Question

Find the area of the region included between y² = 2x and y = 2x.

Answer 1

The vertex of the parabola y² = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,

∴ y² = y

∴ y² – y = 0

∴ y(y – 1) = 0

∴ y = 0 or y = 1

When y = 0, x = `0/2` = 0

When y = 1, x = `(1)/(2)`

∴ the points of intersection are O(0, 0) and `"B"(1/2, 1)`

Required area = area of the region OABCO

= area of the region OABDO – area of the region OCBDO

Now, area of the region OABDO

= area under the parabola y² = 2x between x = 0 and x = `(1)/(2)`

= `int_0^(1//2)y * dx`,  where  `y = sqrt(2)x`

= `int_0^(1//2) sqrt(2) x dx`

= `sqrt(2)[x^(3/2)/(3//2)]_0^(1//2)`

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

= `sqrt(2)[2/3 * 1/(2sqrt(2))]`

= `(1)/(3)`

Area of the region OCBDO

= area under the line y = 2x between x = 0 and x = `(1)/(2)`

= `int_0^(1//2)y * dx`, where y = 2x

= `int_0^(1//2)2x * dx`

= `[(2x^2)/2]_0^(1//2)`

= `(1)/(4) - 0`

= `(1)/(4)`

∴ required area = `1/3 - 1/4`

= `1/12` sq unit.