for example let's compute this elementary definite integral$$I = \int_{-1}^{1}x^2dx = \frac{x^3}{3}|_{-1}^{1} = \frac{2}{3}$$ that's pretty simple. ok.
now let's say I use the following u-substitution $u = x^2$
the integral becomes of the following form : $$I = \int_{1}^{1} \frac{\sqrt{u}}{2}du$$
here the limits of integration are both $1$ so $I = 0$
why am I getting such a contradiction ? is it because $x^2$ isn't bijective on the interval $[-1,1]$ ?
The calculation is incorrect as $\sqrt{x^2}\ne x$ when $x<0$. But we can still use this substitution if we handle the signs carefully.
Note that if $x<0$, $\sqrt{u}=\sqrt{x^2}=|x|=-x$.
If $x\ge0$, $\sqrt{u}=x$
Let $u=x^2$. Then $du=2xdx$.
\begin{align} \int_{-1}^1x^2dx&=\frac{1}{2}\int_{-1}^0x(2x)dx+\frac{1}{2}\int_0^1x(2x)dx\\ &=\frac{1}{2}\int_1^0(-\sqrt{u})du+\frac{1}{2}\int_0^1\sqrt{u}du\\ &=\int_0^1\sqrt{u}du\\ &=\left[\frac{2}{3}u^\frac{3}{2}\right]_0^1\\ &=\frac{2}{3} \end{align}