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

The area (in sq. units) bounded between the curves y = e x cos ⁡ x and y = e x sin ⁡ x from x = π 4 to x = π 2 is

Options

  1. Ae π 2
  2. Be π 4 2
  3. Ce λ 2
  4. De λ 2

Correct answer

B. e π 4 2

Step-by-step solution

As sin ⁡ x > cos ⁡ x ,   ∀ x ∈ π 4 , π 2 ⇒ e x sin ⁡ x > e x cos ⁡ x Thus the required area is A = ∫ π 4 π 2 e x sin ⁡ x - e x cos ⁡ x d x = - ∫ π 4 π 2 e x cos ⁡ x + - sin ⁡ x d x = - e x cos ⁡ x π 4 π 2 = − 0 − e π 4 ⋅ 1 2 = e π 4 2 As ∫ e x f x + f ′ x dx = e x ⋅ f x + C

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