Question

Consider the given function and the given interval. f\(x\) = 2 sin\(x\) - sin\(2 x\) text(, ) [0 text(, ) pi] (a) Find the average value fave of f on the given interval. fave = Correct: Your answer is correct. (b) Find c such that fave = f(c). (Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes. Round the answers to three decimal places.)

Answer 1

a. The average value of $f$ on the given interval is

$\displaystyle f_{\rm ave}=\frac1{\pi-0}\int_0^\pi(2\sin x-\sin2x)\,\mathrm dx=\boxed{\frac4\pi}$

b. Solve for $c$:

$\dfrac4\pi=2\sin c-\sin2c\implies\boxed{c\approx1.238\text{ or }c\approx2.808}$