The value of I = int_{-pi / 2}^{pi / 2} |sin | vedclass

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(B) Given the integral $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$. Since $|\sin x|$ is an even function,we can use the property $\int_{-a}^{a} f(x

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$2$

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(B) Given the integral $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$. Since $|\sin x|$ is an even function,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$. Thus,$I = 2 \int_{0}^{\pi / 2} |\sin x| \, dx$. In the interval $[0, \pi / 2]$,$\sin x \geq 0$,so $|\sin x| = \sin x$. Therefore,$I = 2 \int_{0}^{\pi / 2} \sin x \, dx$. Evaluating the integral: $I = 2 [-\cos x]_{0}^{\pi / 2}$. $I = 2 [-\cos(\pi / 2) - (-\cos(0))]$. $I = 2 [0 - (-1)] = 2(1) = 2$.

The value of $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$ is

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