Let [•] denote the greatest integer function. Then the value of $\int_{0}^{3}\left(\frac{e^{x}+e^{-x}}{[x]!}\right)dx$ is:

(B) $\frac{1}{2}\left(e^{2}+e^{3}-\frac{1}{e^{2}}-\frac{1}{e^{3}}\right)$

(A) $e^{2}+e^{3}-\frac{1}{e^{2}}-\frac{1}{e^{3}}$

(C) $e^{2}+e^{3}-\frac{1}{2e^{2}}-\frac{1}{2e^{3}}$

(D) $\frac{1}{2}\left(e^{2}+e^{3}\right)-\frac{1}{e^{2}}-\frac{1}{e^{3}}$

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Definite Integrals

Let [•] denote the greatest integer function. Then the value of

\int_{0}^{3} (\frac{e ^{x} + e ^{- x}}{[ x ]!}) d x

is:

(A) $e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}}$
(B) $\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}})$ ✓ Correct
(C) $e^{2} + e^{3} - \frac{1}{2 e ^{2}} - \frac{1}{2 e ^{3}}$
(D) $\frac{1}{2} (e^{2} + e^{3}) - \frac{1}{e ^{2}} - \frac{1}{e ^{3}}$

Answer: (B)

[0, 1)

[x] = 0

with

[x]! = 1

; on

[1, 2)

[x]! = 1

; on

[2, 3)

[x]! = 2

. Thus

I = \int_{0}^{1} (e^{x} + e^{- x}) d x + \int_{1}^{2} (e^{x} + e^{- x}) d x + \frac{1}{2} \int_{2}^{3} (e^{x} + e^{- x}) d x

. Since

\int (e^{x} + e^{- x}) d x = e^{x} - e^{- x}

, the first two integrals combine to

[e^{x} - e^{- x}]_{0}^{2} = e^{2} - e^{- 2}

, and the last equals

\frac{1}{2} [e^{x} - e^{- x}]_{2}^{3} = \frac{1}{2} (e^{3} - e^{- 3} - e^{2} + e^{- 2})

. Adding these, the value of the integral is

\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}})

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option