The value of int_0^{npi + v} {|sin x|,dx} is | vedclass

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(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n

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$2n + 1 - \cos v$

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(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n \int_0^{\pi} \sin x \, dx = n [-\cos x]_0^{\pi} = n(1 - (-1)) = 2n$. Now,consider the integral $\int_{n\pi}^{n\pi + v} |\sin x| \, dx$. Let $x = n\pi + t$,then $dx = dt$. When $x = n\pi$,$t = 0$. When $x = n\pi + v$,$t = v$. Since $|\sin(n\pi + t)| = |\sin(n\pi) \cos t + \cos(n\pi) \sin t| = |(-1)^n \sin t| = |\sin t|$. So,$\int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_0^v |\sin t| \, dt$. Assuming $0 \le v \le \pi$,we have $\int_0^v \sin t \, dt = [-\cos t]_0^v = 1 - \cos v$. Therefore,$\int_0^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v$.

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

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