Difference between revisions of "009B Sample Final 2, Problem 2"

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!Foundations:    
 
!Foundations:    
 
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|'''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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&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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&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.
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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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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|&nbsp; &nbsp; &nbsp; &nbsp;<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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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{0} & = & \displaystyle{2x^3-8x}\\
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&&\\
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& = & \displaystyle{2x(x^2-4)}\\
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&&\\
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& = & \displaystyle{2x(x-2)(x+2).}
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\end{array}</math>
 
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|Therefore, we get that these two curves intersect at &nbsp;<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 &nbsp;<math style="vertical-align: 0px">x=-2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Since the curves intersect also intersect at &nbsp;<math style="vertical-align: -4px">x=0,</math>&nbsp; this breaks our region up into two parts,
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|which correspond to the intervals &nbsp;<math style="vertical-align: -5px">[-2,0]</math>&nbsp; and &nbsp;<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 &nbsp;<math style="vertical-align: -4px">y</math>&nbsp; value.
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|To figure this out, we use test points in each interval.
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|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px"> y=3(-1)-(-1)^2=-4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=2(-1)^3-(-1)^2-5(-1)=2.</math>
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|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px"> y=3(1)-(1)^2=2</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=2(1)^3-(1)^2-5(1)=-4.</math>
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|Hence, the area &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; of the region bounded by these two curves is given by
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<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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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
 
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|Now, we integrate to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{A} & = & \displaystyle{\int_{-2}^0 (2x^3-8x)~dx+\int_0^2 (-2x^3+8x)~dx}\\
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&&\\
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& = & \displaystyle{\bigg(\frac{x^4}{2}-4x^2\bigg)\bigg|_{-2}^0+\bigg(\frac{-x^4}{2}+4x^2\bigg)\bigg|_0^2}\\
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&&\\
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& = & \displaystyle{0-\bigg(\frac{(-2)^4}{2}-4(-2)^2\bigg)+\bigg(\frac{-2^4}{2}+4(2)^2\bigg)-0}\\
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&&\\
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& = & \displaystyle{-(8-16)+(-8+16)}\\
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&&\\
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& = & \displaystyle{16.}
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\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>16</math>
 
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[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 16:05, 4 March 2017

Find the area of the region between the two curves  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=3x-x^2}   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=2x^3-x^2-5x.}

Foundations:  
1. You can find the intersection points of two functions, say  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),g(x),}

       by 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 f(x)=g(x)}   and solving 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 x.}

2. The area between 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 f(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 g(x),}   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_a^b f(x)-g(x)~dx}

       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:

Step 1:  
First, we need to find the intersection points of these two curves.
To do this, we set
       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 3x-x^2=2x^3-x^2-5x.}
Getting all the terms on one side of the equation, 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 \begin{array}{rcl} \displaystyle{0} & = & \displaystyle{2x^3-8x}\\ &&\\ & = & \displaystyle{2x(x^2-4)}\\ &&\\ & = & \displaystyle{2x(x-2)(x+2).} \end{array}}
Therefore, we get that these two curves intersect at  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,~x=0,~x=2.}
Hence, the region we are interested in occurs between  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}   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.}
Step 2:  
Since the curves intersect also intersect at  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,}   this breaks our region up into two parts,
which correspond to the intervals  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 [-2,0]}   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 [0,2].}
Now, in each of the regions we need to determine which curve has the higher  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}   value.
To figure this out, we use test points in each interval.
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 x=-1,}   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 y=3(-1)-(-1)^2=-4}   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=2(-1)^3-(-1)^2-5(-1)=2.}
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 x=1,}   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 y=3(1)-(1)^2=2}   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=2(1)^3-(1)^2-5(1)=-4.}
Hence, the area  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}   of the region bounded by these two curves 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 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.}
Step 3:  
Now, 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{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}}


Final Answer:  
       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 16}

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