Difference between revisions of "009B Sample Final 1, Problem 3"

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!Step 1:    
 
!Step 1:    
 
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|First, we graph these two functions.
 
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|Insert graph here
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!Step 2:  
 
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|Setting <math>\sin x=\frac{2}{\pi}x</math>, we get three solutions <math>x=0,\frac{\pi}{2},\frac{-\pi}{2}</math>
 
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|So, the three intersection points are <math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>.
 
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|You can see these intersection points on the graph shown in Step 1.
 
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!Step 1: &nbsp;  
 
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|Using symmetry of the graph, the area bounded by the two functions is given by 
 
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|<math>2\int_0^{\frac{\pi}{2}}\bigg(\sin(x)-\frac{2}{\pi}x\bigg)~dx</math>
 
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!Step 2: &nbsp;
 
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|Lastly, we integrate to get
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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::<math>\begin{array}{rcl}
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\displaystyle{2\int_0^{\frac{\pi}{2}}\bigg(\sin (x)-\frac{2}{\pi}x\bigg)~dx} & {=} & \displaystyle{2\bigg(-\cos (x)-\frac{x^2}{\pi}\bigg)\bigg|_0^{\frac{\pi}{2}}}\\
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&&\\
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& = & \displaystyle{2\bigg(-\cos \bigg(\frac{\pi}{2}\bigg)-\frac{1}{\pi}\bigg(\frac{\pi}{2}\bigg)^2\bigg)}-2(-\cos(0))\\
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&&\\
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& = & \displaystyle{2\frac{-\pi}{4}+2}\\
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&&\\
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& = & \displaystyle{\frac{-\pi}{2}+2}\\
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\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''
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|'''(a)''' <math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>
 
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|'''(b)'''  
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|'''(b)''' <math>\frac{-\pi}{2}+2</math>
 
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[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 18:26, 4 February 2016

Consider the area bounded by the following two functions:

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 y=\sin 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 y=\frac{2}{\pi}x}

a) Find the three intersection points of the two given functions. (Drawing may be helpful.)

b) Find the area bounded by the two functions.

Foundations:  
Review the area between two functions

Solution:

(a)

Step 1:  
First, we graph these two functions.
Insert graph here
Step 2:  
Setting 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 x=\frac{2}{\pi}x} , we get three solutions 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,\frac{\pi}{2},\frac{-\pi}{2}}
So, the three intersection points are 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 (0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)} .
You can see these intersection points on the graph shown in Step 1.

(b)

Step 1:  
Using symmetry of the graph, the area bounded by the two functions is given by
Step 2:  
Lastly, 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{2\int_0^{\frac{\pi}{2}}\bigg(\sin (x)-\frac{2}{\pi}x\bigg)~dx} & {=} & \displaystyle{2\bigg(-\cos (x)-\frac{x^2}{\pi}\bigg)\bigg|_0^{\frac{\pi}{2}}}\\ &&\\ & = & \displaystyle{2\bigg(-\cos \bigg(\frac{\pi}{2}\bigg)-\frac{1}{\pi}\bigg(\frac{\pi}{2}\bigg)^2\bigg)}-2(-\cos(0))\\ &&\\ & = & \displaystyle{2\frac{-\pi}{4}+2}\\ &&\\ & = & \displaystyle{\frac{-\pi}{2}+2}\\ \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 (0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)}
(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 \frac{-\pi}{2}+2}

Return to Sample Exam

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