## Calculus: Early Transcendentals 8th Edition

$f_{avg}=7$
Given: Average value of the function $f(x)=3x^{2}+8x$ over the interval $[-1,2]$ is given by Average value of the function $f(x)$ over the interval $[a.b]$ is given by $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$ $f_{avg}=\frac{1}{2-(-1)}\int_{-1}^{2}3x^{2}+8xdx$ $=\frac{1}{3}[x^{3}+4x^{2}]_{-1}^{2}$ $=\frac{1}{3}[8+16]-\frac{1}{3}[-1+4]$ $=7$ Hence, $f_{avg}=7$