009B Sample Final 2, Problem 2

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

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

by setting  ${\displaystyle f(x)=g(x)}$  and solving for  ${\displaystyle x.}$

2. The area between two functions,  ${\displaystyle f(x)}$  and  ${\displaystyle g(x),}$  is given by  ${\displaystyle \int _{a}^{b}f(x)-g(x)~dx}$

for  ${\displaystyle a\leq x\leq b,}$  where  ${\displaystyle f(x)}$  is the upper function and  ${\displaystyle g(x)}$  is the lower function.

Solution:

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

${\displaystyle 16}$