The area, enclosed by the curves y = x + x and y = | x - x | and the lines x = 0, x = (π)/(2), is :

Correct answer: B. 2 2 ( 2 - 1 ) A = ∫ _ 0 ^ π / 2 ( ( x + x ) - | x - x | ) dx A = ∫ _ 0 ^ π / 4 ( ( x + x ) - ( x - x ) ) dx + ∫ _ π / 4 ^ π / 2 ( ( x + x ) - ( x - x ) ) dx A = 2 ∫ _ 0 ^ π / 4 sinxdx + 2 ∫ _ π / 4 ^ π / 2 cosxdx = 2(- 1 2 + 1) + 2(1 - 1 2 ) = 4 - 2 2 = 2 2 ( 2 - 1 )

JEE Main 20211 SeptemberEvening Shift

Exam: JEE Main
Year: 2021
Date: 1 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2021 · 1 September · Evening Shift · Mathematics · Area Under Curves

The area, enclosed by the curves y = x + x and y = | x - x | and...

The area, enclosed by the curves

y = \sin x + \cos x

and

y = | \cos x - \sin x |

and the lines

x = 0, x = \dfrac{\pi}{2},

is :

Options

A $2\sqrt{2} ( \sqrt{2} + 1 )$
B $2\sqrt{2} ( \sqrt{2} - 1 )$
C $4 ( \sqrt{2} - 1 )$
D $2 ( \sqrt{2} + 1 )$

Correct answer

B. $2\sqrt{2} ( \sqrt{2} - 1 )$

Step-by-step solution

$A = {∫}_{0}^{\pi / 2} ( ( \cos x + \sin x ) - | \cos x - \sin x | ) dx$

$A = {∫}_{0}^{\pi / 4} ( ( \cos x + \sin x ) - ( \cos x - \sin x ) ) dx + {∫}_{\pi / 4}^{\pi / 2} ( ( \cos x + \sin x ) - ( \sin x - \cos x ) ) dx$

$A = 2{∫}_{0}^{\pi / 4}sinxdx + 2{∫}_{\pi / 4}^{\pi / 2}cosxdx$

$= 2\left(- \dfrac{1}{\sqrt{2}} + 1\right) + 2\left(1 - \dfrac{1}{\sqrt{2}}\right) = 4 - 2\sqrt{2} = 2\sqrt{2} ( \sqrt{2} - 1 )$

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