Answer
If $f $is continuous on $[0, 1]$, prove that:
$$
\int_{0}^{1} f(x) d x =\int_{0}^{1} f(1-x) d x
$$
Let $u=1-x $. Then $d u=-d x,$. so$x = a, u = f(a)$; when $x = b, u =f(b)$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S. &= \int_{0}^{1} f(1-x) d x \\
&=\int_{1}^{0} f(u)(-d u) \\
&=\int_{0}^{1} f(u) d u \\
&=\int_{0}^{1} f(x) d x\\
& =L.H.S.
\end{aligned}
$$
Work Step by Step
If $f $is continuous on $[0, 1]$, prove that:
$$
\int_{0}^{1} f(x) d x =\int_{0}^{1} f(1-x) d x
$$
Let $u=1-x $. Then $d u=-d x,$. so$x = a, u = f(a)$; when $x = b, u =f(b)$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S. &= \int_{0}^{1} f(1-x) d x \\
&=\int_{1}^{0} f(u)(-d u) \\
&=\int_{0}^{1} f(u) d u \\
&=\int_{0}^{1} f(x) d x\\
& =L.H.S.
\end{aligned}
$$
