Answer
$\frac{64}{3}$
Work Step by Step
Find the intersection between $y=x^2$ and $y=8x-x^2$:
$y=y$
$x^2=8x-x^2$
$2x^2-8x=0$
$x(2x-8)=0$
$x=0\vee x=4$
So, both parabola curves intersects at $x=0$ and $x=4$.
Describe the region bounded by the curves.
$y=x^2$ is a function for an upward parabola and $y=8x-x^2$ is a function for a downward parabola.
Graphically, the region is bounded above by the graph $y=8x-x^2$ and bounded below by the graph $y=x^2$ between $x=0$ and $x=4$.
Calculate the area of the region:
$A=\int_0^4(8x-x^2)-x^2 dx$
$=\int_0^48x-2x^2 dx$
$=[4x^2-\frac{2x^3}{3}]_0^4$
$=(4\cdot 4^2-\frac{2\cdot 4^3}{3})-(4\cdot 0^2-\frac{2\cdot 0^3}{3})$
$=(64-\frac{128}{3})-0$
$=\frac{64}{3}$