Question

The area bounded by the curves x + 2y² = 0 and x + 3y² = 1 is ______.

Options

• 1 sq. unit

• `1/3` sq. units

• `2/3` sq. units

• `4/3` sq. units

Answer 1

The area bounded by the curves x + 2y² = 0 and x + 3y² = 1 is `underlinebb(4/3 sq. units)`.

Explanation:

We have, x + 2y² = 0

`\implies` y² = `-x/2`  ...(i), a parabola with vertex (0, 0) and x + 3y² = 1

`\implies` y² = `(1 - x)/3 = -((x - 1)/3)` ...(ii), a parabola with vertex (1, 0)

Solving (i) and (ii), we get y = ± 1

A = `int_-1^1 ((1 - 3y^2) - (-2y^2))dy`

= `2int_0^1(1 - y^2)dy`

= `2[y - y^3/3]_0^1`

= `4/3` sq.units