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...
 A: 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 
$$\int_{-2}^1 x^3dx=[\frac{x^4}{4}]_{-2}^1=[\frac{u^2(x)}{4}]_{-2}^{1}=\frac{u^2(1)}{4}-\frac{u^2(-2)}{4}=[\frac{y^2}{4}]_{u(1)}^{u(-2)}=\int_{u(1)}^{u(-2)}\frac{y}{2}dy$$
$$=\frac{1}{2}\int_4^1\frac{y}{2}dy=\int_{1}^{4}-\frac{y}{2}dy=-\frac{1}{2}\int_1^4 ydy.$$
A: 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
A: You have misunderstood the Variable Substitution of integration! 
While substituting $ u $ for $ x $, we need a continuously differentiable function $\varphi: I_u \to I_x $, where $ I_u = [0,4],I_x=[-2,1] $ in your case. Or, if there exists $ \phi:I_x\to I_u $, and it's inverse $ \phi^{-1} $ is continuously differentiable, then $ \phi^{-1} $ can be the $ \varphi $. 
$ u=\phi(x)=x^2 $ doesn't have inverse on the whole $ [-2,1] $, so you are wrong. But it does on $ [-2,0] $ and $ [0,1] $ respectively, namely $ -\sqrt{u} $ and $ \sqrt{u} $. So the following is correct:
$$ \int_{-2}^1 x^3\mathrm dx = \int_{-2}^0 x^3\mathrm dx + \int_{0}^1 x^3\mathrm dx = \frac{1}{2}\int_{4}^0 u\mathrm du + \frac{1}{2}\int_{0}^1 u\mathrm du = \frac{1}{2}\int_{4}^1 u\mathrm du $$
A: $u=x^2$,  if $x \in [-2,1]$ then $u \in [4,1]$.
the rest is OK.
A: If $g:[a,b]\to [c,d]$ is a continuously differentiable function such that $g(a)=c$ and $g(b)=d$ or vice-versa(the orientation is taken care by the sign of the derivative, which does not happen in higher dimensions), then $\int_{g(a)}^{g(b)} f=\int_a^b (f\circ g)g'$.
Just by above,we let $g:[-2,1]\to [1,4]$ and $f(x)=x$, then $\int_{-2}^1x^3=\frac{1}{2}\int_{-2}^1 (x^2)(2x)=\frac{1}{2}\int_{(-2)^2}^{1^2} u=\frac{1}{2}\int_4^1 u$.
