$\int_{-1}^1{x^4}dx$ by substitution My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t 
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't properly explain why this happens. How can I prove that this substitution is wrong?

Since for $r>0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.
Thus,
\begin{align}\int_{-1}^1 x^4\,dx&\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&=\frac{4}{5}\int_0^1 y dy\\
&=\frac{4}{5}\left[\frac{y^2}{2}\right]_0^1\\
&=\frac{4}{5}\times \frac{1}{2}\\
&=\boxed{\frac{2}{5}}
\end{align}
