Definite Integration: Let [cdot] denote the greatest integer function. Then the value of… | JEE Main 2026

JEE Main 2026 (Online) 2nd April Morning ShiftQuestion 1

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

Definite Integration

$e^{2}+e^{3}-\frac{1}{e^{2}}-\frac{1}{e^{3}}$$\frac12\left(e^{2}+e^{3}-\frac{1}{e^{2}}-\frac{1}{e^{3}}\right)$✓ Correct answer$e^{2}+e^{3}-\frac{1}{2e^{2}}-\frac{1}{2e^{3}}$$\frac12\left(e^{2}+e^{3}\right)-\frac{1}{e^{2}}-\frac{1}{e^{3}}$

Solution

Concept. $[x]!$ is constant on each $[k,k+1)$: it is $0!=1$, $1!=1$, $2!=2$. Formulas used. (1) $\int(e^x+e^{-x})\,dx=e^x-e^{-x}$Step 1 - Split. $$I=\int_0^2\left(e^x+e^{-x}\right)dx+\frac12\int_2^3\left(e^x+e^{-x}\right)dx.$$ Step 2 - Use (1). $$I=\left(e^2-e^{-2}\right)+\frac12\left(e^3-e^{-3}-e^2+e^{-2}\right).$$ Step 3 - Simplify. $$I=\frac12\left(e^2+e^3-e^{-2}-e^{-3}\right)=\frac12\left(e^2+e^3-\frac1{e^2}-\frac1{e^3}\right).$$ Answer: (B)

Definite Integration - JEE Mathematics question (JEE Main 2026 (Online) 2nd April Morning Shift)

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