Question:
If $[x]$ denotes the greatest integer less than or equal to $\mathrm{x}$, then the value of the integral $\int_{-\pi / 2}^{\pi / 2}[[x]-\sin x] d x$ is equal to :
Correct Option: 1
Solution:
$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}([x]+[-\sin x]) d x$..(1)
$I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}([-x]+[\sin x]) d x \ldots(2)$
(King property)
$2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\underbrace{[\mathrm{x}]+[-\mathrm{x}]}_{-1})+(\underbrace{[\sin \mathrm{x}]+[-\sin \mathrm{x}]}_{-1}) \mathrm{dx}$
$2 \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(-2) \mathrm{d} \mathrm{x}=-2(\pi)$
$I=-\pi$
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