Evaluate int_{-frac{pi}{2}}^{frac{pi}{2}} f(x | vedclass

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(C) Given the integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) dx$. Since $f(x) = \sin |x| + \cos |x|$ is an even function be

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$4$

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(C) Given the integral $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) dx$. Since $f(x) = \sin |x| + \cos |x|$ is an even function because $f(-x) = \sin |-x| + \cos |-x| = \sin |x| + \cos |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$. Thus,$I = 2 \int_{0}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) dx$. For $x \in [0, \frac{\pi}{2}]$,$|x| = x$,so the integral becomes $I = 2 \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) dx$. Evaluating the integral: $I = 2 [-\cos x + \sin x]_{0}^{\frac{\pi}{2}}$. Substituting the limits: $I = 2 [(-\cos \frac{\pi}{2} + \sin \frac{\pi}{2}) - (-\cos 0 + \sin 0)]$. $I = 2 [(0 + 1) - (-1 + 0)] = 2 [1 + 1] = 2(2) = 4$.

Evaluate $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) dx$,where $f(x) = \sin |x| + \cos |x|$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

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