Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20211 Sep 2021Evening ShiftMathematicsArea Under CurvesActual

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

Options

  1. A2 2 ( 2 + 1 )
  2. B2 2 ( 2 - 1 )
  3. C4 ( 2 - 1 )
  4. D2 ( 2 + 1 )

Correct answer

B. 2 2 ( 2 - 1 )

Step-by-step solution

A = ∫ 0 π / 2 ( ( cos x + sin x ) - | cos x - sin x | ) d x A = ∫ 0 π / 4 ( ( cos x + sin x ) - ( cos x - sin x ) ) d x + ∫ π / 4 π / 2 ( ( cos x + sin x ) - ( sin x - cos x ) ) d x A = 2 ∫ 0 π / 4 sin x d x + 2 ∫ π / 4 π / 2 cos x d x = 2 - 1 2 + 1 + 2 1 - 1 2 = 4 - 2 2 = 2 2 ( 2 - 1 )

Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves

Passage: Consider the curve C₁ given by y = e^ -x for x [0, 10 ] , and the curve C₂ given by y = e^ -x ( x + x) for x [0, 10 ] . Let n be the total number of points of intersection 2026Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1 . Question: If is the area of the common region that lies inside both the given ellipses, then the value o 2026The area of the region (x, y) : x^2 - 8x y -x is : 2026The area of the region (x, y) : 0 y 6 - x, y^2 4x - 3, x 0 is: 2026The area of the region R = (x, y): xy 27, 1 y x^2 is equal to: 2026The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to: 2026The area of the region (x, y): y - |x|, y |x x|, y 0 is: 2026If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + 6 _e(3)) is equal to _______. 2026 Full Area Under Curves list All JEE Main PYQs