Question

If f(x) = sin(x) for all x, then the average value of f on the interval [0, π] is

Answer 1

Solution.

Our function is continuous on the closed interval $[\mathbf{0},\pmb{\pi}]$. Then there exists $c$ in the closed interval $[\mathbf{0},\pmb{\pi}]$ such that

$$\int_{0}^{\pi} f(x) \, dx = f(c) (b - a)$$

Then we have

$$\int_{0}^{\pi} \sin x \, dx = \sin c (\pi - 0)$$

Find the value of the integral

$$\int_{0}^{\pi} \sin x \, dx = - \cos x \big|_{0}^{\pi} = \cos 0 - \cos \pi = 1 - (-1) = 2$$

Then

$$\frac{2}{\pi} = \sin c$$ $$c = \arcsin \frac{2}{\pi}$$

Answer:

$$c = \arcsin \frac{2}{\pi}$$