Answer
True.
Work Step by Step
True
$(g+f)_{\text {ave }}=\frac{1}{-a+b} \int_{a}^{b}(g(x)+f(x)) d x=\frac{1}{-a+b} \int_{a}^{b} g(x) d x+\frac{1}{-a+b} \int_{a}^{b} f(x) d x=$
$g_{\text {ave}}+f_{\text {ave}}$
Thus, the correct answer is $\underline{true}$.