Question

Find the area of the following region using integration ((x, y) : y² ≤ 2x and y ≥ x – 4).

Answer 1

y² ≤ 2x

y ≥ x – 4

y² = 2x  ...(i)

y – x + 4 = 0 or x – y = 4  ...(ii)

Put the value of x from (ii) in (i), we have

y² = 2(y + 4)

y² – 2y – 8 = 0

`\implies` y = 4, – 2

When y = 4, x = 4 + 4 = 8

When y = – 2, x = 4 – 2 = 2

Required area = `int_-2^4 (y + 4) dy - int_-2^4 y^2/2 dy`

= `[(y + 4)^2/2]_-2^4 - 1/2[y^3/3]_-2^4` 

= `1/2 [64 - 4] - 1/6[64 + 8]`

= 30 – 12

= 18 sq. units.