## Calculus: Early Transcendentals 8th Edition

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.