## Calculus: Early Transcendentals 8th Edition

$f_{avg}=\frac{2(e-1)}{\pi}$
Given: Average value of the function $f(x)=e^{sint}cost$ over the interval $[0,\pi/2]$ is given by $f_{avg}=\frac{1}{\pi/2-0}\int_{0}^{\pi/2}e^{sint}costdt$ Substitute $sint=u$ and $costdt=du$ Limits of integration changes $\int_{0}^{\pi/2}$ to $\int_{sin0}^{sin\pi/2}=\int_{0}^{1}$ $f_{avg}=\frac{2}{\pi}\int_{0}^{1}e^{u}du$ $f_{avg}=\frac{2}{\pi}[e^{u}]_{0}^{1}$ Hence, $f_{avg}=\frac{2(e-1)}{\pi}$