Difference between revisions of "Integration by Substitution"

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===Integration by Substitution===
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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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<p>'''Ex. 1'''</p>
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<p><math>\int_{0}^{2} x(x^2+1)^2 \operatorname {d}x</math></p>
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<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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<p><br>Step 3</br></p>
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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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<p>Step 4</p>
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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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<p><br>Step 5</br></p>
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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>
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<p>I'll give you some other problems to work on as well.</p>
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<p>'''Ex. 2'''</p>
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<p><math>\int_{0}^{\frac {\pi}{2}} \sin (x) \cos (x) \operatorname {d}x</math></p>
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<p>'''Ex. 3'''</p>
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<p><math>\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x</math></p>
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==Resources==
 
[https://www.youtube.com/watch?v=uoCW8S-I9Es Example 1]. Produced by Professor Zachary Sharon, UTSA
 
[https://www.youtube.com/watch?v=uoCW8S-I9Es Example 1]. Produced by Professor Zachary Sharon, UTSA
  

Revision as of 12:30, 6 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

Ex. 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.

I'll give you some other problems to work on as well.

Ex. 2

Ex. 3


Resources

Example 1. Produced by Professor Zachary Sharon, UTSA

Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution


Definite Integrals Using Substitution