Answer
$f_{avg}=\frac{2(e-1)}{\pi}$
Work Step by Step
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, $f_{avg}=\frac{2(e-1)}{\pi}$
