Difference between revisions of "Integration by Parts"
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 56: | Line 56: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | | | + | |We have two options when doing integration by parts. |
|- | |- | ||
| − | | | + | |We can let <math style="vertical-align: 0px">u=x</math> or <math style="vertical-align: 0px">u=e^x.</math> |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| + | |In this case, we let <math style="vertical-align: 0px">u</math> be the polynomial. | ||
|- | |- | ||
| − | |<u> | + | |So, we let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> |
|- | |- | ||
| − | | | + | |Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x.</math> |
|- | |- | ||
| − | | | + | |Hence, by integration by parts, we get |
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{\int xe^x~dx} & = & \displaystyle{xe^x-\int e^x~dx}\\ |
| − | |||
| − | & = & \displaystyle{ | ||
| − | |||
| − | |||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{x^ | + | & = & \displaystyle{xe^x-e^x+C.} |
\end{array}</math> | \end{array}</math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 95: | Line 79: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math> | + | | <math>xe^x-e^x+C</math> |
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 108: | Line 88: | ||
|- | |- | ||
| | | | ||
| − | + | We have a choice to make. | |
|- | |- | ||
| − | | | + | |We can let <math style="vertical-align: 0px">u=x</math> or <math style="vertical-align: -5px">u=\cos(2x).</math> |
| − | |||
|- | |- | ||
| − | | | + | |In this case, we let <math style="vertical-align: 0px">u</math> be the polynomial. |
|- | |- | ||
| − | | | + | |So, we let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: -5px">dv=\cos(2x)~dx.</math> |
| − | |||
|- | |- | ||
| − | |< | + | |Then, <math style="vertical-align: 0px">du=dx.</math> |
|- | |- | ||
| − | | | + | |By <math style="vertical-align: 0px">u-</math>substitution, <math style="vertical-align: -13px">v=\int \cos(2x)~dx=\frac{1}{2} \sin(2x).</math> |
|- | |- | ||
| − | | | + | |Hence, by integration by parts, we get |
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{\int x\cos(2x)~dx} & = & \displaystyle{\frac{1}{2}x\sin(2x)-\int \frac{1}{2}\sin(2x)~dx}\\ |
| − | |||
| − | |||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{x+ | + | & = & \displaystyle{\frac{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C,} |
\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | | | + | |where we use <math style="vertical-align: 0px">u-</math> substitution to evaluate the last integral. |
| − | |||
| − | |||
| − | : | ||
|} | |} | ||
| Line 142: | Line 115: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | + | | <math>\frac{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C</math> | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 156: | Line 123: | ||
!Solution: | !Solution: | ||
|- | |- | ||
| − | | | + | |We have a choice to make. |
| + | |- | ||
| + | |We can let <math style="vertical-align: -1px">u=\ln x</math> or <math style="vertical-align: -1px">dv=\ln x~dx.</math> | ||
| + | |- | ||
| + | |In this case, we don't want to let <math style="vertical-align: -1px">dv=\ln x~dx</math> since we don't know how to integrate <math style="vertical-align: -1px">\ln x</math> yet. | ||
| + | |- | ||
| + | |So, we let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: -1px">dv=1~dx.</math> | ||
| + | |- | ||
| + | |Then, <math style="vertical-align: -13px">du=\frac{1}{x}~dx</math> and <math style="vertical-align: 0px">v=x.</math> | ||
| + | |- | ||
| + | |Hence, by integration by parts, we get | ||
|- | |- | ||
| | | | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | \displaystyle{ | + | \displaystyle{\int \ln x~dx} & = & \displaystyle{x\ln (x) -\int x \big(\frac{1}{x}\big)~dx}\\ |
| − | |||
| − | & = & \displaystyle{\ | ||
| − | |||
| − | |||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{\ | + | & = & \displaystyle{x\ln (x)-\int 1~dx}\\ |
&&\\ | &&\\ | ||
| − | & = & \displaystyle{\ | + | & = & \displaystyle{x\ln (x)-x+C.} |
\end{array}</math> | \end{array}</math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 182: | Line 148: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math> | + | | <math>x\ln (x)-x+C</math> |
|- | |- | ||
|} | |} | ||
Revision as of 08:58, 27 October 2017
Introduction
Let's say we want 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^{x^3}~dx.}
Here, we can compute this antiderivative by using 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-} substitution.
While 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-} substitution is an important integration technique, it will not help us evaluate all integrals.
For example, consider 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 xe^x~dx.}
There is no substitution that will allow us to integrate this integral.
We need another integration technique called integration by parts.
The formula for integration by parts comes from the product rule for derivatives.
Recall from the product rule,
- 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(x)g(x))'=f'(x)g(x)+f(x)g'(x).}
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 \begin{array}{rcl} \displaystyle{f(x)g(x)} & = & \displaystyle{\int f'(x)g(x)+f(x)g'(x)~dx}\\ &&\\ & = & \displaystyle{\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.} \end{array}}
If we solve the last equation for the second integral, 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 f(x)g'(x)~dx = f(x)g(x)-\int f'(x)g(x)~dx.}
This formula is the formula for integration by parts.
But, as it is currently stated, it is long and hard to remember.
So, we make a substitution to obtain a nicer formula.
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=f(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=g'(x)~dx.}
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 du=f'(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=g(x).}
Plugging these into our formula, 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 u~dv=uv-\int v~du.}
Warm-Up
Evaluate the following integrals.
1) 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 xe^x~dx}
| Solution: |
|---|
| We have two options when doing integration by parts. |
| We can 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=x} 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 u=e^x.} |
| In this case, 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} be the polynomial. |
| So, 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=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^x~dx.} |
| 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 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=e^x.} |
| Hence, by integration by parts, we get |
|
| Final Answer: |
|---|
| 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 xe^x-e^x+C} |
2) 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\cos (2x)~dx}
| Solution: |
|---|
|
We have a choice to make. |
| We can 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=x} 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 u=\cos(2x).} |
| In this case, 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} be the polynomial. |
| So, 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=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=\cos(2x)~dx.} |
| 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 du=dx.} |
| By 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-} substitution, 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=\int \cos(2x)~dx=\frac{1}{2} \sin(2x).} |
| Hence, by integration by parts, we get |
|
| where we use 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-} substitution to evaluate the last integral. |
| Final Answer: |
|---|
| 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{1}{2}x\sin(2x)+\frac{1}{4}\cos(2x)+C} |
3) 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 \ln x~dx}
| Solution: |
|---|
| We have a choice to make. |
| We can 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=\ln x} 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 dv=\ln x~dx.} |
| In this case, we don't want to 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 dv=\ln x~dx} since we don't know how 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 \ln x} yet. |
| So, 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=\ln 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=1~dx.} |
| 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 du=\frac{1}{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=x.} |
| Hence, by integration by parts, we get |
|
| Final Answer: |
|---|
| 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\ln (x)-x+C} |
Exercise 1
Evaluate 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^3 e^{-2x}~dx.}
First, we need to know the 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 \csc x.} Recall
- 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 \csc x =\frac{1}{\sin x}.}
Now, using the Quotient Rule, 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 \begin{array}{rcl} \displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\ &&\\ & = & \displaystyle{-\csc x \cot x.} \end{array}}
Using the Product Rule and Power Rule, 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.} \end{array}}
So, we have
Exercise 2
Evaluate
Notice that the function is the product of three functions.
We start by grouping two of the functions together. So, we have
Using the Product Rule, we get
Now, we need to use the Product Rule again. So,
So, we have
But, there is another way to do this problem. Notice
Now, you would only need to use the Product Rule once instead of twice.
Exercise 3
Evaluate
Using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we get
Exercise 4
Evaluate
First, using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we have
Exercise 5
Evaluate
First, using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we have
Exercise 6
Evaluate
First, using the Quotient Rule, we have
Now, we need to use the Product Rule. So, we have
So, we have