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...
May be you can see the problem differently. The best thing is to see what happen we you set $u(x)=x^2$. Therefore $x^3= \frac{1}{2}u(x)u'(x)$ and you know that an antiderivative of a function of type $\frac{1}{2}u(x)u'(x)$ is $\frac{u^2(x)}{4}$. 

Since $u(x)=x^2$, you get that the antiderivatives of $x^3$ is $\frac{x^4}{4},$
and thus 
$$\int_{-2}^1 x^3dx=[\frac{x^4}{4}]_{-2}^1=[\frac{u^2(x)}{4}]_{-2}^{1}=\frac{u^2(1)}{4}-\frac{u^2(-2)}{4}=[\frac{y^2}{4}]_{u(1)}^{u(-2)}=\int_{u(1)}^{u(-2)}\frac{y}{2}dy$$
$$=\frac{1}{2}\int_4^1\frac{y}{2}dy=\int_{1}^{4}-\frac{y}{2}dy=-\frac{1}{2}\int_1^4 ydy.$$
$u=x^2$,  if $x \in [-2,1]$ then $u \in [4,1]$.

the rest is OK.