If $[x]$ is the greatest integer $\leq x$, then $\pi^{2} \int_{0}^{2}\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x$ is equal to:

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Definite Integrals

[x]

is the greatest integer

\leq x

, then

π^{2} \int_{0}^{2} (sin \frac{π x}{2}) (x - [x])^{[x]} d x

is equal to:

Answer: (B)

[0, 1)

[x] = 0

and

(x - [x])^{[x]} = 1

; on

[1, 2)

[x] = 1

and

(x - [x])^{[x]} = x - 1

. The integral splits as

π^{2} [\int_{0}^{1} sin \frac{π x}{2} d x + \int_{1}^{2} (x - 1) sin \frac{π x}{2} d x]

. The first integral equals

\frac{2}{π}

; the second, by parts, equals

\frac{2}{π} - \frac{4}{π ^{2}}

. Adding,

π^{2} (\frac{4}{π} - \frac{4}{π ^{2}}) = 4 π - 4 = 4 (π - 1)

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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