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

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Definite Integrals

Consider the integral

I = \int_{0}^{10} \frac{[ x ] e ^{[x]}}{e ^{x - 1}} dx

, where

[x]

denotes the greatest integer less than or equal to x. Then the value of I is equal to :

Answer: (D)

k \leq x < k + 1

the integrand equals

k e^{k} e^{1 - x}

, so

I = k = 0 \sum 9 k e^{k + 1} \int_{k}^{k + 1} e^{- x} d x = k = 0 \sum 9 k e^{k + 1} (e^{- k} - e^{- k - 1})

Each term reduces to

k e^{k + 1} e^{- k} (1 - e^{- 1}) = k (e - 1)

, so

I = (e - 1) k = 0 \sum 9 k = (e - 1) \cdot \frac{9 \cdot 10}{2} = 45 (e - 1)

The integral equals

45 (e - 1)

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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