Let [t] denote the greatest integer less than or equal to t. Then the value of the integral $\int_{- 3}^{101} ([\sin (πx)] + e^{[\cos (2 πx)]}) dx$ is equal to

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$\frac{52 (1 - e)}{e}$

$\frac{52}{e}$

$\frac{52 (2 + e)}{e}$

$\frac{104}{e}$

answer is B.

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Detailed Solution

$\int_{- 3}^{101} ([\sin (πx)] + e^{[\cos (2 πx)]}) dx = \int_{- 3}^{101} [\sin (πx)] dx + \int_{- 3}^{101} e^{[\cos (2 πx)]} dx = 52 \int_{0}^{2} [\sin (πx)] dx + 104 \int_{0}^{1} e^{[\cos (2 πx)]} dx I_{1 =} \int_{0}^{2} [\sin (πx)] dx = \int_{0}^{1} [\sin (πx)] dx + \int_{1}^{2} [\sin (πx)] dx I_{1} = \int_{0}^{1} 0 dx + \int_{1}^{2} (- 1) dx = - 1 I_{2} = \int_{0}^{1} e^{[\cos (2 πx)]} dx = \int_{0}^{\frac{1}{4}} e^{[\cos (2 πx)]} dx + \int_{\frac{1}{4}}^{\frac{3}{4}} e^{[\cos (2 πx)]} dx + \int_{\frac{3}{4}}^{1} e^{[\cos (2 πx)]} dx I_{2} = \int_{0}^{\frac{1}{4}} e^{0} dx + \int_{\frac{1}{4}}^{\frac{3}{4}} e^{- 1} dx + \int_{\frac{3}{4}}^{1} e^{0} dx I_{2} = \frac{1}{4} + \frac{1}{e} (\frac{3}{4} - \frac{1}{4}) + (1 - \frac{3}{4}) = \frac{1}{2} + \frac{1}{e} (\frac{1}{2}) ∴ I = 52 (- 1) + \frac{104}{2} (1 + \frac{1}{e}) = \frac{52}{e}$