The value of the integral int_0^pi(1-|sin 8 x | vedclass

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(C) Let $I = \int_0^\pi (1 - |\sin 8x|) dx$. We can split the integral as: $I = \int_0^\pi 1 dx - \int_0^\pi |\sin 8x| dx$. The first part is $\int_0^

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$\pi-2$

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(C) Let $I = \int_0^\pi (1 - |\sin 8x|) dx$. We can split the integral as: $I = \int_0^\pi 1 dx - \int_0^\pi |\sin 8x| dx$. The first part is $\int_0^\pi dx = \pi$. For the second part,let $f(x) = |\sin 8x|$. The period of $|\sin 8x|$ is $\frac{\pi}{8}$. Since the interval $[0, \pi]$ contains $8$ full periods of the function $|\sin 8x|$,we can write: $\int_0^\pi |\sin 8x| dx = 8 \int_0^{\pi/8} |\sin 8x| dx$. In the interval $[0, \pi/8]$,$\sin 8x \ge 0$,so $|\sin 8x| = \sin 8x$. Thus,$8 \int_0^{\pi/8} \sin 8x dx = 8 \left[ -\frac{\cos 8x}{8} \right]_0^{\pi/8} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. Substituting these values back into the expression for $I$: $I = \pi - 2$.

The value of the integral $\int_0^\pi(1-|\sin 8 x|) d x$ is

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