The graphs of $f(x) \colon= x^2$ and $g(x) \colon= cx^3$, where $c > 0$, intersect at the points $(0,0)$ and $(1/c, 1/c^2)$. What is the value of $c$---and how to compute this value---so that the region which lies between these graphs and over the interval $[0,1/c]$ has area $2/3$?
Now the problem is whether to integrate $f(x) - g(x)$ over the interval $[0,1/c]$ or whether to integrate $g(x) - f(x)$.
Just set the integral equal to $\frac{2}{3}$ and solve.
$$ \int_0^{\frac{1}{c}} \left(x^2-cx^3\right)dx = \frac{2}{3} \\ \Longleftrightarrow \frac{1}{3c^3}-\frac{1}{4c^3}=\frac{2}{3} \\ \Longleftrightarrow \frac{1}{12c^3}=\frac{2}{3} \\ \Longleftrightarrow c^3=\frac{1}{8} \\ \Longleftrightarrow c=\frac{1}{2} $$
By plugging in $\frac{1}{2}$ for $c$, you can check that this is correct.
EDIT: Didn't see your issue on integrating $f\left(x\right)-g\left(x\right)$ or $g\left(x\right)-f\left(x\right)$. However, others posted good hints for this reasoning.