MathAdmin: Created page with " Evaluate the following integrals. (a) $\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx$ (b) &nbs..."

2017-04-10T16:52:49Z

Created page with " Evaluate the following integrals. (a) <math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx</math> (b) &nbs..."

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Evaluate the following integrals.

(a)  <math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx</math>

(b)  <math>\int \frac{x^2}{(1+x^3)^2}~dx</math>

(c)  <math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' Recall
|-
|       <math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
|-
|'''2.''' How would you integrate   <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math>
|-
|
        You could use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|        Let  <math style="vertical-align: -5px">u=\ln(x).</math>
|-
|        Then,  <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
|-
|
        Thus,
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
&&\\
& = & \displaystyle{\frac{u^2}{2}+C}\\
&&\\
& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
\end{array}</math>
|-
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we notice
|-
|       <math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.</math>
|-
|Now, we use  <math>u</math>-substitution.
|-
|Let  <math style="vertical-align: -2px">u=4x.</math>
|-
|Then,  <math style="vertical-align: -1px">du=4dx</math>  and  <math>\frac{du}{4}=dx.</math>
|-
|Also, we need to change the bounds of integration.
|-
|Plugging in our values into the equation  <math style="vertical-align: -5px">u=4x,</math>  we get
|-
|       <math style="vertical-align: -6px">u_1=4(0)=0</math>  and  <math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
|-
|Therefore, the integral becomes
|-
|        <math style="vertical-align: -19px">\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\
&&\\
& = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\
&&\\
& = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\
&&\\
& = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\
&&\\
& = & \displaystyle{\frac{\pi}{12}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use  <math style="vertical-align: 0px">u</math>-substitution.
|-
|Let  <math style="vertical-align: -2px">u=1+x^3.</math>
|-
|Then,  <math style="vertical-align: 0px">du=3x^2dx</math>  and  <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
|-
|Therefore, the integral becomes
|-
|        <math style="vertical-align: -13px">\frac{1}{3}\int \frac{1}{u^2}~du.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \frac{x^2}{(1+x^3)^2}~dx} & = & \displaystyle{\frac{1}{3}\int \frac{1}{u^2}~du}\\
&&\\
& = & \displaystyle{-\frac{1}{3u}+C}\\
&&\\
& = & \displaystyle{-\frac{1}{3(1+x^3)}+C.}
\end{array}</math>
|-
|
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use  <math>u</math>-substitution.
|-
|Let  <math style="vertical-align: -5px">u=\ln(x).</math>
|-
|Then,  <math style="vertical-align: -15px">du=\frac{1}{x}dx.</math>
|-
|Also, we need to change the bounds of integration.
|-
|Plugging in our values into the equation  <math style="vertical-align: -5px">u=\ln(x),</math>
|-
|we get
|-
|       <math style="vertical-align: -6px">u_1=\ln(1)=0</math>  and  <math style="vertical-align: -6px">u_2=\ln(e)=1.</math>
|-
|Therefore, the integral becomes
|-
|        <math style="vertical-align: -19px">\int_0^1 \cos(u)~du.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We now have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\int_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^1 \cos(u)~du}\\
&&\\
& = & \displaystyle{\sin(u)\bigg|_0^1}\\
&&\\
& = & \displaystyle{\sin(1)-\sin(0)}\\
&&\\
& = & \displaystyle{\sin(1).}
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{\pi}{12}</math>
|-
|   '''(b)'''    <math>-\frac{1}{3(1+x^3)}+C</math>
|-
|   '''(c)'''    <math>\sin(1)</math>
|}
[[009B_Sample_Final_3|'''Return to Sample Exam''']]

009B Sample Final 3, Problem 2 - Revision history

MathAdmin: Created page with " Evaluate the following integrals. (a) \int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx (b) &nbs..."

MathAdmin: Created page with " Evaluate the following integrals. (a) $\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx$ (b) &nbs..."