JEE Main 2025 Apr 2 Shift 1, Mathematics Q21: Some Properties of Definite Integrals

Let $f(x)=e^{1-x}$. The integrand is $\lfloor f(x)\rfloor$. Track where $f(x)$ crosses the integer levels on $[0,e^3]$.

$f(0)=e^1\approx 2.718$, so $\lfloor f(0)\rfloor =2$.

$f(x)=2$ when $e^{1-x}=2$, i.e. $x=1-\ln 2\approx 0.307$.

$f(x)=1$ when $1-x=0$, i.e. $x=1$.

$f(x)<1$ for $x>1$, where $\lfloor f(x)\rfloor =0$ (since $f(x)>0$ throughout).

Splitting the integral:

$$I={\int }_0^{1-\ln 2}2dx+{\int }_{1-\ln 2}^{1}1dx+{\int }_1^{e^3}0dx$$

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

$$=2-2\ln 2+\ln 2=2-\ln 2$$

Comparing with $I=\alpha -\ln 2$, we get $\alpha =2$. Hence ${\alpha }^3=8$.