## Calculus: Early Transcendentals 8th Edition

Given: $f(x)\geq g(x)$ This implies $f(x)-g(x)\geq 0$ $\int _{a}^{b}[f(x)-g(x)] dx\geq 0$ $\int _{a}^{b}f(x) dx-\int _{a}^{b}g(x) dx\geq 0$ $\int _{a}^{b}f(x) dx\geq \int _{a}^{b}g(x) dx$ Hence, the statement is true.