Difference between revisions of "009B Sample Final 1, Problem 4"

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(Created page with "<span class="exam"> Compute the following integrals. <span class="exam">a) <math>\int e^x(x+\sin(e^x))~dx</math> <span class="exam">b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</m...")
 
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<span class="exam"> Compute the following integrals.
 
<span class="exam"> Compute the following integrals.
  
−
<span class="exam">a) <math>\int e^x(x+\sin(e^x))~dx</math>
+
::<span class="exam">a) <math>\int e^x(x+\sin(e^x))~dx</math>
  
−
<span class="exam">b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
+
::<span class="exam">b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
  
−
<span class="exam">c) <math>\int \sin^3x~dx</math>
+
::<span class="exam">c) <math>\int \sin^3x~dx</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|Recall:
 
|Recall:
 
|-
 
|-
−
|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du</math>.
+
|
 +
::'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 
|-
 
|-
−
|'''2.''' Through partial fraction decomposition, we can write the fraction &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math> &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
+
|
 +
::'''2.''' Through partial fraction decomposition, we can write the fraction &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math> &nbsp;
 
|-
 
|-
−
|'''3.''' We have the Pythagorean identity <math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x)</math>.
+
|
 +
:::for some constants <math style="vertical-align: -4px">A,B.</math>
 +
|-
 +
|
 +
::'''3.''' We have the Pythagorean identity <math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x).</math>
 
|}
 
|}
  
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|Now, for the first integral on the right hand side of the last equation, we use integration by parts.  
 
|Now, for the first integral on the right hand side of the last equation, we use integration by parts.  
 
|-
 
|-
−
|Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x</math>.
+
|Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
 
|-
 
|-
 
|So, we have
 
|So, we have
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|Now, for the one remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|Now, for the one remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math style="vertical-align: 0px">u=e^x</math>. Then, <math style="vertical-align: 0px">du=e^xdx</math>.
+
|Let <math style="vertical-align: 0px">u=e^x.</math> Then, <math style="vertical-align: 0px">du=e^xdx.</math>  
 
|-
 
|-
 
|So, we have
 
|So, we have
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|Now, we need to use partial fraction decomposition for the second integral.
 
|Now, we need to use partial fraction decomposition for the second integral.
 
|-
 
|-
−
|Since <math style="vertical-align: -5px">2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
+
|Since <math style="vertical-align: -5px">2x^2+x=x(2x+1),</math> we let  
 +
|-
 +
|
 +
::<math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.</math>
 +
|-
 +
|Multiplying both sides of the last equation by <math style="vertical-align: -5px">x(2x+1),</math> we get
 +
|-
 +
|
 +
::<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
 
|-
 
|-
−
|Multiplying both sides of the last equation by <math style="vertical-align: -5px">x(2x+1)</math>,  
+
|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes
 
|-
 
|-
−
|we get <math style="vertical-align: -5px">1-x=A(2x+1)+Bx</math>.
+
|
 +
::<math style="vertical-align: -1px">1=A.</math>
 
|-
 
|-
−
|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
+
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2},</math> then we get &thinsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B.</math> Thus,
 
|-
 
|-
−
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get &thinsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
+
|
 +
::<math style="vertical-align: 0px">B=-3.</math>
 
|-
 
|-
−
|So, in summation, we have&thinsp; <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
+
|So, in summation, we have&thinsp;  
 +
|-
 +
|
 +
::<math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.</math>
 
|}
 
|}
  
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|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2\,dx</math> and&thinsp; <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
+
|Let <math style="vertical-align: -2px">u=2x+1.</math> Then, <math style="vertical-align: 0px">du=2\,dx</math> and&thinsp; <math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
 
|-
 
|-
 
|Thus, our final integral becomes
 
|Thus, our final integral becomes
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
+
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx.</math>
 
|-
 
|-
−
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>.
+
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get  
 +
|-
 +
|
 +
::<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>  
 
|-
 
|-
 
|If we use this identity, we have
 
|If we use this identity, we have
 
|-
 
|-
−
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
+
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx.</math>
−
|-
 
−
|
 
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math style="vertical-align: -1px">du=-\sin x dx</math>.
+
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x.</math> Then, <math style="vertical-align: -1px">du=-\sin x dx.</math>  
 
|-
 
|-
 
|So we have
 
|So we have

Revision as of 12:03, 18 April 2016

Compute the following integrals.

a) 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(x+\sin(e^x))~dx}
b)
c)
Foundations:  
Recall:
1. Integration by parts tells us that
2. Through partial fraction decomposition, we can write the fraction    
for some constants
3. We have the Pythagorean identity

Solution:

(a)

Step 1:  
We first distribute 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 e^x(x+\sin(e^x))~dx\,=\,\int e^xx~dx+\int e^x\sin(e^x)~dx.}
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
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^xdx.} 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.}
So, 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{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{\bigg(xe^x-\int e^x~dx \bigg)+\int e^x\sin(e^x)~dx}\\ &&\\ & = & \displaystyle{xe^x-e^x+\int e^x\sin(e^x)~dx}.\\ \end{array}}
Step 2:  
Now, for the one remaining integral, 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.
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=e^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 du=e^xdx.}
So, 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{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(u)+C}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(e^x)+C}.\\ \end{array}}

(b)

Step 1:  
First, we add and subtract 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} from the numerator.
So, 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{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}.\\ \end{array}}
Step 2:  
Now, we need to use partial fraction decomposition for the second integral.
Since 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 2x^2+x=x(2x+1),} 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 \frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.}
Multiplying both sides of the last equation 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 x(2x+1),} we 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 1-x=A(2x+1)+Bx.}
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 x=0} , the last equation becomes
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 1=A.}
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 x=-\frac{1}{2},} then we 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 \frac{3}{2}=-\frac{1}{2}\,B.} Thus,
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 B=-3.}
So, in summation, 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 \frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.}
Step 3:  
If we plug in the last equation from Step 2 into our final integral in Step 1, 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{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}.\\ \end{array}}
Step 4:  
For the final remaining integral, 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.
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=2x+1.} 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=2\,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 \frac{du}{2}=dx.}
Thus, our final integral becomes
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{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\ &&\\ & = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}.\\ \end{array}}
Therefore, the final answer is
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 \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.}

(c)

Step 1:  
First, we write 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\sin^3x~dx=\int \sin^2 x \sin x~dx.}
Using the identity 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^2x+\cos^2x=1} , we 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 \sin^2x=1-\cos^2x.}
If we use this identity, 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\sin^3x~dx=\int (1-\cos^2x)\sin x~dx.}
Step 2:  
Now, we proceed 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. 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=\cos 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 du=-\sin x dx.}
So 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{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}.\\ \end{array}}
Final Answer:  
(a)  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-\cos(e^x)+C}
(b)  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-\frac{3}{2}\ln (2x+1) +C}
(c)  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+\frac{\cos^3x}{3}+C}

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