# How do you find the average value of the function for f(x)=x+sinx, 0<=x<=2pi?

Jun 13, 2018

The average value is $\pi$.

#### Explanation:

The average value of a function continuous on $\left[a , b\right]$ is given by

$A = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

Thus

$A = \frac{1}{2 \pi - 0} {\int}_{0}^{2 \pi} x + \sin x$

$A = \frac{1}{2 \pi} {\left[\frac{1}{2} {x}^{2} - \cos x\right]}_{0}^{2 \pi}$

$A = \frac{1}{2 \pi} \left(\frac{1}{2} \left(4 {\pi}^{2}\right) - 1 + \cos \left(0\right)\right)$

$A = \frac{1}{4 \pi} \left(4 {\pi}^{2}\right)$

$A = \pi$

Hence, the average value of the given function on the interval 0 ≤ x ≤ 2pi is $\pi$.

Hopefully this helps!