Question

Find the area of the region included between y² = 9x and y = x

Answer 1

Given that, y² = 9x  .....(i)

And y = x  .....(ii)

Solving equation. (i) and (ii)

We have x² = 9x

⇒ x² – 9x = 0

x(x – 9) = 0

∴ x = 0, 9

Required area = `int_0^9 sqrt(9x)  "d"x - int_0^9 x  "d"x`

= `3int_0^9 sqrt(x)  "d"x - int_0^9 x  "d"x`

= `3 * 2/3 [x^(3/2)]_0^9 - 1/2 [x^2]_0^9`

= `2[(9)^(3/2) - 0] - 1/2 [(9)^2 - 0]`

= `2(27) - 1/2 (81)`

= `54 - 81/2`

= `(108 - 81)/2`

= `27/2` sq.units

Hence, the required area = `27/2` sq.units