JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Definite Integrals

Both integrand pieces have period $1$, so the integral over $[-3,101]$, a span of $104$ unit intervals, is $104$ times the per-period behaviour, treated piece by piece. For $[\sin (\pi x)]$, on a unit interval where $\sin (\pi x)$ is positive the floor is $0$ and where it is negative the floor is $-1$; the sine is negative on exactly half of the $104$ intervals, namely the $52$ of the form $(\text{odd},\text{odd}+1)$, so ${\int }_{-3}^{101}[\sin (\pi x)]dx=-52$. For $e^{[\cos (2\pi x)]}$, within each period $\cos (2\pi x)$ is non-negative on half the period, where the floor is $0$ and the integrand is $1$, and negative on the other half, where the floor is $-1$ and the integrand is $\frac{1}{e}$; one period thus contributes $\frac{1}{2}\left(1+\frac{1}{e}\right)$, and $104$ periods give $52\left(1+\frac{1}{e}\right)=52+\frac{52}{e}$. Adding the two parts, the $-52$ cancels the $52$ and leaves $\frac{52}{e}$.