009B Sample Final 1, Problem 4 - Revision history

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MathAdmin: Created page with " Compute the following integrals. a) $\int e^x(x+\sin(e^x))~dx$ b) $\int \frac{2x^2+1}{2x^2+x}~dx$
2016-02-27T05:40:09Z

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">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">c) <math>\int \sin^3x~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''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  <math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</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>.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first distribute to get
|-
|
::<math>\int e^x(x+\sin(e^x))~dx\,=\,\int e^xx~dx+\int e^x\sin(e^x)~dx.</math>
|-
|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>.
|-
|So, we have
|-
|
::<math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>.
|-
|So, we have
|-
|
::<math>\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}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we add and subtract <math style="vertical-align: 0px">x</math> from the numerator.
|-
|So, we have
|-
|
::<math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>.
|-
|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>.
|-
|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  <math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
|-
|So, in summation, we have  <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|If we plug in the last equation from Step 2 into our final integral in Step 1, we have
|-
|
::<math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|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  <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
|-
|Thus, our final integral becomes
|-
|
::<math>\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}</math>
|-
|Therefore, the final answer is
|-
|
::<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>.
|-
|If we use this identity, we have
|-
|     <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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
|-
|
::<math>\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}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>xe^x-e^x-\cos(e^x)+C</math>
|-
|'''(b)'''  <math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
|-
|'''(c)'''  <math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
|}
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]