The area (in sq. units) of the region A = ( x, y ) : ( x - 1 ) [ x ] ≤ y ≤ 2 x , 0 ≤ x ≤ 2 , where [ t ] denotes the greatest integer function, is :

Correct answer: A. (8)/(3) 2 - (1)/(2) y = [ x ] ( x - 1 ) = \ cases 0 & 0 ≤ x Area = ∫ _ 0 ^ 2 2 x · dx - (1)/(2) ( 1 ) ( 1 ) = ( 4 x ^ 3 / 2 3 ) _ 0 ^ 2 - (1)/(2) = 8 2 3 - (1)/(2)

JEE Main 20205 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 5 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Area Under Curves

The area (in sq. units) of the region A = ( x, y ) : ( x - 1 ) [ x...

The area (in sq. units) of the region

A = { ( x, y ) : ( x - 1 ) [ x ] \le y \le 2\sqrt{x}, 0 \le x \le 2 } ,

where

[ t ]

denotes the greatest integer function, is : The area (in sq. units) of the region A = ( x, y ) : ( x - 1 ) [ x ] ≤ y ≤ 2 x , 0 ≤ x ≤ 2 , where [ t ] denotes the

Options

A $\dfrac{8}{3}\sqrt{2} - \dfrac{1}{2}$
B $\dfrac{4}{3}\sqrt{2} + 1$
C $\dfrac{8}{3}\sqrt{2} - 1$
D $\dfrac{4}{3}\sqrt{2} - \dfrac{1}{2}$

Correct answer

A. $\dfrac{8}{3}\sqrt{2} - \dfrac{1}{2}$

Step-by-step solution

y = [ x ] ( x - 1 ) $<br />$ = \left\{ $\begin{cases}0 & 0 \le x < 1 \\ x - 1 & 1 \le x < 2\end{cases}$ \right)

The area (in sq. units) of the region A = ( x, y ) : ( x - 1 ) [ x ] ≤ y ≤ 2 x , 0 ≤ x ≤ 2 , where [ t ] denotes the

Area $= {∫}_{0}^{2}2\sqrt{x} \cdot dx - \dfrac{1}{2} ( 1 ) ( 1 ) = {\left(\dfrac{4{x}^{3 / 2}}{3}\right)}_{0}^{2} - \dfrac{1}{2} = \dfrac{8\sqrt{2}}{3} - \dfrac{1}{2}$

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