009B Sample Final 1, Problem 3 - Revision history

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MathAdmin: Undo revision 1239 by MathAdmin (talk)

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Undo revision 1239 by MathAdmin (talk)

MathAdmin at 05:50, 27 February 2016

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MathAdmin: Created page with "Consider the area bounded by the following two functions: :::::: $y=\sin x$ and $y=\frac{2}{\pi}x.$
2016-02-27T05:39:16Z

Created page with "Consider the area bounded by the following two functions: ::::::<math>y=\sin x</math> and <math style="vertical-align: -13px">y=\frac{2}{\pi}x.</math> <sp..."

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Consider the area bounded by the following two functions:
::::::<math>y=\sin x</math> and <math style="vertical-align: -13px">y=\frac{2}{\pi}x.</math>

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

b) Find the area bounded by the two functions.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math>
|-
|
::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>
|-
|
::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.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we graph these two functions.
|-
|Insert graph here
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>
|-
|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>.
|-
|You can see these intersection points on the graph shown in Step 1.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using symmetry of the graph, 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>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Lastly, we integrate to get
|-
|
::<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{2\bigg(-\cos \bigg(\frac{\pi}{2}\bigg)-\frac{1}{\pi}\bigg(\frac{\pi}{2}\bigg)^2\bigg)}-2(-\cos(0))\\
&&\\
& = & \displaystyle{2\bigg(-\frac{\pi}{4}\bigg)+2}\\
&&\\
& = & \displaystyle{-\frac{\pi}{2}+2}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>
|-
|'''(b)'''  <math>-\frac{\pi}{2}+2</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 30: / Line 30: @@
 |First, we graph these two functions.
 |-
-|[[File:009A_SF1_3.png |center|500px]]
+|[[File:009B_SF1_3.png |center|500px]]
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the area bounded by the following two functions:
-::::::<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>
-::<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.
+<span class="exam">(b) Find the area bounded by the two functions.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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>
 |-
+|
-::'''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>
 |-
+|'''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>
 |-
+|
-:::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:'''
@@ Line 33: / Line 30: @@
 |First, we graph these two functions.
 |-
-|[[File:9BF1 3 GP.png|center|800px]]
 |}
@@ Line 39: / Line 36: @@
 !Step 2: &nbsp;
 |-
-|Setting <math style="vertical-align: -14px">\sin x=\frac{2}{\pi}x,</math> we get three solutions:
+|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
-::<math>x=0,\frac{\pi}{2},\frac{-\pi}{2}.</math>
+|-
 |-
-|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.
@@ Line 54: / Line 58: @@
 !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>
 |-
+|
@@ Line 68: / Line 72: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&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))\\
+& = & \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>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(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)
 |-
-|&nbsp;&nbsp; '''(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>''']]

@@ Line 31: / Line 31: @@
 |First, we graph these two functions.
 |-
-|Insert graph here
+|[[File:9BF1 3 GP.png|center|800px]]|}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"