Answer
False
Work Step by Step
Consider $f(x)=2x-1$
Note that $f(x)$ is not equal to zero for all values of $x$ such as $0\leq x\leq 1$.
But
$\int _{0}^{1}f(x) dx=\int _{0}^{1}(2x-1) dx$
$=[x^{2}-x]_{0}^{1}$
$=0$
Hence, the statement is false.