{displaymath}$ $\begin{displaymath} A_k=\lim_{x\to x_k} \displaystyle\frac{P_n(x)(x-x_k)}{Q_m(x)},\quad k=1,2,...,m. \end{displaymath}$ (10)
Teorema 5 (Primitivas de las fracciones simples más elementales) ¹

$\begin{displaymath} \begin{array}{l} 1)\quad \displaystyle\int \frac{A}{x-a} d... ...qrt{4q-p^2}}\arctan \frac{2x+p}{\sqrt{4q-p^2}} +C . \end{array}\end{displaymath}$ (11)

Ejemplos:
a) Calcular $\displaystyle\int \frac{x}{x^2-3x+2} dx$ . Primero encontraremos las fracciones simples mas elementales:
$\begin{displaymath} \frac{x}{x^2-3x+2}= \frac{x}{(x-1)(x-2)}= \frac{A}{x-1}+\frac{B}{x-2}. \end{displaymath}$

Luego, utilizando (10) obtenemos
$\begin{displaymath} A=\lim_{x\to 1} \frac{x(x-1)}{(x-1)(x-2)}=-1,\quad B=\lim_{x\to 2} \frac{x(x-2)}{(x-1)(x-2)}=2. \end{displaymath}$

Finalmente, utilizando (11) obtenemos
$\begin{displaymath} \displaystyle\int \frac{x}{x^2-3x+2} dx = \displaystyle\in... ...2}{x-2}\right) dx = -\ln\vert x-1\vert+2\ln\vert x-2\vert+C. \end{displaymath}$

a) Calcular $\displaystyle\int \frac{x}{(x-1)^2 (x^2+1)} dx$ . Primero encontraremos las fracciones simples mas elementales:
$\begin{displaymath} \frac{x}{(x-1)^2(x^2+1)}= \frac{A}{(x-1)^2}+ \frac{B}{x-1}+\... ...= \frac{A(x^2+1)+B(x^2+1)(x-1)+(Cx+D)(x-1)^2}{(x-1)^2(x^2+1)}. \end{displaymath}$

Para encontrar los coeficientes igualamos los polinomios de los numeradores:
$\begin{displaymath} x=A(x^2+1)+B(x^2+1)(x-1)+(Cx+D)(x-1)^2. \end{displaymath}$

Dos polinomios de grado 3 son iguales si los coeficientes de las potencias , , y son iguales, por lo que igualando dichos coeficientes obtenemos el sistema de ecuaciones:
$\begin{displaymath}\begin{array}{cc} \begin{array}{lr} x^3: & B+C=0 x^2: & A-... ...\frac{1}2 B= 0 C=0 D=-\frac{1}2 \end{array}\end{array}\end{displaymath}$

También es posible utilizar otra propiedad de los polinomios: dos polinomios de grado que toman valores iguales en puntos dados son identicamente iguales, es decir, si para ciertos $x_1,...,x_{n+1}$ (distintos entre si), entonces $P_n(x)\equiv Q_n(x)$ para todo $x\in \mbox{${\mathbb{R}}$}$ . En nuestro ejemplo es conveniente tomar como los los ceros de los polinomios denominadores y luego el resto de los valores tomarlos los más sencillos posibles:
$\begin{displaymath} \begin{array}{cc} \begin{array}{lr} x=1: & A= \frac{1}2 x... ...\frac{1}2 B= 0 C=0 D=-\frac{1}2 \end{array}\end{array}\end{displaymath}$

que coincide con la encontrada por el método anterior. Luego,
$\begin{displaymath} \displaystyle\int \frac{x}{(x-1)^2 (x^2+1)} dx =\frac{1}2 ... ...2+1} \right] dx = - \frac{1}{2(x-1)}-\frac{1}2 \arctan x +C. \end{displaymath}$

Integrales trigonométricas. En este apartado vamos a estudiar las integrales de la forma $\displaystyle \int f(\mbox{sen }x,\cos x) dx$ las cuales se convierten en integrales racionales mediante la sustitución trigonométrica $t=\tan \frac{x}2$ ,
$\begin{displaymath} \int f(\mbox{sen }x,\cos x) dx = \left\{ \begin{array}{l... ...\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right) \frac{2}{1+t^2}dt, \end{displaymath}$

que es un integral de una función racional.
Ejemplo. Calcular la integral $\displaystyle\int\frac{ dx }{\mbox{sen }x}$ .
$\begin{displaymath} \displaystyle\int\frac{ dx }{\mbox{sen }x} = \left\{ \beg... ...t = \ln\vert t\vert+C=\ln\left\vert\tan\frac{x}2\right\vert+C. \end{displaymath}$

Existen varios tipos de integrales trigonométricas que se pueden racionalizar con cambios más sencillos. Ellas son las siguientes:
$\displaystyle \int f(\mbox{sen }x,\cos x) dx$ , donde $f(-\mbox{sen }x,\cos x)=-f(\mbox{sen }x,\cos x)$ , cambio $t=\cos x$
$\displaystyle \int f(\mbox{sen }x,\cos x) dx$ , donde $f(\mbox{sen }x,-\cos x)=-f(\mbox{sen }x,\cos x)$ , cambio $t=\mbox{sen }x$

También es posible utilizar otra propiedad de los polinomios: dos polinomios de grado que toman valores iguales en puntos dados son identicamente iguales, es decir, si para ciertos $x_1,...,x_{n+1}$ (distintos entre si), entonces $P_n(x)\equiv Q_n(x)$ para todo $x\in \mbox{${\mathbb{R}}$}$ . En nuestro ejemplo es conveniente tomar como los los ceros de los polinomios denominadores y luego el resto de los valores tomarlos los más sencillos posibles:

$\begin{displaymath} \begin{array}{cc} \begin{array}{lr} x=1: & A= \frac{1}2 x... ...\frac{1}2 B= 0 C=0 D=-\frac{1}2 \end{array}\end{array}\end{displaymath}$

que coincide con la encontrada por el método anterior. Luego,

$\begin{displaymath} \displaystyle\int \frac{x}{(x-1)^2 (x^2+1)} dx =\frac{1}2 ... ...2+1} \right] dx = - \frac{1}{2(x-1)}-\frac{1}2 \arctan x +C. \end{displaymath}$

Integrales trigonométricas. En este apartado vamos a estudiar las integrales de la forma $\displaystyle \int f(\mbox{sen }x,\cos x) dx$ las cuales se convierten en integrales racionales mediante la sustitución trigonométrica $t=\tan \frac{x}2$ ,

$\begin{displaymath} \int f(\mbox{sen }x,\cos x) dx = \left\{ \begin{array}{l... ...\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right) \frac{2}{1+t^2}dt, \end{displaymath}$

que es un integral de una función racional.

Ejemplo. Calcular la integral $\displaystyle\int\frac{ dx }{\mbox{sen }x}$ .

$\begin{displaymath} \displaystyle\int\frac{ dx }{\mbox{sen }x} = \left\{ \beg... ...t = \ln\vert t\vert+C=\ln\left\vert\tan\frac{x}2\right\vert+C. \end{displaymath}$

Existen varios tipos de integrales trigonométricas que se pueden racionalizar con cambios más sencillos. Ellas son las siguientes:

$\displaystyle \int f(\mbox{sen }x,\cos x) dx$ , donde $f(-\mbox{sen }x,\cos x)=-f(\mbox{sen }x,\cos x)$ , cambio $t=\cos x$
$\displaystyle \int f(\mbox{sen }x,\cos x) dx$ , donde $f(\mbox{sen }x,-\cos x)=-f(\mbox{sen }x,\cos x)$ , cambio $t=\mbox{sen }x$ , donde $f(-\mbox{sen }x,-\cos x)=f(\mbox{sen }x,\cos x)$ , cambio $t=\tan x$

Ejemplos.

a) Calcular la integral $\displaystyle\int\frac{ dx }{\mbox{sen }x}$ . Esta integral es del tipo 1. Luego,

$\begin{displaymath} \displaystyle\int\frac{ dx }{\mbox{sen }x} = \left\{ \beg... ...}{1-t}\right\vert+C =\ln\left\vert\tan\frac{x}2 \right\vert+C. \end{displaymath}$

que coincide con el resultado obtenido al utilizar la sustitución $t=\tan \frac{x}2$

b) Calcular la integral $\displaystyle\int\cos^3 x { dx }$ . Esta integral es del tipo 2. Luego,

$\begin{displaymath} \displaystyle\int\cos^3 x dx = \left\{ \begin{array}{l... ...\frac{1}3 t^3+C= \mbox{sen }x -\frac{\mbox{sen }^3 x}3 + C. \end{displaymath}$

c) Calcular la integral $\displaystyle\int\tan^3 x { dx }$ . Esta integral es del tipo 3. Luego,

$\begin{displaymath}\begin{array}{l} \displaystyle\int\tan^3 x dx = \left\{... ...2 x)+C = \frac{\tan^2 x}2 + \ln\vert\cos x\vert+C. \end{array}\end{displaymath}$

Integrales irracionales. En este apartado vamos a estudiar las integrales de la forma $\displaystyle \int f(x,\sqrt{x^2 \pm a^2} ) dx$ , $\displaystyle \int f(x,\sqrt{a^2 - x^2} ) dx$ y $\displaystyle \int f\left( x, \sqrt[n]{\frac{ax+b}{cx+d}} \right) dx$ .

Las integrales $\displaystyle \int f(x,\sqrt{x^2 \pm a^2} ) dx$ y $\displaystyle \int f(x,\sqrt{a^2 - x^2} ) dx$ .

Estas integrales irracionales se convierten en integrales trigonométricas mediante los cambios:

$\displaystyle \int f(x,\sqrt{a^2 - x^2} ) dx$ , cambio $x=a\mbox{sen }t$
$\displaystyle \int f(x,\sqrt{x^2 - a^2} ) dx$ , cambio $\displaystyle x=\frac{a}{\mbox{sen }t}$
$\displaystyle \int f(x,\sqrt{x^2 + a^2} ) dx$ , cambio $x=a \tan t$

Ejemplos.

a) Cal" width=65 align=bottom border=0>

$\displaystyle \int f(\mbox{sen }x,\cos x) dx$ , donde $f(-\mbox{sen }x,-\cos x)=f(\mbox{sen }x,\cos x)$ , cambio $t=\tan x$