Difference between revisions of "Integration by Parts"

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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.
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==Integration by parts==
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If <math>y=uv</math> where <math>u</math> and <math>v</math> are functions of <math>x</math> , then
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:<math>y'=(uv)'=u'v+uv'</math>
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Rearranging,
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:<math>uv'=(uv)'-u'v</math>
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Therefore,
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:<math>\int uv'dx=\int (uv)'\ dx-\int u'v\ dx</math>
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Therefore,
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:<math>\int uv'\ dx=uv-\int vu'\ dx</math>
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or
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:<math>\int u\ dv=uv-\int v\ du</math>
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This is the integration by parts formula. It is very useful in many integrals involving products of functions, as well as others.
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 +
For instance, to treat
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:<math>\int x\sin(x)dx</math>
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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
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:<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>
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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>
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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>
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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>
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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>
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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
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:<math>\int x^nf(x)dx</math>
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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> .
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===Example===
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Use integration by parts to evaluate the integral
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:<math>\int e^x\sin(x)dx</math>
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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
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:<math>\int e^x\sin(x)dx=e^x\sin(x)-\int e^x\cos(x)dx</math>
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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
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:<math>\int e^x\sin(x)dx=e^x\sin(x)-e^x\cos(x)-\int e^x\sin(x)dx</math>
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We may solve this identity to find the anti-derivative of <math>\sin(x)e^x</math> and obtain
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:<math>\int e^x\sin(x)dx=\frac{e^x\big(\sin(x)-\cos(x)\big)}{2}+C</math>
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===With definite integral===
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For definite integrals the rule is essentially the same, as long as we keep the endpoints.
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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'''Integration by parts for definite integrals'''
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Suppose ''f'' and ''g'' are differentiable and their derivatives are continuous. Then
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:<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>
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:<math>=f(b)g(b)-f(a)g(a)-\int\limits_a^b f'(x)g(x)dx</math> .
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</blockquote>
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This can also be expressed in Leibniz notation.
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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:<math>\int\limits_a^b u\ dv=(uv)\Big|_a^b-\int\limits_a^b v\ du.</math>
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</blockquote>
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===More Examples===
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[http://en.wikiversity.org/wiki/Application_of_Integration_by_Parts Examples Set 1: Integration by Parts ]
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==Exercises==
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Evaluate the following using integration by parts.
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{{question-answer|question=1. <math>\int -4\ln(x)dx</math>|answer={{noprint|<math>4x-4x\ln(x)+C</math>}}}}
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{{question-answer|question=2. <math>\int (38-7x)\cos(x)dx</math>|answer={{noprint|<math>(38-7x)\sin(x)-7\cos(x)+C</math>}}}}
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{{question-answer|question=3. <math>\int\limits_0^\tfrac{\pi}{2} (-6x+45)\cos(x)dx</math>|answer={{noprint|<math>51-3\pi</math>}}}}
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{{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>}}}}
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{{question-answer|question=5. <math>\int\limits_{-1}^1 (2x+8)^3(2-x)dx</math>|answer={{noprint|<math>1916.8</math>}}}}
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{{noprint|[[Calculus/Integration_techniques/Integration_by_Parts/Solutions|Solutions]]}}
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==Resources==
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* [https://www.khanacademy.org/math/integral-calculus/integration-techniques/integration_by_parts/ Videos and exercises on integration by parts at Khan Academy]
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===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
  

Revision as of 13:25, 6 October 2021

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 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 y=uv} where and 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 v} are functions 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} , 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 y'=(uv)'=u'v+uv'}

Rearranging,

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 uv'=(uv)'-u'v}

Therefore,

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 uv'dx=\int (uv)'\ dx-\int u'v\ dx}

Therefore,

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 uv'\ dx=uv-\int vu'\ dx}

or

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 u\ dv=uv-\int v\ du}

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

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\sin(x)dx}

we choose 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} and 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 dv=\sin(x)dx} . With these choices, 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 du=dx} and 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 v=-\cos(x)} , and 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 \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}

Note that the choice 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} and 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 dv} was critical. Had we chosen the reverse, so that 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=\sin(x)} and 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 dv=x\ dx} , the result would have been

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{x^2\sin(x)}{2}-\int\frac{x^2\cos(x)}{2}dx}

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 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} to be the factor of the integrand which becomes simpler when we differentiate it. In the last example, we see that 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 \sin(x)} does not become simpler when we differentiate it: 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 \cos(x)} is no simpler than 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 \sin(x)} .

An important feature of the integration by parts method is that we often need to apply it more than once. For instance, to integrate

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^2e^xdx}

we start by choosing 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} and 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 dv=e^xdx} to get

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^2e^xdx=x^2e^x-2\int xe^xdx}

Note that we still have an integral to take care of, and we do this by applying integration by parts again, 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 u=x} and 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 dv=e^xdx} , which gives us

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^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}

So, two applications of integration by parts were necessary, owing to the power 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^2} 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

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^nf(x)dx}

one of our first thoughts ought to be to consider using integration by parts 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 u=x^n} . Of course, in order for it to work, we need to be able to write down an antiderivative for 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 dv} .

Example

Use integration by parts to 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 \int e^x\sin(x)dx}

Solution: If we 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=\sin(x)} and 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 v'=e^xdx} , then 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 u'=\cos(x)dx} and 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 v=e^x} . Using our rule for integration by parts gives

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 e^x\sin(x)dx=e^x\sin(x)-\int e^x\cos(x)dx}

We do not seem to have made much progress.

But if we integrate by parts again 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 u=\cos(x)} and 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 v'=e^xdx} and hence 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'=-\sin(x)dx} and 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 v=e^x} , we obtain

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 e^x\sin(x)dx=e^x\sin(x)-e^x\cos(x)-\int e^x\sin(x)dx}

We may solve this identity to find the anti-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 \sin(x)e^x} and obtain

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 e^x\sin(x)dx=\frac{e^x\big(\sin(x)-\cos(x)\big)}{2}+C}

With definite integral

For definite integrals the rule is essentially the same, as long as we keep the endpoints.

Integration by parts for definite integrals Suppose f and g are differentiable and their derivatives are continuous. 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 \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}
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 =f(b)g(b)-f(a)g(a)-\int\limits_a^b f'(x)g(x)dx} .

This can also be expressed in Leibniz notation.

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_a^b u\ dv=(uv)\Big|_a^b-\int\limits_a^b v\ du.}

More Examples

Examples Set 1: Integration by Parts

Exercises

Evaluate the following using integration by parts. Template:Question-answer

Template:Question-answer

Template:Question-answer

Template:Question-answer

Template:Question-answer Template:Noprint


Resources

Videos

Integration by Parts: The Basics by James Sousa

Integration by Parts (After Integration by Parts Basics) by James Sousa

Integration by Parts - Additional Examples by James Sousa

Integration by Parts: The Basics by James Sousa

Deriving the Integration by Parts Formula - Easy! by patrickJMT

Integration by Parts Made Easy! by patrickJMT

Integration by Parts - Indefinite Integral by patrickJMT

Integration By Parts - Using IBP's Twice by patrickJMT

Integration by Parts - A Loopy Example! by patrickJMT

Integration by Parts - Definite Integral by patrickJMT

How does integration by parts work? by Krista King

Integration by Parts by Krista King

Integration by Parts Example 2 by Krista King

Integration by Parts Example 3 by Krista King

Integration by Parts Example 4 by Krista King

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