If the integral
$$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $$, 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 :
Solution
Given integral<br><br>$$\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } $$ (using property of definite in.)<br><br>$$ = 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]$$<br><br>= $$ - 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]$$<br><br>$= 10{e^{ - 1}} - 10{e^{ - 1/2}}$<br><br>comparing with the given relation,<br><br>$\alpha$ = 10, $\beta$ = $-$10, $\gamma$ = 0<br><br>$\alpha$ + $\beta$ + $\gamma$ = 0<br><br>Therefore, the correct answer is (A).
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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