JEE Main 2026 — Mathematics Calculus

The given integral can be split at integer values of $x$x because of the greatest integer function $[x]$[x].

$$\displaystyle I = \int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$$I=∫03​([x]!ex+e−x​)dx

$$\displaystyle I = \int_0^1 \dfrac{e^x + e^{-x}}{0!} dx + \int_1^2 \dfrac{e^x + e^{-x}}{1!} dx + \int_2^3 \dfrac{e^x + e^{-x}}{2!} dx$$I=∫01​0!ex+e−x​dx+∫12​1!ex+e−x​dx+∫23​2!ex+e−x​dx

Since $0! = 1$0!=1, $1! = 1$1!=1, and $2! = 2$2!=2, we get:

$$\displaystyle I = \int_0^2 (e^x + e^{-x}) dx + \dfrac{1}{2} \int_2^3 (e^x + e^{-x}) dx$$I=∫02​(ex+e−x)dx+21​∫23​(ex+e−x)dx

Integrating the terms:

$$\displaystyle I = \left[ e^x - e^{-x} \right]_0^2 + \dfrac{1}{2} \left[ e^x - e^{-x} \right]_2^3$$I=[ex−e−x]02​+21​[ex−e−x]23​

$$\displaystyle I = (e^2 - e^{-2} - (1 - 1)) + \dfrac{1}{2} (e^3 - e^{-3} - (e^2 - e^{-2}))$$I=(e2−e−2−(1−1))+21​(e3−e−3−(e2−e−2))

$$\displaystyle I = e^2 - e^{-2} + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-3} - \dfrac{1}{2} e^2 + \dfrac{1}{2} e^{-2}$$I=e2−e−2+21​e3−21​e−3−21​e2+21​e−2

$$\displaystyle I = \dfrac{1}{2} e^2 + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-2} - \dfrac{1}{2} e^{-3}$$I=21​e2+21​e3−21​e−2−21​e−3

$$\displaystyle I = \dfrac{1}{2} \left( e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3} \right)$$I=21​(e2+e3−e21​−e31​)

Answer: $\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)$21​(e2+e3−e21​−e31​)