Integration by Parts - Revision history

Khanh at 21:29, 28 October 2021

2021-10-28T21:29:06Z

Lila: /* Exercises */

2021-10-06T19:29:04Z

Exercises

Lila at 19:25, 6 October 2021

2021-10-06T19:25:40Z

James.kercheville: Added video links

2020-08-23T01:13:38Z

Added video links

New page

[https://youtu.be/815s-YB9X44 Integration by Parts: The Basics] by James Sousa

[https://youtu.be/hfJ1zPizj_s Integration by Parts (After Integration by Parts Basics)] by James Sousa

[https://youtu.be/_kJNhoTJ4D4 Integration by Parts - Additional Examples] by James Sousa

[https://youtu.be/815s-YB9X44 Integration by Parts: The Basics] by James Sousa

[https://youtu.be/q_BP4LyvdOA Deriving the Integration by Parts Formula - Easy!] by patrickJMT

[https://youtu.be/dqaDSlYdRcs Integration by Parts Made Easy! ] by patrickJMT

[https://youtu.be/97RasvYIvl4 Integration by Parts - Indefinite Integral] by patrickJMT

[https://youtu.be/_MuKOm8IHBA Integration By Parts - Using IBP's Twice] by patrickJMT

[https://youtu.be/L8HoALsls-E Integration by Parts - A Loopy Example!] by patrickJMT

[https://youtu.be/zGGI4PkHzhI Integration by Parts - Definite Integral] by patrickJMT

[https://youtu.be/Eb5EDMHDnx8 How does integration by parts work?] by Krista King

[https://youtu.be/SlBp9hZBqaQ Integration by Parts] by Krista King

[https://youtu.be/s8l1vs4gjtY Integration by Parts Example 2] by Krista King

[https://youtu.be/0clmZGVl0ss Integration by Parts Example 3] by Krista King

[https://youtu.be/CxV5HE6chSY Integration by Parts Example 4 ] by Krista King

← Older revision		Revision as of 21:29, 28 October 2021
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	[https://youtu.be/CxV5HE6chSY Integration by Parts Example 4 ] by Krista King		[https://youtu.be/CxV5HE6chSY Integration by Parts Example 4 ] by Krista King
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Integration_by_Parts Integration by Parts, Wikibooks] under a CC BY-SA license

@@ Line 69: / Line 69: @@
 ==Exercises==
 Evaluate the following using integration by parts.
-{{question-answer|question=1. <math>\int -4\ln(x)dx</math>|answer={{noprint|<math>4x-4x\ln(x)+C</math>}}}}
+# <math>\int -4\ln(x)dx</math>
+# <math>\int (38-7x)\cos(x)dx</math>
-{{question-answer|question=2. <math>\int (38-7x)\cos(x)dx</math>|answer={{noprint|<math>(38-7x)\sin(x)-7\cos(x)+C</math>}}}}
+# <math>\int\limits_0^\tfrac{\pi}{2} (-6x+45)\cos(x)dx</math>
+# <math>\int(5x+1)(x-6)^4 dx</math>
-{{question-answer|question=3. <math>\int\limits_0^\tfrac{\pi}{2} (-6x+45)\cos(x)dx</math>|answer={{noprint|<math>51-3\pi</math>}}}}
+# <math>\int\limits_{-1}^1 (2x+8)^3(2-x)dx</math>
 ==Resources==

← Older revision		Revision as of 19:25, 6 October 2021
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		+	Continuing on the path of reversing derivative rules in order to make them useful for integration, we reverse the product rule.
		+
		+	==Integration by parts==
		+	If <math>y=uv</math> where <math>u</math> and <math>v</math> are functions of <math>x</math> , then
		+	:<math>y'=(uv)'=u'v+uv'</math>
		+	Rearranging,
		+	:<math>uv'=(uv)'-u'v</math>
		+	Therefore,
		+	:<math>\int uv'dx=\int (uv)'\ dx-\int u'v\ dx</math>
		+	Therefore,
		+	:<math>\int uv'\ dx=uv-\int vu'\ dx</math>
		+	or
		+	:<math>\int u\ dv=uv-\int v\ du</math>
		+	This is the integration by parts formula. It is very useful in many integrals involving products of functions, as well as others.
		+
		+	For instance, to treat
		+	:<math>\int x\sin(x)dx</math>
		+	we choose <math>u=x</math> and <math>dv=\sin(x)dx</math> . With these choices, we have <math>du=dx</math> and <math>v=-\cos(x)</math> , and we have
		+	:<math>\int x\sin(x)dx=-x\cos(x)-\int\big(-\cos(x)\big)dx=-x\cos(x)+\int\cos(x)dx=\sin(x)-x\cos(x)+C</math>
		+	Note that the choice of <math>u</math> and <math>dv</math> was critical. Had we chosen the reverse, so that <math>u=\sin(x)</math> and <math>dv=x\ dx</math> , the result would have been
		+	:<math>\frac{x^2\sin(x)}{2}-\int\frac{x^2\cos(x)}{2}dx</math>
		+	The resulting integral is no easier to work with than the original; we might say that this application of integration by parts took us in the wrong direction.
		+
		+	So the choice is important. One general guideline to help us make that choice is, if possible, to choose <math>u</math> to be the factor of the integrand which ''becomes simpler'' when we differentiate it. In the last example, we see that <math>\sin(x)</math> does not become simpler when we differentiate it: <math>\cos(x)</math> is no simpler than <math>\sin(x)</math> .
		+
		+	An important feature of the integration by parts method is that we often need to apply it more than once. For instance, to integrate
		+	:<math>\int x^2e^xdx</math>
		+	we start by choosing <math>u=x^2</math> and <math>dv=e^xdx</math> to get
		+	:<math>\int x^2e^xdx=x^2e^x-2\int xe^xdx</math>
		+	Note that we still have an integral to take care of, and we do this by applying integration by parts again, with <math>u=x</math> and <math>dv=e^xdx</math> , which gives us
		+	:<math>\int x^2e^xdx=x^2e^x-2\int xe^xdx=x^2e^x-2(xe^x-e^x)+C=x^2e^x-2xe^x+2e^x+C</math>
		+	So, two applications of integration by parts were necessary, owing to the power of <math>x^2</math> in the integrand.
		+
		+	Note that ''any power of x'' does become simpler when we differentiate it, so when we see an integral of the form
		+	:<math>\int x^nf(x)dx</math>
		+	one of our first thoughts ought to be to consider using integration by parts with <math>u=x^n</math> . Of course, in order for it to work, we need to be able to write down an antiderivative for <math>dv</math> .
		+
		+	===Example===
		+	Use integration by parts to evaluate the integral
		+	:<math>\int e^x\sin(x)dx</math>
		+	Solution: If we let <math>u=\sin(x)</math> and <math>v'=e^xdx</math> , then we have <math>u'=\cos(x)dx</math> and <math>v=e^x</math> . Using our rule for integration by parts gives
		+	:<math>\int e^x\sin(x)dx=e^x\sin(x)-\int e^x\cos(x)dx</math>
		+	We do not seem to have made much progress.
		+
		+	But if we integrate by parts again with <math>u=\cos(x)</math> and <math>v'=e^xdx</math> and hence <math>u'=-\sin(x)dx</math> and <math>v=e^x</math> , we obtain
		+	:<math>\int e^x\sin(x)dx=e^x\sin(x)-e^x\cos(x)-\int e^x\sin(x)dx</math>
		+	We may solve this identity to find the anti-derivative of <math>\sin(x)e^x</math> and obtain
		+	:<math>\int e^x\sin(x)dx=\frac{e^x\big(\sin(x)-\cos(x)\big)}{2}+C</math>
		+
		+	===With definite integral===
		+	For definite integrals the rule is essentially the same, as long as we keep the endpoints.
		+
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	'''Integration by parts for definite integrals'''
		+	Suppose ''f'' and ''g'' are differentiable and their derivatives are continuous. Then
		+	:<math>\int\limits_a^b f(x)g'(x)dx=\big(f(x)g(x)\big)\bigg\|_a^b-\int\limits_a^b f'(x)g(x)dx</math>
		+	:<math>=f(b)g(b)-f(a)g(a)-\int\limits_a^b f'(x)g(x)dx</math> .
		+	</blockquote>
		+
		+	This can also be expressed in Leibniz notation.
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	:<math>\int\limits_a^b u\ dv=(uv)\Big\|_a^b-\int\limits_a^b v\ du.</math>
		+	</blockquote>
		+
		+	===More Examples===
		+
		+	[http://en.wikiversity.org/wiki/Application_of_Integration_by_Parts Examples Set 1: Integration by Parts ]
		+
		+	==Exercises==
		+	Evaluate the following using integration by parts.
		+	{{question-answer\|question=1. <math>\int -4\ln(x)dx</math>\|answer={{noprint\|<math>4x-4x\ln(x)+C</math>}}}}
		+
		+	{{question-answer\|question=2. <math>\int (38-7x)\cos(x)dx</math>\|answer={{noprint\|<math>(38-7x)\sin(x)-7\cos(x)+C</math>}}}}
		+
		+	{{question-answer\|question=3. <math>\int\limits_0^\tfrac{\pi}{2} (-6x+45)\cos(x)dx</math>\|answer={{noprint\|<math>51-3\pi</math>}}}}
		+
		+	{{question-answer\|question=4. <math>\int(5x+1)(x-6)^4 dx</math>\|answer={{noprint\|<math>\frac{(5x+1)(x-6)^5}{5}-\frac{(x-6)^6}{6}+C</math>}}}}
		+
		+	{{question-answer\|question=5. <math>\int\limits_{-1}^1 (2x+8)^3(2-x)dx</math>\|answer={{noprint\|<math>1916.8</math>}}}}
		+	{{noprint\|[[Calculus/Integration_techniques/Integration_by_Parts/Solutions\|Solutions]]}}
		+
		+
		+
		+	==Resources==
		+	* [https://www.khanacademy.org/math/integral-calculus/integration-techniques/integration_by_parts/ Videos and exercises on integration by parts at Khan Academy]
		+
		+	===Videos===
	[https://youtu.be/815s-YB9X44 Integration by Parts: The Basics] by James Sousa		[https://youtu.be/815s-YB9X44 Integration by Parts: The Basics] by James Sousa