Use the substitution method to evaluate the integral. 1. \displaystyle \int \frac { 1 } { ( 4 x...

Question:

Use the substitution method to evaluate the integral.

1. {eq}\displaystyle \int \frac { 1 } { ( 4 x + 1 ) ^ { 3 } } d x {/eq}

2. {eq}\displaystyle \int e ^ { 100 x } d x {/eq}

3. {eq}\displaystyle \int ( 1.0003 ) ^ { 12 t } d t {/eq}

4. {eq}\displaystyle \int \frac { e ^ { \frac{ 10 }{ x } } } { x ^ { 2 } } d x {/eq}

Integration by "u"-substitution

There will be integrals that at first glance seem difficult to solve because you can't find them in any table of integrals. Integration by "u"-substitution is a technique where you rewrite the integrand in terms of the new variable "u" such that the new integral's form is easily solvable.

Answer and Explanation:


In integration using "u"-substitution, first you need to rewrite the integral in terms of the new variable and evaluate the new integral. After evaluating, you revert back to the original variables.


1. {eq}\displaystyle \int \frac { 1 } { ( 4 x + 1 ) ^ { 3 } } d x \\ \text{ let } u = ( 4 x + 1 ) ; du = 4 dx; dx = \frac{du} {4} \\ \displaystyle \int \frac { 1 } { u ^ { 3 } } \frac{du} {4} = \frac{1}{4}\int u^{-3}du {/eq}

Using {eq}\displaystyle \int u^n du = \frac{u^{n+1}}{n+1} + C {/eq}

{eq}\displaystyle \frac{1}{4}\int u^{-3}du = \frac{1}{4} \frac{u^{-2}}{-2} + C {/eq}


Reverting back to the original variables.

{eq}u = ( 4 x + 1 ) \\ \displaystyle \int \frac { 1 } { ( 4 x + 1 ) ^ { 3 } } d x = \frac{1}{4} \frac{( 4 x + 1 )^{-2}}{-2} + C \\ \displaystyle \int \frac { 1 } { ( 4 x + 1 ) ^ { 3 } } d x = \frac{-1}{4} \frac{1}{2 \bigg [( 4 x + 1 )^{2} \bigg ] } + C \\ \displaystyle \int \frac { 1 } { ( 4 x + 1 ) ^ { 3 } } d x = - \frac{1}{8( 4 x + 1 )^{2}} + C \\ {/eq}


2.) {eq}\displaystyle \int e ^ { 100 x } d x {/eq}

{eq}\text{let } u = 100 x; du = 100 dx, dx = \frac{du}{100} \\ \displaystyle \int e ^ { 100 x } d x = \int e ^ { u } \frac{du}{100} {/eq}

We use the {eq}\displaystyle \int e^u du = e^u + C \\ \displaystyle \int e ^ { u } \frac{du}{100} = \frac{1}{100}e^u + C {/eq}


Reverting back into the original variables.

{eq}u = 100x \\ \displaystyle \int e ^ { 100 x } d x = \frac{1}{100}e^{100x} + C {/eq}


3.) {eq}\displaystyle \int ( 1.0003 ) ^ { 12 t } d t {/eq}

{eq}\text{ let } u = 12t , du = 12 dt, dt = \frac{du}{12} \\ \displaystyle \int ( 1.0003 ) ^ { 12 t } d t = \frac{1}{12} \int ( 1.0003 ) ^ {u } du {/eq}


Using {eq}\displaystyle \int a^u du = \frac{a^u}{\ln a} + C \\ \displaystyle \frac{1}{12} \int ( 1.0003 ) ^ {u } du = \frac{1}{12}\frac{1.0003^u}{\ln a} + C {/eq}


Reverting back to the original variables.

{eq}u = 12t \\ \displaystyle \int ( 1.0003 ) ^ { 12 t } d t = \bigg( \frac{1}{12}\bigg ) \frac{1.0003^{12t}}{\ln 1.0003 } + C {/eq}


4.) {eq}\displaystyle \int \frac { e ^ { \frac{ 10 }{ x } } } { x ^ { 2 } } d x {/eq}

{eq}\displaystyle \text{ let } u = \frac{10}{x}, du = \frac{-10}{x^2} dx , dx = -\frac{x^2}{10}du \\ \displaystyle \int \frac { e ^ { \frac{ 10 }{ x } } } { x ^ { 2 } } d x = \int \frac { e ^ { u} } { x ^ { 2 } } \bigg ( -\frac{x^2}{10}\bigg ) du \\ \displaystyle \int \frac { e ^ { \frac{ 10 }{ x } } } { x ^ { 2 } } d x = - \int \frac { e ^ { u} } { 10} du {/eq}

We use {eq}\displaystyle \int e^u du = e^u + C \\ \displaystyle - \int \frac { e ^ { u} } { 10} du =- \frac{1}{10}e^u + C {/eq}


Reverting back to the original variables.

{eq}\displaystyle u = \frac{10}{x} \\ \displaystyle \int \frac { e ^ { \frac{ 10 }{ x } } } { x ^ { 2 } } d x = -\frac{1}{10}e^{\frac{10}{x}} + C {/eq}


Learn more about this topic:

Loading...
How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
6.6K

Related to this Question

Explore our homework questions and answers library