For example, I have the problem:
$$
\int_{-1}^1x^4\left(1-x^2\right)^2dx
$$
I set $\,u=1-x^2\,$, which gives $\,x^2=1-u\,$, $\,x=\pm\sqrt{1-u}\,$ and $\,-\frac{1}{2}du=xdx$.
For the upper and lower bounds of the integral:
$$
x=-1 \rightarrow u=1-\left(-1\right)^2=1-1=0
$$
$$
x=1 \rightarrow u=1-\left(1\right)^2=1-1=0
$$
By using u-substitution, I arrive at
$$
-\frac{1}{2}\int_0^0\pm\sqrt{1-u}\left(1-u\right)u^2du
$$
We know that
$$
\int_a^af(x)dx = 0
$$
I came to the conclusion that this integral is equal to zero as the upper and lower bounds are equal, however I know that that is not true.
I found other questions which address this subject, but none seem to make since to me as to why this is not true.
I am aware that u-substitution is unnecessary for this solution, but I am unclear as to why it doesn't seem to work for $u=1-x^2$. I suspect that the issue arises in the square root function, but I'm not sure, and would like to know how to avoid incorrect solutions like this in the future.
Thank-you.
Best Answer
To use the substitution $u=1-x^2$ to evaluate, $$ \int_{-1}^1x^4\left(1-x^2\right)^2\,\mathrm{d}x $$ one has to remember that for $x\in[0,1]$, we have $x=\sqrt{1-u}$ giving $\mathrm{d}x=\frac{-\mathrm{d}u}{2\sqrt{1-u}}$ and for $x\in[-1,0]$, we have $x=-\sqrt{1-u}$ giving $\mathrm{d}x=\frac{\mathrm{d}u}{2\sqrt{1-u}}$. Thus, we get $$ \int_{-1}^0x^4\left(1-x^2\right)^2\,\mathrm{d}x=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}} $$ and $$ \begin{align} \int_0^1x^4\left(1-x^2\right)^2\,\mathrm{d}x &=\int_{\color{#C00}{1}}^{\color{#C00}{0}}(1-u)^2u^2\frac{\color{#C00}{-}\mathrm{d}u}{2\sqrt{1-u}}\\ &=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}} \end{align} $$