Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
99 Percentile Qs Bank for JEE MainMathematicsArea Under Curves

The area (in square units) bounded by the curves y=2 x^2 and y= x-[x], x+|x| in between the lines x=0 and x=2 is

Options

  1. A4 3
  2. B1 2
  3. C1
  4. D2

Correct answer

D. 2

Step-by-step solution

aligned & y= x-[x], x+|x| & y= cases 2 x, & x 0 x , & x aligned & = | ₀^1 2 x d x- ₀^1 2 x^2 d x |+ | ₁^2 2 x^2 d x- ₁^2 2 x d x | & = | [ 2 x^2 2 ]₀^1- [ 2 x^3 3 ]₀^1 |+ | [ 2 x^3 3 ]₁^2- [ 2 x^2 2 ]₁^2 | & = |1- 2 3 |+ | 16 3 - 2 3 -4+1 |= 1 3 + 5 3 =2 sq. units. aligned

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 99 Percentile Qs Bank for JEE Main PYQs