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Compute the average value of the function on the interval

Compute the average value of

    \[ f(x) = \sin^2 x, \qquad 0 \leq x \leq \frac{\pi}{2}. \]


Recalling that \sin^2 x = \frac{1}{2} (1 - \cos 2x), we compute the average A(f),

    \begin{align*}  A(f) = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \sin^2 x \, dx &= \frac{1}{\pi} \int_0^{\frac{\pi}{2}} (1 - \cos 2x) \, dx \\  &= \frac{1}{2} - \frac{1}{2 \pi} \int_0^{\pi} \cos x \, dx \\  &= \frac{1}{2}. \end{align*}

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