If [x] is the greatest integer ≤ x, then π ^ 2 ∫ _ 0 ^ 2 (sin( )/(2)) (x - [x]) ^ [ x ] dx is equal to :

Correct answer: B. 4 ( π - 1 ) Let, I = π ^ 2 ∫ _ 0 ^ 2 (( )/(2)) (x - [x]) ^ [ x ] dx I = π ^ 2 ∫ _ 0 ^ 1 (( )/(2)) (x - [x]) ^ [ x ] dx + π ^ 2 ∫ _ 1 ^ 2 (( )/(2)) (x - [x]) ^ [ x ] dx I = π ^ 2 ∫ _ 0 ^ 1 (( )/(2)) x ^ 0 dx + π ^ 2 ∫ _ 1 ^ 2 (( )/(2)) (x - 1) ^ 1 dx I = π ^ 2 ∫ _ 0 ^ 1 (( )/(2))dx + π ^ 2 ∫ _ 1 ^ 2 (( )/(2))(x - 1)dx I = π ^ 2 ∫ _ 0 ^ 1 (( )/(2))dx + π ^ 2 [x ∫ _ 1 ^ 2 (( )/(2)) d x - ∫ _ 1 ^ 2 \ (d)/(dx)x ∫ _ 1 ^ 2 (( )/(2))dx\ dx] - π ^ 2 ∫ _ 1 ^ 2 (( )/(2))dx I = π ^ 2 [- (2)/(π) (( )/(2))] _ 0 ^ 1 + π ^ 2 [(x)(2)/(π)(- (( )/(2))) _ 1 ^ 2 - ∫ _ 1 ^ 2 (2)/(π)(- cos( )/(2))dx] + π ^ 2 × (2)/(π) [ ((π x)/(2))] _ 1 ^ 2 I = π ^ 2 [- (2)/(π) (( )/(2))] _ 0 ^ 1 + π ^ 2 [(x)(2)/(π)(- (( )/(2))) _ 1 ^ 2 + ((2)/(π)) ^ 2 [ ((π x)/(2))] _ 1 ^ 2 ] + π ^ 2 × (2)/(π) [ ((π x)/(2))] _ 1 ^ 2 I = - π ^ 2 [(2)/(π) ((π)/(2)) - (2)/(π) (0)] + π ^ 2 [2 × (2)/(π)(- cosπ) + 1 × (2)/(π) × (cos(π)/(2))]+ 4[ (π) - sin(π)/(2)]] + π ^ 2 × (2)/(π)[ (π) - cos(π)/(2)] I = 2π + 4π + 4 ( 0 - 1 ) - 2π I = 4π - 4 I = 4(π - 1)

JEE Main 202131 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2021 · 31 August · Evening Shift · Mathematics · Definite Integration

If [x] is the greatest integer ≤ x, then π ^ 2 ∫ _ 0 ^ 2 (sin(...

\left[x\right]

is the greatest integer

\le x,

then

{\pi}^{2}{∫}_{0}^{2}\left(sin\dfrac{\pix}{2}\right){\left(x - \left[x\right]\right)}^{[ x ]}dx

is equal to :

Options

A $2 ( \pi + 1 )$
B $4 ( \pi - 1 )$
C $2 ( \pi - 1 )$
D $4 ( \pi + 1 )$

Correct answer

B. $4 ( \pi - 1 )$

Step-by-step solution

Let, $I = {\pi}^{2}{∫}_{0}^{2}\sin \left(\dfrac{\pix}{2}\right){\left(x - \left[x\right]\right)}^{[ x ]}dx$

$I = {\pi}^{2}{∫}_{0}^{1}\sin \left(\dfrac{\pix}{2}\right){\left(x - \left[x\right]\right)}^{[ x ]}dx + {\pi}^{2}{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right){\left(x - \left[x\right]\right)}^{[ x ]}dx$

$I = {\pi}^{2}{∫}_{0}^{1}\sin \left(\dfrac{\pix}{2}\right){x}^{0}dx + {\pi}^{2}{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right){\left(x - 1\right)}^{1}dx$

$I = {\pi}^{2}{∫}_{0}^{1}\sin \left(\dfrac{\pix}{2}\right)dx + {\pi}^{2}{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right)\left(x - 1\right)dx$

$I = {\pi}^{2}{∫}_{0}^{1}\sin \left(\dfrac{\pix}{2}\right)dx + {\pi}^{2}\left[x{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right) d x - {∫}_{1}^{2}\left\{\dfrac{d}{dx}x{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right)dx\right\}dx\right] - {\pi}^{2}{∫}_{1}^{2}\sin \left(\dfrac{\pix}{2}\right)dx$

$I = {\pi}^{2}{\left[- \dfrac{2}{\pi}\cos \left(\dfrac{\pix}{2}\right)\right]}_{0}^{1} + {\pi}^{2}{\left[\left(x\right)\dfrac{2}{\pi}\right \left(- \cos \left(\dfrac{\pix}{2}\right)\right)}_{1}^{2}\left- {∫}_{1}^{2}\dfrac{2}{\pi}\left(- cos\dfrac{\pix}{2}\right)dx\right] + {\pi}^{2} \times \dfrac{2}{\pi}{\left[\cos \left(\dfrac{\pi x}{2}\right)\right]}_{1}^{2}$

$I = {\pi}^{2}{\left[- \dfrac{2}{\pi}\cos \left(\dfrac{\pix}{2}\right)\right]}_{0}^{1} + {\pi}^{2}{\left[\left(x\right)\dfrac{2}{\pi}\right \left(- \cos \left(\dfrac{\pix}{2}\right)\right)}_{1}^{2}\left+ {\left(\dfrac{2}{\pi}\right)}^{2}{\left[\sin \left(\dfrac{\pi x}{2}\right)\right]}_{1}^{2}\right] + {\pi}^{2} \times \dfrac{2}{\pi}{\left[\cos \left(\dfrac{\pi x}{2}\right)\right]}_{1}^{2}$

$I = - {\pi}^{2}\left[\dfrac{2}{\pi}\cos \left(\dfrac{\pi}{2}\right) - \dfrac{2}{\pi}\cos \left(0\right)\right] + {\pi}^{2}\left[2 \times \dfrac{2}{\pi}\left(- cos\pi\right) + 1 \times \dfrac{2}{\pi} \times \left(cos\dfrac{\pi}{2}\right)\right]\left+ 4\left[\sin \left(\pi\right) - sin\dfrac{\pi}{2}\right]\right] + {\pi}^{2} \times \dfrac{2}{\pi}\left[\cos \left(\pi\right) - cos\dfrac{\pi}{2}\right]$

$I = 2\pi + 4\pi + 4 ( 0 - 1 ) - 2\pi$

$I = 4\pi - 4$

$I = 4\left(\pi - 1\right)$

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