Area under the curveHard
Question
The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to
Options
A.

B.

C.

D.

Solution

Solving the equations we get the points of intersection (-2, 1) and (-2, -1)
The bounded region is shown as shaded region.
The required area = 2
(1 - 3y2) - (-2y2) = 2
(1 - y2) dy = 2
Create a free account to view solution
View Solution FreeMore Area under the curve Questions
The area of the closed figure bounded by y = cos x, y = 1 + (2/π) x and x = π/2 is-...The area enclosed by the curves y = cos x, y = 1 + sin 2x and x = as x varies from 0 to , is -...Let ƒ : [0, ∞)→ℝ) be a continuous function such thatƒ(x) =1- 2x + ∫0xex-t ƒ(t)dt ...Area bounded by the curve y = xeα2 x-axis and the ordinates x = 0, x = α is-...Given f(x) = x4 + αx2 + βx + γ (where α, β, γ are real numbers) and roots of f(x) = 0 are ...