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?

Actually problem part(a) is not different from part (b), as it is asking to just check your work.. so getting zero between $ 0, \sqrt 2 $ shows your result is in fact correct.
