Answer
Let $u=1-x$. Then $x=1-u $ and $dx=-du$. When $x = 0, u = 1$; when $x = 1, u =0$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S.=\int_{0}^{1} x^{a}(1-x)^{b} d x &=\int_{1}^{0}(1-u)^{a} u^{b}(-d u) \\
&=\int_{0}^{1} u^{b}(1-u)^{a} d u \\
&=\int_{0}^{1} x^{b}(1-x)^{a} d x\\
& =L.H.S.
\end{aligned}
$$
Work Step by Step
If $a$ and $b$ are positive numbers, show that:
$$\int_{0}^{1} x^{a}(1-x)^{b} d xu=\int_{0}^{1} x^{b}(1-x)^{a} d x$$
Let $u=1-x$. Then $x=1-u $ and $dx=-du$. When $x = 0, u = 1$; when $x = 1, u =0$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S.=\int_{0}^{1} x^{a}(1-x)^{b} d x &=\int_{1}^{0}(1-u)^{a} u^{b}(-d u) \\
&=\int_{0}^{1} u^{b}(1-u)^{a} d u \\
&=\int_{0}^{1} x^{b}(1-x)^{a} d x\\
& =L.H.S.
\end{aligned}
$$
