$\int_{-1}^1{x^4}dx$ 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?
You write in a comment that the formula
$$\int_{\varphi (a)}^{\varphi (b)}{g(x)}dx = \int_{a}^{b} {g(\varphi (x))\varphi '(x)dx}$$
doesn't require $\varphi$ to be injective. This is true. But if you try and use this formula to evaluate your integral with $\varphi(x)=x^2$, you need to choose $g$ so that
$$g(\varphi(x))=\frac12 x^3$$
on $[a,b]=[-1,1]$. But this would require that $g(\varphi(-1))=-\frac12$ and $g(\varphi(1))=\frac12$, which is impossible, because $\varphi(-1)=\varphi(1)$.
So $\varphi$ has to be invertible on $[a,b]$. Otherwise no suitable $g$ can exist.
Taken together with the requirement that $\varphi$ be continuously differentiable on $[a,b]$, this implies that $\varphi$ is $monotonic$ on $[a,b]$. And this is where your substitution breaks down. As others have explained, you can get round this by integrating separately over $[-1,0]$ and $[0,1]$, because $\varphi$ is monotonic on each of these intervals.
This monotonicity requirement seems to be ignored by the various web pages which state the appropriate theorem. True, it is a logical consequence of the theorem, but not, I would say, an obvious consequence.
Since for $r&gt;0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.
Thus,
\begin{align}\int_{-1}^1 x^4\,dx&amp;\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&amp;\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&amp;=\frac{4}{5}\int_0^1 y dy\\
&amp;=\frac{4}{5}\left[\frac{y^2}{2}\right]_0^1\\
&amp;=\frac{4}{5}\times \frac{1}{2}\\
&amp;=\boxed{\frac{2}{5}}
\end{align}