Using integration, find the area bounded by the curve *x*^{2} = 4y and the line *x* = 4y − 2.

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#### Solution

The area bounded by the curve, *x*^{2} = 4*y*, and line, *x* = 4*y *− 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point A are (-1, 1/4).

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to *x*-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC − Area OMBO

`=int_0^2(x+2)/4dx-int_0^2x^2/4dx`

`=1/4[x^2/2+2x]_0^2-1/4[x^3/3]_0^2`

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

`=7/24`

Therefore, required area = `(5/6+7/24)=9/8 units`

Concept: Area of the Region Bounded by a Curve and a Line

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