009B Sample Final 2, Problem 2 - Revision history

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Created page with "<span class="exam"> Find the area of the region between the two curves <math style="vertical-align: -4px">y=3x-x^2</math> and <math style="vertical-align: -4..."

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<span class="exam"> Find the area of the region between the two curves  <math style="vertical-align: -4px">y=3x-x^2</math>  and  <math style="vertical-align: -4px">y=2x^3-x^2-5x.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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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>
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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>
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|'''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.
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'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First, we need to find the intersection points of these two curves.
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|To do this, we set
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|       <math>3x-x^2=2x^3-x^2-5x.</math>
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|Getting all the terms on one side of the equation, we get
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|        <math>\begin{array}{rcl}
\displaystyle{0} & = & \displaystyle{2x^3-8x}\\
&&\\
& = & \displaystyle{2x(x^2-4)}\\
&&\\
& = & \displaystyle{2x(x-2)(x+2).}
\end{array}</math>
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|Therefore, we get that these two curves intersect at  <math style="vertical-align: -4px">x=-2,~x=0,~x=2.</math>
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|Hence, the region we are interested in occurs between  <math style="vertical-align: 0px">x=-2</math>  and  <math style="vertical-align: 0px">x=2.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Since the curves intersect also intersect at  <math style="vertical-align: -4px">x=0,</math>  this breaks our region up into two parts,
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|which correspond to the intervals  <math style="vertical-align: -5px">[-2,0]</math>  and  <math style="vertical-align: -5px">[0,2].</math>
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|Now, in each of the regions we need to determine which curve has the higher  <math style="vertical-align: -4px">y</math>  value.
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|To figure this out, we use test points in each interval.
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|For  <math style="vertical-align: -5px">x=-1,</math>  we have
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|       <math style="vertical-align: -5px"> y=3(-1)-(-1)^2=-4</math>  and  <math style="vertical-align: -5px">y=2(-1)^3-(-1)^2-5(-1)=2.</math>
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|For  <math style="vertical-align: -5px">x=1,</math>  we have
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|       <math style="vertical-align: -5px"> y=3(1)-(1)^2=2</math>  and  <math style="vertical-align: -5px">y=2(1)^3-(1)^2-5(1)=-4.</math>
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|Hence, the area  <math style="vertical-align: 0px">A</math>  of the region bounded by these two curves is given by
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|       <math>A=\int_{-2}^0 (2x^3-x^2-5x)-(3x-x^2)~dx+\int_0^2 (3x-x^2)-(2x^3-x^2-5x)~dx.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
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|Now, we integrate to get
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|        <math>\begin{array}{rcl}
\displaystyle{A} & = & \displaystyle{\int_{-2}^0 (2x^3-8x)~dx+\int_0^2 (-2x^3+8x)~dx}\\
&&\\
& = & \displaystyle{\bigg(\frac{x^4}{2}-4x^2\bigg)\bigg|_{-2}^0+\bigg(\frac{-x^4}{2}+4x^2\bigg)\bigg|_0^2}\\
&&\\
& = & \displaystyle{0-\bigg(\frac{(-2)^4}{2}-4(-2)^2\bigg)+\bigg(\frac{-2^4}{2}+4(2)^2\bigg)-0}\\
&&\\
& = & \displaystyle{-(8-16)+(-8+16)}\\
&&\\
& = & \displaystyle{16.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|       <math>16</math>
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[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]