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(B) Let $I = \int_{-3}^{101} ([\sin(\pi x)] + e^{[\cos(2\pi x)]}) dx$.Since $[\sin(\pi x)]$ has a period of $2$ and $e^{[\cos(2\pi x)]}$ has a period

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$\frac{52}{e}$

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(B) Let $I = \int_{-3}^{101} ([\sin(\pi x)] + e^{[\cos(2\pi x)]}) dx$.Since $[\sin(\pi x)]$ has a period of $2$ and $e^{[\cos(2\pi x)]}$ has a period of $1$,the integrand $f(x) = [\sin(\pi x)] + e^{[\cos(2\pi x)]}$ has a period of $2$.The interval length is $101 - (-3) = 104$. Since the period is $2$,the integral over the interval of length $104$ is $52$ times the integral over one period $[0, 2]$.$I = 52 \int_{0}^{2} ([\sin(\pi x)] + e^{[\cos(2\pi x)]}) dx$.For $[\sin(\pi x)]$: In $[0, 1]$,$\sin(\pi x) \in [0, 1]$,so $[\sin(\pi x)] = 0$ except at $x=0, 1$ where it is $0$. In $[1, 2]$,$\sin(\pi x) \in [-1, 0]$,so $[\sin(\pi x)] = -1$ except at $x=1, 2$ where it is $0$. Thus,$\int_{0}^{2} [\sin(\pi x)] dx = \int_{1}^{2} -1 dx = -1$.For $e^{[\cos(2\pi x)]}$: $\cos(2\pi x) \ge 0$ when $x \in [0, 1/4] \cup [3/4, 5/4] \cup [7/4, 2]$,where $[\cos(2\pi x)] = 0$,so $e^0 = 1$. When $\cos(2\pi x) $\int_{0}^{2} e^{[\cos(2\pi x)]} dx = (1/4 + 1/2 + 1/4) 	imes 1 + (1/2 + 1/2) 	imes (1/e) = 1 + 1/e$.$I = 52 	imes (-1 + 1 + 1/e) = 52/e$.

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

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