JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $[x]$ is the greatest integer $\le x$, then $\pi^{2}\int_{0}^{2}\left(\sin\frac{\pi x}{2}\right)(x-[x])^{[x]}dx$ is equal to :
Definite Integration
Solution
Concept. $(x-[x])^{[x]}$ is $1$ on $[0,1)$ and $x-1$ on $[1,2)$. Integrate each piece; the second needs integration by parts. Formulas used. (1) $[x]=0$ on $[0,1)$, $[x]=1$ on $[1,2)$(2) $\sin\frac{\pi(t+1)}2=\cos\frac{\pi t}2$; $\int_0^1t\cos\frac{\pi t}2dt=\frac2\pi-\frac4{\pi^2}$ (by parts)Step 1 - The integrand. By (1):Step 2 - First piece. $$\int_0^1\sin\frac{\pi x}2dx=\frac2\pi\Big[-\cos\frac{\pi x}2\Big]_0^1=\frac2\pi.$$ Step 3 - Second piece. Put $t=x-1$. By (2), $$\int_1^2(x-1)\sin\frac{\pi x}2dx=\int_0^1t\cos\frac{\pi t}2dt=\frac2\pi\Big[t\sin\frac{\pi t}2\Big]_0^1-\frac2\pi\int_0^1\sin\frac{\pi t}2dt=\frac2\pi-\frac4{\pi^2}.$$ Step 4 - Total. $\frac4\pi-\frac4{\pi^2}$; times $\pi^2$: $4\pi-4=4(\pi-1)$. Answer: (D)