Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020MathematicsArea Under CurvesPractice

The area bounded by the curve a 2 y = x 2 ( x + a ) and the x -axis is

Options

  1. Aa 2 3 s q . u n i t s
  2. Ba 2 4 s q . u n i t s
  3. C3 a 2 4 s q . u n i t s
  4. Da 2 12 s q . u n i t s

Correct answer

D. a 2 12 s q . u n i t s

Step-by-step solution

The curve is y = x 2 ( x + a ) a 2 which is a cubic polynomial. Since x 2 ( x + a ) a 2 = 0 has a repeated root x = 0 , it touches x -axis at ( 0 ,   0 ) and intersects at ( - a ,   0 ) . Hence, required area = ∫ - a 0 y   d x = ∫ - a 0 x 2 x + a a 2 d x = 1 a 2 x 4 4 + a x 3 3 - a 0 = a 2 12   s q .   u n i t s

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 NTA Abhyas JEE Main PYQs