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

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations:    
 
!Foundations:    
−
|-
 
−
|Recall:
 
 
|-
 
|-
 
|'''1.''' You can find the intersection points of two functions, say &nbsp;<math style="vertical-align: -5px">f(x),g(x),</math>
 
|'''1.''' You can find the intersection points of two functions, say &nbsp;<math style="vertical-align: -5px">f(x),g(x),</math>
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Setting <math style="vertical-align: -14px">\sin x=\frac{2}{\pi}x</math>, we get three solutions: <math>x=0,\frac{\pi}{2},\frac{-\pi}{2}.</math>
+
|Setting &nbsp;<math style="vertical-align: -4px">\cos x=2-\cos x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">2\cos x=2.</math>
 +
|-
 +
|Therefore, we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos x=1.</math>
 +
|-
 +
|In the interval &nbsp;<math style="vertical-align: -4px">0\le x\le 2\pi,</math>&nbsp; the solutions to this equation are
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2\pi.</math>
 +
|-
 +
|Plugging these values into our equations,
 
|-
 
|-
−
|So, the three intersection points are <math style="vertical-align: -15px">(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>.
+
|we get the intersection points &nbsp;<math style="vertical-align: -4px">(0,1)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2\pi,1).</math>
 
|-
 
|-
 
|You can see these intersection points on the graph shown in Step 1.
 
|You can see these intersection points on the graph shown in Step 1.
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|Using symmetry of the graph, the area bounded by the two functions is given by   
+
|The area bounded by the two functions is given by   
 
|-
 
|-
 
|
 
|
−
::<math>2\int_0^{\frac{\pi}{2}}\bigg(\sin(x)-\frac{2}{\pi}x\bigg)~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^{2\pi} (2-\cos x)-\cos x~dx.</math>
 
|-
 
|-
 
|
 
|
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\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{\int_0^{2\pi} (2-\cos x)-\cos x~dx} & {=} & \displaystyle{\int_0^{2\pi} 2-2\cos x~dx}\\
 
&&\\
 
&&\\
−
& = & \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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& = & \displaystyle{(2x-2\sin x)\bigg|_0^{2\pi}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{2\bigg(-\frac{\pi}{4}\bigg)+2}\\
+
& = & \displaystyle{(4\pi-2\sin(2\pi))-(0-2\sin(0))}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{-\frac{\pi}{2}+2}.\\
+
& = & \displaystyle{4\pi.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|'''(a)''' &nbsp;<math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>
+
|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>(0,1),(2\pi,1)</math>&nbsp; (See Step 1 above for graph)
 
|-
 
|-
−
|'''(b)''' &nbsp;<math>-\frac{\pi}{2}+2</math>  
+
|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>4\pi</math>  
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 13:55, 20 May 2017

Consider the area bounded by the following two functions:

  and  

(a) Sketch the graphs and find their points of intersection.

(b) Find the area bounded by the two functions.

Foundations:  
1. You can find the intersection points of two functions, say  

       by setting  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=g(x)}   and solving for  

2. The area between two functions,    and    is given by  

       for    where    is the upper function and    is the lower function.


Solution:

(a)

Step 1:  
First, we graph these two functions.
Insert graph here
Step 2:  
Setting    we get  
Therefore, we have
       
In the interval    the solutions to this equation are
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=0}   and  
Plugging these values into our equations,
we get the intersection points  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 (0,1)}   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 (2\pi,1).}
You can see these intersection points on the graph shown in Step 1.

(b)

Step 1:  
The area bounded by the two functions is given 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 \int_0^{2\pi} (2-\cos x)-\cos x~dx.}

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{\int_0^{2\pi} (2-\cos x)-\cos x~dx} & {=} & \displaystyle{\int_0^{2\pi} 2-2\cos x~dx}\\ &&\\ & = & \displaystyle{(2x-2\sin x)\bigg|_0^{2\pi}}\\ &&\\ & = & \displaystyle{(4\pi-2\sin(2\pi))-(0-2\sin(0))}\\ &&\\ & = & \displaystyle{4\pi.}\\ \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 (0,1),(2\pi,1)}   (See Step 1 above for graph)
   (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 4\pi}

Return to Sample Exam

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