Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202330 Jan 2023Evening ShiftMathematicsArea Under CurvesActual

Let A be the area of the region ( x , y ) : y ≥ x 2 , y ≥ ( 1 - x ) 2 , y ≤ 2 x ( 1 - x ) . Then 540 A is equal to

Correct answer

0

Step-by-step solution

Given, A be the area of the region ( x , y ) : y ≥ x 2 , y ≥ ( 1 - x ) 2 , y ≤ 2 x ( 1 - x ) Now solving y = x 2   &   y = 2 x 1 - x we get, x = 0 ,   2 3 And solving y = 1 - x 2   &   y = 2 x 1 - x we get, ⇒ 1 + x 2 - 2 x = 2 x - 2 x 2 ⇒ 3 x 2 - 4 x + 1 = 0 ⇒ x = 1 ,   1 3 Now on plotting the diagram of the above region we get, Now from the above diagram area of the shaded region will be, A = ∫ 1 3 2 3 2 x - 2 x 2 d x - ∫ 1 3 1 2

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