JEE Main 2026 Apr 2 Shift 1, Mathematics Q19: Definite Integration And Fundamental Theorem

JEE Main2026Apr 2, Shift 1Mathematics

Let $[\cdot]$ 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
$\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}})$
C
$e^{2} + e^{3} - \frac{1}{2 e ^{2}} - \frac{1}{2 e ^{3}}$
D
$\frac{1}{2} (e^{2} + e^{3}) - \frac{1}{e ^{2}} - \frac{1}{e ^{3}}$

Solution

Split the integral by the floor function: on $[0, 1)$ and $[1, 2)$ the denominator $[x]!$ equals $0! = 1! = 1$ , while on $[2, 3)$ it equals $2! = 2$ .

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

$\int (e^{x} + e^{- x}) d x = e^{x} - e^{- x}$ . So:

$\int_{0}^{2} = (e^{2} - e^{- 2}) - 0 = e^{2} - e^{- 2}$

$\frac{1}{2} \int_{2}^{3} = \frac{1}{2} [(e^{3} - e^{- 3}) - (e^{2} - e^{- 2})]$

Adding: $I = \frac{1}{2} (e^{2} - e^{- 2}) + \frac{1}{2} (e^{3} - e^{- 3}) = \frac{1}{2} (e^{2} + e^{3} - \frac{1}{e ^{2}} - \frac{1}{e ^{3}})$ .

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JEE Main 2026 Apr 2 Shift 1, Mathematics Q19: Definite Integration And Fundamental Theorem

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