I am asked to find the area between three curves for x greater than or equal to 0 (there is no upper limit.)
The equation for the functions are as follows: y = 8x^2, y = 2x^2, y = 5 - 3x.
Since there is no upper limit, I am not sure what to do.
Draw the graphs and that will help. Here they are.
You need to find the area enclosed by the three curves. First, note intersection points. $$5-3x=8x^2\to x=\frac58\to (\frac58,\frac{25}{8})$$ $$5-3x=2x^2\to x=1\to(1,2)$$
Use area subtraction with integration to find the desired area:
The large triangle's area is the area bounded by $5-3x$ and the co-ordinate axes and therefore is $\frac12\cdot5\cdot\frac53=\frac{25}{6}$
Now, the desired area is $\frac{25}{6}-(A+B+C)$, so we simply need to find those three areas.
$C$ is very easy. It's simply another triangle, this time with vertices $(1,2), (1,0), (\frac 53, 0)$ and thus has area $\frac12\cdot2\cdot\frac23=\frac23$
$B$ is simply $$\int^{1}_{0}{2x^2dx}=\bigg[\frac23x^3\bigg]^{1}_{0}=\frac23$$ (It's the area under the curve between $x=0$ and the intercept point with $5-3x$)
$A$ is slightly more challenging. It is found by $$\int_{0}^{\frac58}{(5-3x)-8x^2dx}=\bigg[5x-\frac32x^2-\frac83x^3\bigg]^{\frac 58}_0=\frac{725}{384}$$
Thus the desired area is $\frac{25}{6}-(\frac23+\frac23+\frac{725}{384})=\frac{121}{128}$