009B Sample Final 2, Problem 2

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Find the area of the region between the two curves    and  

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    where    is the upper function and    is the lower function.


Solution:

Step 1:  
First, we need to find the intersection points of these two curves.
To do this, we set
       
Getting all the terms on one side of the equation, we get
       
Therefore, we get that these two curves intersect at  
Hence, the region we are interested in occurs between    and  
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 interval    and  
Now, in each of the regions we need to determine which curve has the higher 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
          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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