Find the area bounded by the curve x² = 4y and the line x = 4y – 2.  [CBSE 2010, 2013, 2014(C)]

It is given that the area of the region bounded by the parabola x² = 4y and x = 4y - 2.

Let A and B be the points of intersection of the line and parabola.
Let us calculate the point of intersection of both the curves,
Given: x² = 4y and x = 4y - 2
Therefore, putting the value of x, equation of parabola, we get,
(4y - 2)² = 4y
16y² + 4 - 16y = 4y
16y² - 20y + 4 = 0'
4y² - 5y + 1 = 0
4y² - 4y - y + 1 = 0
4y(y - 1) - 1(y - 1) = 0
y = 1 or y = 1/4 
Corresponding values of x are, x = 2 or x = -1
Therefore,

Coordinates of point A are

Coordinates of point B are (2, 1).

Now, draw AL and BM perpendicular to x axis.

We can see that,

Area OBCA = Area of line - Area of Parabola …(1)

Hence, the area bounded by the x² = 4y and the line x = 4y – 2 is  square units.