009B Sample Final 1, Problem 1

Consider the region bounded by the following two functions:

$y=2(-x^{2}+9)$ and $y=0$

a) Using the lower sum with three rectangles having equal width , approximate the area.

b) Using the upper sum with three rectangles having equal width, approximate the area.

c) Find the actual area of the region.

Foundations:
Link to Riemann sums page

Solution:

(a)

Step 1:
We need to set these two equations equal in order to find the intersection points of these functions.
So, we let $2(-x^{2}+9)=0$. Solving for $x$, we get $x=-3,3$.
This means that we need to calculate the Riemann sums over the interval $[-3,3]$.

Step 2:
Since the length of our interval is $6$ and we are using $3$ rectangles,
each rectangle will have width $2$.
Thus, the lower Riemann sum is
$2(f(-3)+f(-1)+f(3))=2(0+16+0)=32$.

(b)

Step 1:
As in Part (a), the length of our inteval is $6$ and
each rectangle will have width $2$. (See Step 1 and 2 for part (a))

Step 2:
Thus, the upper Riemann sum is
$2(f(-1)+f(-1)+f(1))=2(16+16+16)=96$

(c)

Step 1:
To find the actual area of the region, we need to calculate
$\int _{-3}^{3}2(-x^{2}+9)~dx$

Step 2:
We integrate to get
${\begin{array}{rcl}\displaystyle {\int _{-3}^{3}2(-x^{2}+9)~dx}&=&\displaystyle {2{\bigg (}{\frac {-x^{3}}{3}}+9x{\bigg )}{\bigg |}_{-3}^{3}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {-3^{3}}{3}}+9\times 3{\bigg )}-2{\bigg (}{\frac {-(-3)^{3}}{3}}+9(-3){\bigg )}}\\&&\\&=&\displaystyle {2(-9+27)-2(9-27)}\\&&\\&=&\displaystyle {2(18)-2(-18)}\\&&\\&=&\displaystyle {72}\\\end{array}}$

Final Answer:
(a) $64$
(b) $64$
(c) $72$

Return to Sample Exam