Question

Area lying between the curves `y^2 = 4x` and `y = 2x`

Options

• `2/3`

• `1/3`

• `1/4`

• `3/4`

Answer 1

`1/3`

Explanation:

The curve is `y^2 = 4x`  ......(1)

And the line is `y = 2x`  .......(2)

From (1) and (2)

`(4x^2) = 4x`

∴ `(x - 1)x = 0` or `x = 0, x = 1`

Curves (1) and (2) intersect at 0(0, 0), A (1, 2)

Area of region OACO

= Area of region OMACO – Area of OMA  ......(3)

Now Area of region OMACO

= Area of region bounded by curve OCA

`y = sqrt(4x)`, `x`-axis and `x` = 1

= `int_0^1 sqrt(4x)dx = 2 * 2/3 [x^(3/2)]_0^1`

= `4/3 xx 1 = 4/3` sq.units

Area of ΔOMA Area of region bounded by OA,

y = `2x, x = 1` and `x`-axis

= `int_0^1 2xdx = [x^2]_0^1` = 1

Putting the there value in (3)

∴ Area of region OMACO = `4/3 - 1 = 1/3` sq.units