Difference between revisions of "Integration by Substitution"

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<p>Instead of making this a big polynomial we will just use the substitution method.</p>
 
<p>Instead of making this a big polynomial we will just use the substitution method.</p>
  
<p>Step 1 </p>
+
<p>'''Step 1'''</p>
<p>Identify your <i>u</i></p>
+
: <p>Identify your <i>u</i></p>
  
<p>Let <math> u = x^2+1</math></p>
+
: <p>Let <math> u = x^2+1</math></p>
  
<p>Step 2</p>
+
<p>'''Step 2''</p>
<p><br>Identify <math>\operatorname {d}u</math></br></p>
+
: <p><br>Identify <math>\operatorname {d}u</math></br></p>
<p> <br><math>\operatorname {d}u = 2x\operatorname {d}x</math></br></p>
+
: <p> <br><math>\operatorname {d}u = 2x\operatorname {d}x</math></br></p>
  
<p><br>Step 3</br></p>
+
'''Step 3'''
<p>Now we plug in our limits of integration to our <i>u</i> to find our new limits of integration</p>
+
: <p>Now we plug in our limits of integration to our <i>u</i> to find our new limits of integration</p>
  
<p>When <math> x = 0, u =0^2 + 1 = 1</math></p>
+
: <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>
+
: <p>and when <math>x = 2, u = 2^2 + 1 = 5</math></p>
  
<p>Now our integration problem looks something like this</p>
+
: <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>
+
: <p><center><math>\frac {1}{2} \int_{0}^{5} (x^2 + 1)^2 (2x)\operatorname {d}x</math></center></p>
  
<p>Step 4</p>
+
'''Step 4'''
  
<p>write your new integration problem</p>
+
: <p>write your new integration problem</p>
  
<p><br>When we plug in our <i>u</i> it looks like </br></p>
+
: <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>
 
<p><center><math>\frac {1}{2} \int_{0}^{5} (u)^2 \operatorname {d}u</math></center></p>
  
<p><br>Step 5</br></p>
+
'''Step 5'''
  
<p>Evaluate the Integral</p>
+
: <p>Evaluate the Integral</p>
  
<p><center><math>\frac {1}{2} \left[\frac {1}{3} u^3 \right]_{0}^{5}</math></center></p>
+
: <p><center><math>\frac {1}{2} \left[\frac {1}{3} u^3 \right]_{0}^{5}</math></center></p>
  
<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>
+
: <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>
  
<p><br><center><math>\frac {1}{2} \left[\frac {1}{3} * 125 \right]</math></center></br></p>
+
: <p><br><center><math>\frac {1}{2} \left[\frac {1}{3} * 125 \right]</math></center></br></p>
  
<p><br><center><math>\frac {1}{2} \left[\frac {125}{3}\right]</math></center></br></p>
+
: <p><br><center><math>\frac {1}{2} \left[\frac {125}{3}\right]</math></center></br></p>
  
<p><br><center><math>\frac {125}{6}</math></center></br></p>
+
: <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>
+
: <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===
 
===Example 2===

Revision as of 12:35, 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

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 1. Produced by Professor Zachary Sharon, UTSA

Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution


Definite Integrals Using Substitution