JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 7)
Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x$$ is equal to
$$\frac{52(1-e)}{e}$$
$$\frac{52}{e}$$
$$\frac{52(2+e)}{e}$$
$$\frac{104}{e}$$
Explanation
$$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $$
$$[\sin \pi x]$$ is periodic with period 2 and $${{e^{[\cos (2\pi x)]}}}$$ is periodic with period 1.So,
$$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $$
$$ = 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}$$
$$ = {{52} \over e}$$
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