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>
 +
 
 +
<p>what the theorem looks like is this</p>
 +
 
 +
<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>
 +
 
 +
<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>
 +
 
 +
====Steps====
 +
:{|
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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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|<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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|<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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|<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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|<math>=\int\limits_{u=c}^{u=d}h(u)du</math>
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|style="padding-left: 20px"|(7)
 +
|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>
 +
 
 +
<p>'''Step 1'''</p>
 +
: <p>Identify your <i>u</i></p>
 +
 
 +
: <p>Let <math> u = x^2+1</math></p>
 +
 
 +
<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>
 +
 
 +
'''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>
 +
 
 +
: <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>
 +
 
 +
: <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'''
 +
 
 +
: <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>
 +
 
 +
<p><center><math>\frac {1}{2} \int_{0}^{5} (u)^2 \operatorname {d}u</math></center></p>
 +
 
 +
'''Step 5'''
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: <p>Evaluate the Integral</p>
 +
 
 +
: <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>
 +
 
 +
: <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===
 +
 
 +
:<math>\int 3x^2(x^3+1)^5dx</math>
 +
 
 +
we see that <math>3x^2</math> is the derivative of <math>x^3+1</math> . Letting
 +
 
 +
:<math>u=x^3+1</math>
 +
 
 +
we have
 +
 
 +
:<math>\frac{du}{dx}=3x^2</math>
 +
 
 +
or, in order to apply it to the integral,
 +
 
 +
:<math>du=3x^2dx</math>
 +
 
 +
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.
 +
 
 +
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>
 +
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>
 +
 
 +
==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
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* [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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{a}^{b} f(x)\operatorname {d}x = \int_{\alpha}^{\beta} f(g(u))g\prime (u)\operatorname {d}u}


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.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int\limits_{x=a}^{x=b}h(g(x))g'(x)dx} (4) i.e. Now equate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(f(x)\ \frac{dx}{du}\right)} with
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int\limits_{x=a}^{x=b}h(u)g'(x)dx} (5) i.e. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=u}
(6) i.e. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=\frac{du}{dx}dx=g'(x)dx}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int\limits_{u=c}^{u=d}h(u)du} (7) i.e. We have achieved our desired result

Example 1

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} x(x^2+1)^2 \operatorname {d}x}

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

Step 1

Identify your u

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = x^2+1}

Step 2


Identify


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname {d}u = 2x\operatorname {d}x}

Step 3

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

When

and when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = 2, u = 2^2 + 1 = 5}

Now our integration problem looks something like this

Step 4

write your new integration problem


When we plug in our u it looks like

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {1}{2} \int_{0}^{5} (u)^2 \operatorname {d}u}

Step 5

Evaluate the Integral


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {1}{2} \left[\left(\frac {1}{3} * 5^3 \right) - \left(\frac {1}{3} * 0^3 \right)\right]}


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {1}{2} \left[\frac {1}{3} * 125 \right]}




Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {125}{6}}


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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int 3x^2(x^3+1)^5dx}

we see that is the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^3+1} . Letting

we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{dx}=3x^2}

or, in order to apply it to the integral,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=3x^2dx}

With this we may write

Note that it was not necessary that we had exactly the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^5} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=5x^4dx}

and so

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^4\sin(x^5)dx=\int\frac{\sin(u)}{5}du=-\frac{\cos(u)}{5}+C=-\frac{\cos(x^5)}{5}+C}

Resources

  • Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution


Definite Integrals Using Substitution

Licensing

Content obtained and/or adapted from: