Question

Consider the given function and the given interval.

f(x) = 8 sin x - 4 sin 2x, [0,pi]

(a) Find the average value f ave of f on the given interval.

(b) Find c such that f ave = f(c).

Answer 1

(a) The average value of f(x) on the closed interval [0, π] is

$\displaystyle\frac1{\pi-0}\int_0^\pi f(x)\,\mathrm dx = \frac1\pi\int_0^\pi(8\sin(x)-4\sin(2x))\,\mathrm dx = \boxed{\frac{16}\pi}$

(b) By the mean value theorem, there is some c in the open interval (0, π) such that f(c) = 16/π. Solve for c :

8 sin(c) - 4 sin(2c) = 16/π

8 sin(c) - 8 sin(c) cos(c) = 16/π

sin(c) - sin(c) cos(c) = 2/π

Use a calculator to solve this. You should get two solutions, c ≈ 1.2382 and c ≈ 2.8081.