Integration by Substitution - Revision history

Lila: /* Resources */

2021-10-28T19:21:22Z

Resources

Lila: /* Resources */

2021-10-06T18:37:11Z

Resources

Lila: /* Example 1 */

2021-10-06T18:35:41Z

Example 1

Lila at 18:35, 6 October 2021

2021-10-06T18:35:19Z

Lila: /* Integration by Substitution */

2021-10-06T18:33:00Z

Integration by Substitution

Lila: /* Integration by Substitution */

2021-10-06T18:31:06Z

Integration by Substitution

Lila at 18:30, 6 October 2021

2021-10-06T18:30:03Z

Glenn.lahodny at 18:52, 4 December 2020

2020-12-04T18:52:34Z

Glenn.lahodny at 18:34, 4 December 2020

2020-12-04T18:34:36Z

James.kercheville: minor edit

2020-08-19T15:34:53Z

minor edit

← Older revision		Revision as of 19:21, 28 October 2021
Line 178:		Line 178:

	* [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

@@ Line 133: / Line 133: @@
 ==Resources==
-[https://www.youtube.com/watch?v=uoCW8S-I9Es Example 1]. Produced by Professor Zachary Sharon, UTSA
+* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Recognizing_Derivatives_and_the_Substitution_Rule Recognizing Derivatives and Substitution Rules], WikiBooks: Calculus
-[https://www.youtube.com/watch?v=zqMxMtjbaBE Example 2]. Produced by TA Catherine Sporer, UTSA
+* [https://www.youtube.com/watch?v=zqMxMtjbaBE Example 2]. Produced by TA Catherine Sporer, UTSA
 <strong>Indefinite Integrals Using Substitution</strong>

@@ Line 60: / Line 60: @@
 : <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><math>\operatorname {d}u = 2x\operatorname {d}x</math></br></p>

@@ Line 55: / Line 55: @@
 <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><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===

@@ Line 49: / Line 49: @@
 |}
-<p>'''Ex. 1'''</p>
+===Example 1===
 <p><math>\int_{0}^{2} x(x^2+1)^2 \operatorname {d}x</math></p>
@@ Line 99: / Line 99: @@
 <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>I'll give you some other problems to work on as well.</p>
+===Example 2===
-<p>'''Ex. 2'''</p>
+:<math>\int 3x^2(x^3+1)^5dx</math>
-<p><math>\int_{0}^{\frac {\pi}{2}} \sin (x) \cos (x) \operatorname {d}x</math></p>
+we see that <math>3x^2</math> is the derivative of <math>x^3+1</math> . Letting
-<p>'''Ex. 3'''</p>
+:<math>u=x^3+1</math>
-<p><math>\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x</math></p>
+we have
 ==Resources==

← Older revision		Revision as of 18:30, 6 October 2021
Line 1:		Line 1:
		+	===Integration by Substitution===
		+
		+	<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>
		+
		+	<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>
		+
		+	<p>'''Ex. 1'''</p>
		+
		+	<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>
		+	<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>Step 3</br></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>and when <math>x = 2, u = 2^2 + 1 = 5</math></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>Step 4</p>
		+
		+	<p>write your new integration problem</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><br>Step 5</br></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><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 {125}{3}\right]</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>I'll give you some other problems to work on as well.</p>
		+
		+	<p>'''Ex. 2'''</p>
		+
		+	<p><math>\int_{0}^{\frac {\pi}{2}} \sin (x) \cos (x) \operatorname {d}x</math></p>
		+
		+	<p>'''Ex. 3'''</p>
		+
		+	<p><math>\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x</math></p>
		+
		+
		+
		+	==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

@@ Line 3: / Line 3: @@
 [https://www.youtube.com/watch?v=zqMxMtjbaBE Example 2]. Produced by TA Catherine Sporer, UTSA
-[https://www.youtube.com/watch?v=568hUU4beuk Indefinite Integration Using Substitution] by James Sousa, Math is Power 4U
+<strong>Indefinite Integrals Using Substitution</strong>
-[https://www.youtube.com/watch?v=hjr27pv8pHQ Definite Integration Using Subsitution] by James Sousa, Math is Power 4U
+* [https://youtu.be/qclrs-1rpKI Integration using U-Substitution] by patrickJMT
-[https://youtu.be/VfvJ5C9FjcA Integration by Substitution Part 1] by James Sousa, Math is Power 4U
+* [https://youtu.be/HHSepCzP56M U-Substitution Example 1] by Krista King
-[https://youtu.be/pmekyFVIorE Integration by Substitution Part 2] by James Sousa, Math is Power 4U
+* [https://youtu.be/IAh00vU3FSY How To Integrate Using U-Substitution] by The Organic Chemistry Tutor
-[https://youtu.be/qclrs-1rpKI Integration using U-Substitution]  by patrickJMT
+* [https://youtu.be/3XOzg2-DdTA Integration by U-Substitution, Definite Integral] by patrickJMT
-[https://youtu.be/x06V9xuLdqg U-Substitution - More Complicated Examples]  by patrickJMT
+* [https://youtu.be/0A2RlnutO8U U-Substitution Integration, Indefinite & Definite Integral] by The Organic Chemistry Tutor

@@ Line 19: / Line 19: @@
 [https://youtu.be/FJoyIAIC1Ag U-Substitution : When Do I Have to Change the Limits of Integration?]  by patrickJMT
-[https://youtu.be/HHSepCzP56M U-Substitution U-Substitution] by Krista King
+[https://youtu.be/HHSepCzP56M U-Substitution] by Krista King
 [https://youtu.be/iRIhO-w46L0 U-Substitution Example 2] by Krista King