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

Let the area of the region A = (x,y) : 2^x y 2 + x - x^2, x 0 be p q - ₂(e) , where p and q are coprime natural numbers. Then the value of p + q is equal to

Options

  1. A17
  2. B21
  3. C19
  4. D23

Correct answer

C. 19

Step-by-step solution

The bounding curves are y = 2^x and y = 2 + x - x^2 . To find their intersection in the domain x 0 , we equate them: 2^x = 2 + x - x^2 By observation, x = 1 satisfies the equation since 2^1 = 2 + 1 - 1^2 = 2 . In the interval (0, 1) , the parabola y = 2 + x - x^2 lies above the exponential curve y = 2^x . The required area A is given by the definite integral: A = ₀¹ (2 + x - x^2 - 2^x) dx Integrating the terms: A = [ 2x + x^2 2 - x^3 3 - 2^x 2 ]₀¹ Evaluating at the upper bound x = 1 : ( 2 + 1 2 - 1 3 - 2 2 ) = 13 6

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