Q37 / 90 · JEE Main 2021 (26 Feb Shift 1) · Mathematics

The value of $\sum_{n = 1}^{100} \int_{n - 1}^{n} e^{x - [x]} d x$ , where $[x]$ is the greatest integer $\leq x,$ is:

Correct answer
A
$100 (e - 1)$
B
$100 e$
C
$100 (1 - e)$
D
$100 (1 + e)$

Solution

\sum_{n = 1}^{100} \int_{n - 1}^{n} e^{\{x\}} d x,

period of

\{x\} = 1

$\sum_{n = 1}^{100} \int_{0}^{1} e^{\{x\}} d x = \sum_{n = 1}^{100} \int_{0}^{1} e^{x} d x$

$\sum_{n = 1}^{100} (e - 1) = 100 (e - 1)$

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

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