Question

Find the area of the region included between: y² = 4x, and y = x

Answer 1

The vertex of the parabola y² = 4x is at the origin O = (0, 0).

Points of intersection of parabola and line are 

∴ x² = 4x

x² = 4x = 0

∴ x(x - 4) = 0

x = 0 or x = 4

∴ y = x

points are (0, 0) & (4, 4)

Area bounded by parabola and line is = Area (OABCO)

= A (OCBCO) - A (OCBAO)

`int_0^4 2sqrtx*dx - int_0^4x*dx`

= `2 [x^(3/2)/(3/2)]_0^4 - [x^2/2]_0^4`

= `2 xx 2/3 [(4)^(3/2) - (0)^(3/2)] - 1/2 [4^2 - 0^2]`

= `4/3[8 - 0]- 1/2 [16 - 0]`

= `32/3 - 8`

= `(32 - 24)/3`

Area = `8/3` square units.