JEE Main 2025 — Definite Integration Question with Solution

To solve this, we start by evaluating the integral:

$I={\int }_0^{e^3}\left[\frac{1}{e^{x-1}}\right]dx$

The greatest integer function $[\cdot ]$ returns the largest integer less than or equal to the input value. Here's how we can approach the problem:

Determine the function inside the integral:

$\frac{1}{e^{x-1}}=e^{1-x}$.

Identifying the intervals:

When $e^{1-x}\ge 2$, which simplifies to $x\le 1-\ln 2$, we have $\left[e^{1-x}\right]=2$.

When $1\le e^{1-x}<2$, simplifying gives $1-\ln 2<x\le 1$, and thus $\left[e^{1-x}\right]=1$.

When $0\le e^{1-x}<1$, which holds for $x>1$, thus $\left[e^{1-x}\right]=0$ from $x=1$ to $x=e^3$.

Evaluate the integral on these intervals:

${\int }_0^{1-\ln 2}2dx=2(1-\ln 2)$

${\int }_{1-\ln 2}^{1}1dx=1-(1-\ln 2)=\ln 2$

${\int }_1^{e^3}0dx=0$

Combine these results:

$I=2(1-\ln 2)+\ln 2+0=2-\ln 2$

Thus, we are given that:

$\alpha -\ln 2=2-\ln 2$

This implies that:

$\alpha =2$

Therefore, ${\alpha }^3=2^3=8$.