Find the area bounded by the curve `x^2=4y` and the straight line `x=4y-2.`

Explanation

The area bounded by the curve, `x^2=4y`, and line, `x=4y−2`, is represented by the shaded area `OBAO`.Let A and B be the points of intersection of the line and parabola.
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).
It can be observed that,
Area `OBAO`=Area `OBCO`+Area `OACO` ..(i)
Then, Area `OBCO`=Area `OMBC`−Area `OMBO`
`= int_0^2( (x+2)/4)dx -int_0^2 (x^2)/4 dx`
`= 1/4[x^2/(2) +2x]_0^2 - 1/4 [ x^3/3]_0^2`
`=1/4[2+4]-1/4[8/3]`
`=5/6`
Similarly, Area `OACO` = Area `OLAC` - Area `OLAO`
= `int_-1^0 (x+2)/(4) dx - int_-1^0 (x^2)/(4) dx`
`= 1/4[(x^2)/(2) +2x]_-1^0 - 1/4[x^3/3]_-1^0`
by solving the limits we get
`=1/(2) - 1/(8) -1/(12) = 7/24`
therefor required area
`= 5/(6) +7/(24) = 9/8` sq.units