What's wrong in $\int_{-2}^1 x^3dx=-\frac{1}{2}\int_0^4 udu$ What's wrong in $$\int_{-2}^1 x^3\mathrm dx=-\frac{1}{2}\int_0^4 u\mathrm du \ ?$$
I set $u=x^2$, therefore $du=2xdx$. But when $x$ traverse $[-2,1]$, then $u$ traverse $[0,4]$ and thus 
$$\int_{-2}^1 x^3\mathrm dx=\frac{1}{2}\int_{4}^0 u\mathrm du=-\frac{1}{2}\int_0^4 u\mathrm du.$$
But $\int_{)2}^1 x^3 \mathrm dx=-\frac{15}{4}$ whereas $-\frac{1}{2}\int_{0}^4u\mathrm du=-4$, therefore something is wrong here, but I don't see what's wrong...

Its your limits where you have made the mistake:
$$\int_{-2}^1 x^3\mathrm dx=\frac{1}{2}\int_{4}^{\color{red}{0}} u\mathrm du$$
The part marked red should be 1
