Question

Find the area of the region bounded by the curve x² = 4y and the line x = 4y – 2.

Answer 1

Curve x² = 4y   ...(1)

Line x = 4y – 2   ...(2)

Solving (1) and (2)

x = x² – 2

`\implies` x² – x – 2 = 0

(x – 2)(x + 1) = 0

∴ x = 2, – 1

∴ y = `x^2/4 = 1, 1/4`

Points of intersection are `A(2, 1), B(-1, 1/4)`

The shaded region is required.

Let PQ to be the elementary strip.

So area of shaded region

= `int_-1^2 (y_"line" - y_"curve")dx`

= `int_-1^2 (x + 2)/4 - x^2/4dx`

= `1/4[x^2/2 + 2x - x^3/3]_-1^2`

= `1/4[(4/2 + 4 - 8/3)] - [(1/2 - 2 + 1/3)]`

= `1/4[6 - 8/3 - 1/2 + 2 - 1/3]`

= `1/4[6 - 9/3 - 1/2 + 2]`

= `1/4[6 - 3 - 1/2 + 2]`

= `1/4 xx 9/2`

= `9/8` sq.units