MathematicsEasy210×since 2002Q3400Consider the integral I=∫010[x]e[x]ex−1dxI = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx}I=∫010ex−1[x]e[x]dx, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :A45 (e −-− 1)B45 (e + 1)C9 (e + 1)D9 (e −-− 1)Check answerSkip