Integrate $\int^a_0x^3 - x \mathop{\mathrm{d}x}$ to find the area 
(a) Solve the equation: $$\int^a_0x^3 - x \mathop{\mathrm{d}x} = 0, a > 0$$
(b) For this value of $a$, find the total area enclosed between the $x$-axis  and the curve $y=x^3 - x$ for $0 \leq x \leq a$.

I can quite easily solve the first part but for part (b) when I substitute $\sqrt{2}$ I just get $0$, what am I doing wrong?

In general, you can determine the area between the curve and the $x$-axis as follows :
You determine the roots of $f(x)$. You only need the roots inside the given interval.
To get the intervals :
Go from the left end of the interval to the smallest root. From here to the next larger root , continue in this manner, and finally from the largest root to the right end of the interval. 
Take the absolute values of the integrals over the intervals. Add these values and you are done.
