If the value of the integral \(\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} dx=\alpha e^{-1}+\beta\), where \(\alpha, \beta \in R\), \(5\alpha+6\beta=0\), and \([x]\) denotes the greatest integer less than or equal to \(x\); then the value of \((\alpha+\beta)^{2}\) is equal to

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Definite Integrals

If the value of the integral

\int_{0}^{5} \frac{x + [ x ]}{e ^{x - [x]}} d x = α e^{- 1} + β

, where

α, β \in R

5 α + 6 β = 0

, and

[x]

denotes the greatest integer less than or equal to

x

; then the value of

(α + β)^{2}

is equal to

Answer: (C)

Split the range at the integers. On

[n, n + 1)

with

n = 0, 1, 2, 3, 4

put

t = x - n

, so that

[x] = n

x - [x] = t

and

x + [x] = t + 2 n

. Then

\int_{0}^{5} \frac{x + [ x ]}{e ^{x - [x]}} d x = n = 0 \sum 4 \int_{0}^{1} (t + 2 n) e^{- t} d t = 5 \int_{0}^{1} t e^{- t} d t + 2 (0 + 1 + 2 + 3 + 4) \int_{0}^{1} e^{- t} d t

Integrating by parts,

\int_{0}^{1} t e^{- t} d t = [- (t + 1) e^{- t}]_{0}^{1} = 1 - 2 e^{- 1}

, while

\int_{0}^{1} e^{- t} d t = 1 - e^{- 1}

. Hence the integral equals

5 (1 - 2 e^{- 1}) + 20 (1 - e^{- 1}) = 25 - 30 e^{- 1}

α = - 30

and

β = 25

, a pair that satisfies

5 α + 6 β = 0

. Therefore

(α + β)^{2} = (- 5)^{2} = 25

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option