So I'm giving this problem as an extra exercise:
$$\int_{0}^{1} x^a(1-x)^bdx = \int_{0}^{1} x^b(1-x)^adx$$
What I tried is let $u = 1 - x$, $x = 1 - u$ and $du = -dx$ and then substituted in and got
$$\int_{x = 0}^{x = 1} x^a(1-x)^bdx = \int_{u = 0}^{u = 1}(1 - u)^au^bdu$$
after some simplification. However, I don't know how to continue to prove that the original integrals in terms of x are equal. Although this doesn't count for marks I'm really curious as to how to complete this question. I've tried looking up this question but it seems that people did similar things as what I have done but I'm not seeing a way to equate the original integrals. Any help would be appreciated.
In the presentation \begin{align}\int_{0}^{1} x^a(1-x)^b\,\mathrm dx = &\ldots = \int_{0}^{1}u^b(1 - u)^a\,\mathrm du\\ =&\int_{0}^{1}x^b(1 - x)^a\,\mathrm dx,\end{align} each expression has a definite value and contains no free variable; in other words, the scope of each variable does not extend beyond the (definite) integral in which it occurs; in yet other words, each variable is a dummy variable.
(On the other hand, the $x$ and $y$ in the indefinite integrals $\,\int f(x)\,\mathrm dx\,$ and $\,\int f(y)\,\mathrm dy\,$ are not dummy variables.)
Thus, there is no conflict between the variable change in the first line (integration by substitution) and that in the second line (renaming); the working is valid.