009B Sample Final 2, Problem 2

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

Foundations:
1. You can find the intersection points of two functions, say  $f(x),g(x),$
by setting  $f(x)=g(x)$  and solving for  $x.$
2. The area between two functions,  $f(x)$  and  $g(x),$  is given by  $\int _{a}^{b}f(x)-g(x)~dx$
for  $a\leq x\leq b,$  where  $f(x)$  is the upper function and  $g(x)$  is the lower function.

Solution:

Step 1:
First, we need to find the intersection points of these two curves.
To do this, we set
$3x-x^{2}=2x^{3}-x^{2}-5x.$
Getting all the terms on one side of the equation, we get
${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {2x^{3}-8x}\\&&\\&=&\displaystyle {2x(x^{2}-4)}\\&&\\&=&\displaystyle {2x(x-2)(x+2).}\end{array}}$
Therefore, we get that these two curves intersect at  $x=-2,~x=0,~x=2.$
Hence, the region we are interested in occurs between  $x=-2$  and  $x=2.$

Step 2:
Since the curves intersect also intersect at  $x=0,$  this breaks our region up into two parts,
which correspond to the interval  $[-2,0]$  and  $[0,2].$
Now, in each of the regions we need to determine which curve has the higher $y$ value.
To figure this out, we use test points in each interval.
For $x=-1,$ we have
$y=3(-1)-(-1)^{2}=-4$  and  $y=2(-1)^{3}-(-1)^{2}-5(-1)=2.$
For $x=1,$ we have
$y=3(1)-(1)^{2}=2$  and  $y=2(1)^{3}-(1)^{2}-5(1)=-4.$
Hence, the area $A$ of the region bounded by these two curves is given by
$A=\int _{-2}^{0}(2x^{3}-x^{2}-5x)-(3x-x^{2})~dx+\int _{0}^{2}(3x-x^{2})-(2x^{3}-x^{2}-5x)~dx.$

Step 3:
Now, we integrate to get
${\begin{array}{rcl}\displaystyle {A}&=&\displaystyle {\int _{-2}^{0}(2x^{3}-8x)~dx+\int _{0}^{2}(-2x^{3}+8x)~dx}\\&&\\&=&\displaystyle {{\bigg (}{\frac {x^{4}}{2}}-4x^{2}{\bigg )}{\bigg |}_{-2}^{0}+{\bigg (}{\frac {-x^{4}}{2}}+4x^{2}{\bigg )}{\bigg |}_{0}^{2}}\\&&\\&=&\displaystyle {0-{\bigg (}{\frac {(-2)^{4}}{2}}-4(-2)^{2}{\bigg )}+{\bigg (}{\frac {-2^{4}}{2}}+4(2)^{2}{\bigg )}-0}\\&&\\&=&\displaystyle {-(8-16)+(-8+16)}\\&&\\&=&\displaystyle {16.}\end{array}}$

Final Answer:
$16$

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