If the integral int_{0}^{10} frac{[sin 2 pi x | vedclass

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(A) Let $I = \int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{ x -[ x ]}} dx = \int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{\{ x \}}} dx$.Since the function $f(

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(A) Let $I = \int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{ x -[ x ]}} dx = \int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{\{ x \}}} dx$.Since the function $f(x) = \frac{[\sin 2 \pi x ]}{ e ^{\{ x \}}}$ is periodic with period $1$,we have $I = 10 \int_{0}^{1} \frac{[\sin 2 \pi x ]}{ e ^{ x }} dx$.We split the integral into two parts: $I = 10 \left( \int_{0}^{1/2} \frac{[\sin 2 \pi x ]}{ e ^{ x }} dx + \int_{1/2}^{1} \frac{[\sin 2 \pi x ]}{ e ^{ x }} dx \right)$.For $0 \le x For $1/2 \le x Thus,$I = 10 \left( 0 + \int_{1/2}^{1} \frac{-1}{ e ^{ x }} dx \right) = -10 \int_{1/2}^{1} e ^{-x} dx$.$I = -10 \left[ -e ^{-x} \right]_{1/2}^{1} = 10 (e ^{-1} - e ^{-1/2}) = 10 e ^{-1} - 10 e ^{-1/2} + 0$.Comparing this with $\alpha e ^{-1} + \beta e ^{-1/2} + \gamma$,we get $\alpha = 10, \beta = -10, \gamma = 0$.Therefore,$\alpha + \beta + \gamma = 10 - 10 + 0 = 0$.

If the integral $\int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{ x -[ x ]}} dx =\alpha e ^{-1}+\beta e ^{-\frac{1}{2}}+\gamma$,where $\alpha, \beta, \gamma$ are integers and $[ x ]$ denotes the greatest integer less than or equal to $x$,then the value of $\alpha+\beta+\gamma$ is equal to ........ .

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