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$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} f(x) d x=$
Where $f(x)=\sin |x|+\cos |x|, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
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Where $f(x)=\sin |x|+\cos |x|, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
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The correct answer is:
4
$\begin{aligned} & \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} f(x) d x=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x \\ & \quad=2 \int_0^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x \quad[\because \text { even function }] \\ & \quad \frac{\pi}{2} \\ & =2 \int_0(\sin x+\cos x) d x \quad[\because x>0] \\ & =2[-\cos x+\sin x]_0^{\frac{\pi}{2}}=4\end{aligned}$
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