Consider the integral
$I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx}$,
where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
Solution
$I = \int_0^{10} {[x]\,.\,{e^{[x] + 1 - x}}} dx$<br><br>$$ = \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx} $$<br><br>$= - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\}$<br><br>$= 45(e - 1)$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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