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Q.

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 equal to :

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a

$45 (e - 1)$

b

$9 (e - 1)$

c

$45 (e + 1)$

d

$9 (e + 1)$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{10} \frac{[x] e^{[x]}}{e^{x - 1}} d x \\ = \sum_{n = 0}^{9} \int_{n}^{n + 1} \frac{n e^{n}}{e^{x - 1}} d x \\ = \sum_{n = 0}^{9} \int_{n}^{n + 1} n e^{n} e^{1 - x} d x \\ = - \sum_{n = 0}^{9} n e^{n} (e^{- n} - e^{1 - n}) \\ = - \sum_{n = 0}^{9} n (1 - e) = 45 (e - 1) \end{array}$

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