Difference between revisions of "009B Sample Midterm 3, Problem 3"

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&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
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&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
 
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|&nbsp; &nbsp; Let <math style="vertical-align: -3px">u=x^2+1.</math> Then, <math style="vertical-align: -1px">du=2x~dx.</math> Thus,
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|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -3px">u=x^2+1.</math>  
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=2x~dx.</math>  
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|&nbsp; &nbsp; &nbsp; &nbsp; Thus,
 
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&nbsp; &nbsp; <math>\begin{array}{rcl}
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&nbsp; &nbsp;&nbsp;&nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\
 
\displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\
 
&&\\
 
&&\\
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|We proceed using <math style="vertical-align: 0px">u</math>-substitution.  
 
|We proceed using <math style="vertical-align: 0px">u</math>-substitution.  
 
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|Let <math style="vertical-align: -1px">u=x^3.</math> Then, <math style="vertical-align: -1px">du=3x^2~dx</math> and <math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
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|Let <math style="vertical-align: -1px">u=x^3.</math>  
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|Then, <math style="vertical-align: -1px">du=3x^2~dx</math> and <math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
 
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|Therefore, we have
 
|Therefore, we have
 
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&nbsp; &nbsp;<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
 
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&nbsp;&nbsp;<math>\begin{array}{rcl}
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\
 
\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\
 
&&\\
 
&&\\
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|We proceed using u substitution.  
 
|We proceed using u substitution.  
 
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|-
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|Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx.</math>  
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|Let <math style="vertical-align: -5px">u=\cos(x).</math>  
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|-
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|Then, <math style="vertical-align: -5px">du=-\sin(x)~dx.</math>  
 
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|Since this is a definite integral, we need to change the bounds of integration.  
 
|Since this is a definite integral, we need to change the bounds of integration.  
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&nbsp; &nbsp;<math>\begin{array}{rcl}
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\
 
\displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\
 
&&\\
 
&&\\
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' &nbsp; <math>\frac{-1}{3}\cos(x^3)+C</math>
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{-1}{3}\cos(x^3)+C</math>
 
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|'''(b)''' &nbsp; <math>0</math>
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>0</math>
 
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|}
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:38, 7 February 2017

Compute the following integrals:

a)  
b)  


Foundations:  
How would you 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 2x(x^2+1)^3~dx?}

        You could 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=x^2+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=2x~dx.}
        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 \begin{array}{rcl} \displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\ &&\\ & = & \displaystyle{\frac{u^4}{4}+C}\\ && \\ & = & \displaystyle{\frac{(x^2+1)^4}{4}+C.}\\ \end{array}}


Solution:

(a)

Step 1:  
We proceed 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.
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^3.}
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=3x^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}{3}=x^2~dx.}
Therefore, 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^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.}

Step 2:  
We integrate 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 \begin{array}{rcl} \displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\ &&\\ & = & \displaystyle{\frac{-1}{3}\cos(x^3)+C.}\\ \end{array}}

(b)

Step 1:  
We proceed using 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.}
Since this is a definite integral, we need to change the bounds of integration.
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_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{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 u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.}
Step 2:  
Therefore, 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 \begin{array}{rcl} \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\ &&\\ & = & \displaystyle{\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}}\\ &&\\ & = & \displaystyle{0.} \\ \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 \frac{-1}{3}\cos(x^3)+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 0}

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