Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainMathematicsArea Under Curves

The area (in sq. units) of the region R = (x, y) : |y| x 2 - y^2 is

Options

  1. A7 6
  2. B10 3
  3. C8 3
  4. D7 3

Correct answer

D. 7 3

Step-by-step solution

The given region is bounded by the curves x = |y| and x = 2 - y^2 . To find the points of intersection, we equate the two expressions for x : |y| = 2 - y^2 |y|^2 + |y| - 2 = 0 (|y| + 2)(|y| - 1) = 0 Since |y| 0 , we have |y| = 1 , which gives y = -1 and y = 1 . The area of the region is given by the integral with respect to y : Area = _ -1 ¹ (x_ right - x_ left ) dy Area = _ -1 ¹ (2 - y^2 - |y|) dy Since the integrand is an even function, we can simplify this to: Area = 2 ₀¹ (2 - y^2 - y) dy Area = 2 [ 2y - y^3 3 -

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