int_{,pi }^{,10pi } {|sin x|dx} is | vedclass

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(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,w

Vedclass · Accepted Answer

$18$

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(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,where $T$ is the period,we have: $\int_{\,\pi }^{\,10\pi } {|\sin x|dx} = \int_{\,0}^{\,9\pi } {|\sin x|dx}$ (by shifting the limits by $-\pi$). Since the period is $\pi$,we can write this as: $9 \int_{\,0}^{\,\pi } {|\sin x|dx} = 9 \int_{\,0}^{\,\pi } {\sin x dx}$ (since $\sin x \ge 0$ in $[0, \pi]$). Evaluating the integral: $9 [-\cos x]_{0}^{\pi} = 9 (-(\cos \pi) + \cos 0) = 9 (-(-1) + 1) = 9(1 + 1) = 18$.

$\int_{\,\pi }^{\,10\pi } {|\sin x|dx}$ is

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