Answer
$\int_{0}^{1} f(x)~dx = \int_{0}^{1} f(1-x)~dx$
Work Step by Step
The function $f$ is continuous on $[0,1]$.
Let $g(x) = 1-x$
Then $g'(x) = -1$
$g'(x)$ is continuous for all values of $x$
According to the Substitution Rule:
$\int_{g(0)}^{g(1)} f(x)~dx = \int_{0}^{1} f(g(x))~g'(x)~dx$
$\int_{1}^{0} f(x)~dx = \int_{0}^{1} f(1-x)~(-1)~dx$
$-\int_{0}^{1} f(x)~dx = -\int_{0}^{1} f(1-x)~dx$
$\int_{0}^{1} f(x)~dx = \int_{0}^{1} f(1-x)~dx$
