Difference between revisions of "Integration by Substitution"

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[https://www.youtube.com/watch?v=uoCW8S-I9Es Example 1]. Produced by Professor Zachary Sharon, UTSA
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===Integration by Substitution===
  
[https://www.youtube.com/watch?v=zqMxMtjbaBE Example 2]. Produced by TA Catherine Sporer, UTSA
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<p>There is a theorem that will help you with substitution for integration.  It is called '''Change of Variables for Definite Integrals'''.</p>
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<p>what the theorem looks like is this</p>
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 +
<p><center><math>\int_{a}^{b} f(x)\operatorname {d}x = \int_{\alpha}^{\beta} f(g(u))g\prime (u)\operatorname {d}u</math></center></p>
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<p><br>In order to get <math>\alpha</math> you must plug '''<i>a</i>''' into the function '''g''' and to get <math>\beta</math> you must plug '''<i>b</i>''' into the function '''g'''.</br></p>
 +
 
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<p>The tricky part is trying to identify what you want to make your '''<i>u</i>''' to be.  Some times substitution will not be enough and you will have to use the rules for integration by parts.  That will be covered in a different section</p>
 +
 
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====Steps====
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:{|
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|<math>\int\limits_{x=a}^{x=b}f(x)dx</math>
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|<math>=\int\limits_{x=a}^{x=b} f(x)\ \frac{du}{du}\ dx</math>
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|style="padding-left: 20px"|(1)
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|i.e. <math>\frac{du}{du}=1</math>
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|-
 +
|
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|<math>=\int\limits_{x=a}^{x=b}{\left(f(x)\ \frac{dx}{du}\right)\left(\frac{du}{dx}\right)}\ dx</math>
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|style="padding-left: 20px"|(2)
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|i.e. <math>\frac{dx}{du}\cdot\frac{du}{dx}=1</math>
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|-
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|
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|<math>=\int\limits_{x=a}^{x=b}\left(f(x)\ \frac{dx}{du}\right)g'(x)\ dx</math>
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|style="padding-left: 20px"|(3)
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|i.e. <math>\frac{du}{dx}=g'(x)</math>
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|-
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|
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|<math>=\int\limits_{x=a}^{x=b}h(g(x))g'(x)dx</math>
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|style="padding-left: 20px"|(4)
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|i.e. Now equate <math>\left(f(x)\ \frac{dx}{du}\right)</math> with <math>h(g(x))</math>
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|-
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|
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|<math>=\int\limits_{x=a}^{x=b}h(u)g'(x)dx</math>
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|style="padding-left: 20px"|(5)
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|i.e. <math>g(x)=u</math>
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|-
 +
|
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|<math>=\int\limits_{u=g(a)}^{u=g(b)}h(u)du</math>
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|style="padding-left: 20px"|(6)
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|i.e. <math>du=\frac{du}{dx}dx=g'(x)dx</math>
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|-
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|
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|<math>=\int\limits_{u=c}^{u=d}h(u)du</math>
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|style="padding-left: 20px"|(7)
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|i.e. We have achieved our desired result
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|}
 +
 
 +
===Example 1===
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<p><math>\int_{0}^{2} x(x^2+1)^2 \operatorname {d}x</math></p>
 +
 
 +
<p>Instead of making this a big polynomial we will just use the substitution method.</p>
 +
 
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<p>'''Step 1'''</p>
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: <p>Identify your <i>u</i></p>
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: <p>Let <math> u = x^2+1</math></p>
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<p>'''Step 2'''</p>
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: <p><br>Identify <math>\operatorname {d}u</math></br></p>
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: <p> <br><math>\operatorname {d}u = 2x\operatorname {d}x</math></br></p>
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'''Step 3'''
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: <p>Now we plug in our limits of integration to our <i>u</i> to find our new limits of integration</p>
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: <p>When <math> x = 0, u =0^2 + 1 = 1</math></p>
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: <p>and when <math>x = 2, u = 2^2 + 1 = 5</math></p>
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: <p>Now our integration problem looks something like this</p>
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: <p><center><math>\frac {1}{2} \int_{0}^{5} (x^2 + 1)^2 (2x)\operatorname {d}x</math></center></p>
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'''Step 4'''
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: <p>write your new integration problem</p>
 +
 
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: <p><br>When we plug in our <i>u</i> it looks like </br></p>
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<p><center><math>\frac {1}{2} \int_{0}^{5} (u)^2 \operatorname {d}u</math></center></p>
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'''Step 5'''
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: <p>Evaluate the Integral</p>
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: <p><center><math>\frac {1}{2} \left[\frac {1}{3} u^3 \right]_{0}^{5}</math></center></p>
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: <p><br><center><math>\frac {1}{2} \left[\left(\frac {1}{3} * 5^3 \right) - \left(\frac {1}{3} * 0^3 \right)\right]</math></center></br></p>
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: <p><br><center><math>\frac {1}{2} \left[\frac {1}{3} * 125 \right]</math></center></br></p>
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: <p><br><center><math>\frac {1}{2} \left[\frac {125}{3}\right]</math></center></br></p>
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: <p><br><center><math>\frac {125}{6}</math></center></br></p>
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: <p><br>As you can see this all simplified fairly nice.  Using substitution will be hard, for most people, at first.  Once you get the hang of doing this it should come to you faster and faster each time.</br></p>
 +
 
 +
===Example 2===
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:<math>\int 3x^2(x^3+1)^5dx</math>
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we see that <math>3x^2</math> is the derivative of <math>x^3+1</math> . Letting
 +
 
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:<math>u=x^3+1</math>
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 +
we have
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:<math>\frac{du}{dx}=3x^2</math>
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or, in order to apply it to the integral,
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:<math>du=3x^2dx</math>
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With this we may write
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:<math>\int 3x^2(x^3+1)^5dx=\int u^5du=\frac{u^6}{6}+C=\frac{(x^3+1)^6}{6}+C</math>
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Note that it was not necessary that we had <i>exactly</i> the derivative of <math>u</math> in our integrand. It would have been sufficient to have any constant multiple of the derivative.
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For instance, to treat the integral
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:<math>\int x^4\sin(x^5)dx</math>
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we may let <math>u=x^5</math> . Then
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:<math>du=5x^4dx</math>
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and so
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:<math>\frac{du}{5}=x^4dx</math>
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the right-hand side of which is a factor of our integrand. Thus,
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:<math>\int x^4\sin(x^5)dx=\int\frac{\sin(u)}{5}du=-\frac{\cos(u)}{5}+C=-\frac{\cos(x^5)}{5}+C</math>
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==Resources==
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* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Recognizing_Derivatives_and_the_Substitution_Rule Recognizing Derivatives and Substitution Rules], WikiBooks: Calculus
 +
* [https://en.wikibooks.org/wiki/High_School_Calculus/Integration_by_Substitution Integration by Substitution], WikiBooks: High School Calculus
 +
* [https://www.youtube.com/watch?v=uoCW8S-I9Es Example 1]. Produced by Professor Zachary Sharon, UTSA
 +
 
 +
* [https://www.youtube.com/watch?v=zqMxMtjbaBE Example 2]. Produced by TA Catherine Sporer, UTSA
  
 
<strong>Indefinite Integrals Using Substitution</strong>
 
<strong>Indefinite Integrals Using Substitution</strong>
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* [https://youtu.be/0A2RlnutO8U U-Substitution Integration, Indefinite & Definite Integral] by The Organic Chemistry Tutor
 
* [https://youtu.be/0A2RlnutO8U U-Substitution Integration, Indefinite & Definite Integral] by The Organic Chemistry Tutor
 +
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Recognizing_Derivatives_and_the_Substitution_Rule Recognizing Derivatives and Substitution Rules, WikiBooks: Calculus] under a CC BY-SA license
 +
 +
* [https://en.wikibooks.org/wiki/High_School_Calculus/Integration_by_Substitution Integration by Substitution, WikiBooks: High School Calculus] under a CC BY-SA license

Latest revision as of 13:21, 28 October 2021

Integration by Substitution

There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.

what the theorem looks like is this


In order to get you must plug a into the function g and to get you must plug b into the function g.

The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section

Steps

(1) i.e.
(2) i.e.
(3) i.e.
(4) i.e. Now equate with
(5) i.e.
(6) i.e.
(7) i.e. We have achieved our desired result

Example 1

Instead of making this a big polynomial we will just use the substitution method.

Step 1

Identify your u

Let

Step 2


Identify



Step 3

Now we plug in our limits of integration to our u to find our new limits of integration

When

and when

Now our integration problem looks something like this

Step 4

write your new integration problem


When we plug in our u it looks like

Step 5

Evaluate the Integral










As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.

Example 2

we see that is the derivative of . Letting

we have

or, in order to apply it to the integral,

With this we may write

Note that it was not necessary that we had exactly the derivative of in our integrand. It would have been sufficient to have any constant multiple of the derivative.

For instance, to treat the integral

we may let . Then

and so

the right-hand side of which is a factor of our integrand. Thus,

Resources

  • Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution


Definite Integrals Using Substitution

Licensing

Content obtained and/or adapted from: