int_0^{2npi } {left( {|sin x| - left| {frac{1 | vedclass

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(B) Given integral is $I = \int_0^{2n\pi } {\left( {|\sin x| - \frac{1}{2}|\sin x|} \right)} \;dx$. Simplifying the integrand,we get $I = \int_0^{2n\p

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$2n$

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(B) Given integral is $I = \int_0^{2n\pi } {\left( {|\sin x| - \frac{1}{2}|\sin x|} ight)} \;dx$. Simplifying the integrand,we get $I = \int_0^{2n\pi } {\frac{1}{2}|\sin x|} \;dx$. Since $|\sin x|$ is a periodic function with period $\pi$,we use the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$. Here $T = \pi$,so $I = \frac{1}{2} 	imes n \int_0^{\pi} |\sin x| \;dx$. Since $\sin x \ge 0$ for $x \in [0, \pi]$,$|\sin x| = \sin x$. $I = \frac{n}{2} \int_0^{\pi} \sin x \;dx = \frac{n}{2} [-\cos x]_0^{\pi}$. $I = \frac{n}{2} (-(-1) - (-1)) = \frac{n}{2} (1 + 1) = \frac{n}{2} (2) = 2n$ is incorrect,let us re-evaluate: $I = \frac{n}{2} [-\cos x]_0^{\pi} = \frac{n}{2} (-(-1) - (-1)) = \frac{n}{2} (2) = n$. Wait,the integral $\int_0^{\pi} \sin x dx = 2$. So $I = \frac{n}{2} 	imes 2 = n$. Re-checking the provided solution: The step $2n \int_0^{\pi/2} \sin x dx = 2n(1) = 2n$ is used. Actually,$\int_0^{2n\pi} |\sin x| dx = 2n \int_0^{\pi} \sin x dx = 2n(2) = 4n$. Then $I = \frac{1}{2} (4n) = 2n$. The correct answer is $2n$.

$\int_0^{2n\pi } {\left( {|\sin x| - \left| {\frac{1}{2}\sin x} \right|} \right)} \;dx$ equals

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