Definite IntegrationmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Integral with Greatest Integer Factorial [x]!: Value | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let [⋅][\cdot] denote the greatest integer function. Then the value of ∫03ex+e−x[x]! dx\displaystyle\int_0^{3}\dfrac{e^x+e^{-x}}{[x]!}\,dx is
Ae2+e3−1e2−1e3e^2+e^3-\dfrac{1}{e^2}-\dfrac{1}{e^3}
B12(e2+e3−1e2−1e3)\dfrac12\left(e^2+e^3-\dfrac{1}{e^2}-\dfrac{1}{e^3}\right)correct
Ce2+e3−12e2−12e3e^2+e^3-\dfrac{1}{2e^2}-\dfrac{1}{2e^3}
D12(e2+e3)−1e2−1e3\dfrac12\left(e^2+e^3\right)-\dfrac{1}{e^2}-\dfrac{1}{e^3}
Solution
Step 1: Split by the value of [x][x]: on [0,1)[0,1), [x]=0[x]=0 so [x]!=1[x]!=1; on [1,2)[1,2), [x]=1[x]=1 so [x]!=1[x]!=1; on [2,3)[2,3), [x]=2[x]=2 so [x]!=2[x]!=2. Thus
I=∫01(ex+e−x) dx+∫12(ex+e−x) dx+12∫23(ex+e−x) dx.I=\int_0^1(e^x+e^{-x})\,dx+\int_1^2(e^x+e^{-x})\,dx+\frac12\int_2^3(e^x+e^{-x})\,dx.
 Step 2: Combine the first two:
I=∫02(ex+e−x) dx+12∫23(ex+e−x) dx.I=\int_0^2(e^x+e^{-x})\,dx+\frac12\int_2^3(e^x+e^{-x})\,dx.
 Step 3: Since ∫(ex+e−x) dx=ex−e−x\displaystyle\int(e^x+e^{-x})\,dx=e^x-e^{-x}:
∫02=(e2−e−2)−(1−1)=e2−e−2,\int_0^2=(e^2-e^{-2})-(1-1)=e^2-e^{-2},
12∫23=12[(e3−e−3)−(e2−e−2)].\frac12\int_2^3=\frac12\big[(e^3-e^{-3})-(e^2-e^{-2})\big].
 Step 4: Add:
I=(e2−e−2)+12(e3−e−3)−12(e2−e−2)=12(e2−e−2)+12(e3−e−3).I=(e^2-e^{-2})+\frac12(e^3-e^{-3})-\frac12(e^2-e^{-2})=\frac12(e^2-e^{-2})+\frac12(e^3-e^{-3}).
 Step 5: Therefore
I=12(e2+e3−1e2−1e3).I=\frac12\left(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\right).
 Correct answer: (2)
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