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Chapter 4: Problem 47

Find the average value of the function over the given interval and all values of \(x\) in the interval for which the function equals its average value. $$ f(x)=\sin x, \quad[0, \pi] $$

Short Answer

Expert verified
The average value of the function \(f(x) = \sin x\) over the interval `[0, pi]` is \(\mu = \frac{2}{\pi}\). The function equals its average value at \(x = \arcsin(\frac{2}{\pi})\) and \(x = \pi - \arcsin(\frac{2}{\pi})\) within the interval \([0, \pi]\).

Step by step solution

01

Calculate the average value

Use the formula for the average value of a function: \(\mu = \frac{1}{\pi - 0}\int_{0}^{\pi} \sin x dx =\frac{1}{\pi}\int_{0}^{\pi} \sin x dx\). This integral can be calculated with the fundamental theorem of calculus. The integral of \(\sin x\) from 0 to \(\pi\) is \(-\cos \pi + \cos 0 = -(-1) + 1 = 2\), so \(\mu = \frac{1}{\pi} * 2 = \frac{2}{\pi}\).
02

Solve the equation for x

Set the function \(\sin x\) equal to the average value \(\mu = \frac{2}{\pi}\) and solve for \(x\): \(\sin x = \frac{2}{\pi}\). The solution for this equation in the interval \([0, \pi]\) can be found by using the inverse sine function: \(x = \arcsin(\frac{2}{\pi})\) or \(x = \pi - \arcsin(\frac{2}{\pi})\).

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