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

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<span class="exam">Consider the area bounded by the following two functions:  
 
<span class="exam">Consider the area bounded by the following two functions:  
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::::::<math>y=\sin x</math> and <math style="vertical-align: -13px">y=\frac{2}{\pi}x.</math>
+
::<span class="exam"><math style="vertical-align: -4px">y=\cos x</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=2-\cos x,~0\le x\le 2\pi.</math>
  
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::<span class="exam">a) Find the three intersection points of the two given functions. (Drawing may be helpful.)
+
<span class="exam">(a) Sketch the graphs and find their points of intersection.
  
−
::<span class="exam">b) Find the area bounded by the two functions.
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<span class="exam">(b) Find the area bounded by the two functions.
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|Recall:
+
|'''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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::'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;by setting &nbsp;<math style="vertical-align: -5px">f(x)=g(x)</math>&nbsp; and solving for &nbsp;<math style="vertical-align: 0px">x.</math>
 
|-
 
|-
−
|
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|'''2.''' The area between two functions, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x),</math>&nbsp; is given by &nbsp;<math>\int_a^b f(x)-g(x)~dx</math>  
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:::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x.</math>
 
−
|-
 
−
|
 
−
::'''2.''' The area between two functions, <math style="vertical-align: -5px">f(x)</math> and <math style="vertical-align: -5px">g(x),</math> is given by <math>\int_a^b f(x)-g(x)~dx</math>  
 
 
|-
 
|-
 
|
 
|
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:::for <math style="vertical-align: -3px">a\leq x\leq b,</math> where <math style="vertical-align: -5px">f(x)</math> is the upper function and <math style="vertical-align: -5px">g(x)</math> is the lower function.  
+
&nbsp; &nbsp; &nbsp; &nbsp;for &nbsp;<math style="vertical-align: -3px">a\leq x\leq b,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is the upper function and &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is the lower function.  
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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|First, we graph these two functions.
 
|First, we graph these two functions.
 
|-
 
|-
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|[[File:9BF1 3 GP.png|center|800px]]
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|
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Setting <math style="vertical-align: -14px">\sin x=\frac{2}{\pi}x,</math> we get three solutions:  
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|Setting &nbsp;<math style="vertical-align: -4px">\cos x=1-\cos x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">2\cos x=2.</math>
 
|-
 
|-
−
|
+
|Therefore, we have
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::<math>x=0,\frac{\pi}{2},\frac{-\pi}{2}.</math>
+
|-
 +
|&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,
 
|-
 
|-
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|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>
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|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;  
 
|-
 
|-
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|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   
 
|-
 
|-
 
|
 
|
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::<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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|-
 
|-
 
|
 
|
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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}\\
 
&&\\
 
&&\\
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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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& = & \displaystyle{(2x-2\sin x)\bigg|_0^{2\pi}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{2\bigg(-\frac{\pi}{4}\bigg)+2}\\
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& = & \displaystyle{(4\pi-2\sin(2\pi))-(0-2\sin(0))}\\
 
&&\\
 
&&\\
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& = & \displaystyle{-\frac{\pi}{2}+2}.\\
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& = & \displaystyle{4\pi.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|&nbsp;&nbsp; '''(a)''' &nbsp;<math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>
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|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>(0,1),(2\pi,1)</math>&nbsp; (See Step 1 above for graph)
 
|-
 
|-
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|&nbsp;&nbsp; '''(b)''' &nbsp;<math>-\frac{\pi}{2}+2</math>  
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|&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>''']]

Revision as of 08:43, 10 April 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    and solving for  

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

       for  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 a\leq x\leq b,}   where  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 f(x)}   is the upper function 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 g(x)}   is the lower function.


Solution:

(a)

Step 1:  
First, we graph these two functions.
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 \cos x=1-\cos x,}   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 2\cos x=2.}
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 \cos x=1.}
In the interval  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\le x\le 2\pi,}   the solutions to this equation 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 x=0}   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 x=2\pi.}
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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