If I = int_0^{100pi} sqrt{1 - cos 2x} , dx,th | vedclass

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(B) Given $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$. Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have $\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x

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$200\sqrt{2}$

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(B) Given $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$. Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have $\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$. So,$I = \sqrt{2} \int_0^{100\pi} |\sin x| \, dx$. Since $|\sin x|$ is a periodic function with period $\pi$,we can write $\int_0^{100\pi} |\sin x| \, dx = 100 \int_0^{\pi} |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. Thus,$I = 100\sqrt{2} \int_0^{\pi} \sin x \, dx$. Evaluating the integral: $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. Therefore,$I = 100\sqrt{2} 	imes 2 = 200\sqrt{2}$.

If $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$,then the value of $I$ is

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