Question

Find the area of the region included between: y = x² and the line y = 4x

Answer 1

The vertex of the parabola y = x² is at the origin O(0, 0)
To find the points of the intersection of the line and the parabola.

Equating the values of y from the two equations, we get
x² = 4x
∴ x² – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of inteersection are O(0, 0) and B(4, 16)
Required area = area of the region OABCO
= (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO
= area under the line y = 4x between x = 0 and x = 4

= `int_0^4y*dx,  "where"  y = 4x`

= `int_0^4 4x*dx`

= `4int_0^4x*dx`

= `4[x^2/2]_0^4`

= 2(16 – 0)
= 32
Area of the region ODBAO
= area under the parabola y = x² between x = 0 and x = 4

= `int_0^4y*dx,  "where"  y = x^2`

= `int_0^4 x^2*dx`

= `[x^3/3]_0^4`

= `(1)/(3)(64 - 0)`

= `(64)/(3)`

∴ required area

= `32 - (64)/(3)`

= `(32)/(3)"sq units"`.