Q35 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

Consider the integral $I = \int_{0}^{10} \frac{[x] e^{[x]}}{e^{x - 1}} d x$ where $[x]$ denotes the greatest integer less than or equal to $x$ . Then the value of $I$ is equal to :

Solution

$I = \int_{0}^{10} [x] \cdot e^{[x] - x + 1}$

$I = \int_{0}^{1} 0 d x + \int_{1}^{2} 1 \cdot e^{2 - x} + \int_{2}^{3} 2 \cdot e^{3 - x} + \dots . . + \int_{9}^{10} 9 \cdot e^{10 - x} d x$

$\Rightarrow I = \sum_{n = 0}^{9} \int_{n}^{n + 1} n \cdot e^{n + 1 - x} d x$

$= - \sum_{n = 0}^{9} n {(e^{n + 1 - x})}_{n}^{n + 1}$

$= - \sum_{n = 0}^{9} n \cdot (e^{0} - e^{1})$

$= (e - 1) \sum_{n = 0}^{9} n$

$= (e - 1) \cdot \frac{9 \cdot 10}{2}$

$= 45 (e - 1)$

Hence, the correct answer is option: 'C'.