int_{pi}^{16pi} |sin x| dx = | vedclass

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(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$. We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for

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$30$

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(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$. We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for a periodic function with period $T$. Here, the interval length is $16\pi - \pi = 15\pi$. Thus, $\int_{\pi}^{16\pi} |\sin x| dx = 15 \int_{0}^{\pi} |\sin x| dx$. Since $\sin x \ge 0$ for $x \in [0, \pi]$, we have $|\sin x| = \sin x$. Therefore, $15 \int_{0}^{\pi} \sin x dx = 15 [-\cos x]_{0}^{\pi}$. $= 15 [-\cos(\pi) - (-\cos(0))] = 15 [-(-1) - (-1)] = 15 [1 + 1] = 15 	imes 2 = 30$.

$\int_{\pi}^{16\pi} |\sin x| dx = $

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