Hello, gp3!
Find the area bounded by: .$\displaystyle y\,=\,4x^2,\;y=0,\;x=0,\;x=4$
The answer given to me was 16.
I believe the area looks like this: Code:

4*..
* :::*. 4
*+*+
*  2 *:::
*  *::
 '::
*  *:
 ':
 :
*  *

We want the area above the xaxis from 0 to 2:
. . $\displaystyle \int^2_0(4x^2)\,dx \;\;=\;\;4x  \tfrac{1}{3}x^3\,\bigg]^2_0 \;\;=\;\;\left(8  \tfrac{8}{3}\right)  (00) \;\;=\;\;\frac{16}{3}$
We want the area below the xaxis from 2 to 4:
. . $\displaystyle \int^4_2(x^24)\,dx \;\;=\;\;\tfrac{1}{3}x^34x\,\bigg]^4_2 \;\;=\;\;\left(\tfrac{64}{3}  32\right)  \left(\tfrac{8}{3}  8\right) \;\;=\;\;\tfrac{16}{3}  \left(\tfrac{16}{3}\right) \;\;=\;\;\frac{32}{3}$
Total area: . $\displaystyle \frac{16}{3} + \frac{32}{3} \;\;=\;\;\frac{48}{3} \;\;=\;\;16$