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

${\displaystyle \Rightarrow }$ y = 4, – 2

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

When y = – 2, x = 4 – 2 = 2

Required area = ${\displaystyle {\int }_{-2}^4(y+4)dy-{\int }_{-2}^4\frac{y^2}{2}dy}$

= ${\displaystyle {\left[\frac{{(y+4)}^2}{2}\right]}_{-2}^4-\frac{1}{2}{\left[\frac{y^3}{3}\right]}_{-2}^4}$ 

= ${\displaystyle \frac{1}{2}[64-4]-\frac{1}{6}[64+8]}$

= 30 – 12

= 18 sq. units.