Using integration, find the area bounded by the curve x²=4y and the line x=4y-2

Given x²=4y is a parabola with vertex (0,0) and it opens upwards.
Let the interior of the parabola be the required region R1
The line x = 4y- 2 represents a straight line.
By putting x = 0 and y = 0 we obtain y=1/2 and x=-2 respectively. Hence the straight line meets the x-axis at (-2, 0) and at (0, 1/2) on the y-axis respectively.
Hence y=(x+2)/4
In order to find the points of intersection. Let us equate the equation of parabola and the equation of the straight line.
x+2/4 = x²/4
=> x² - x - 2 = 0
(x-2)(x+1) = 0
Hence x = 2 and x = -1
If x = 2, y = 1 and if x=-1, y=1/4
Hence the points of intersection are (2, 1) and (-1,1/4).
Thus the area of the required region is bounded between the straight line and the parabola. This is shown in the figure.
Clearly the curve moves from the point -1 to 0 and from 0 to 2.
The required area is given by,
A = ∫₋₁⁰(y₂-y₁)dx + ∫₀²(y₂-y₁)dx
= ∫₋₁⁰((x+2)/4)dx - ∫₋₁⁰(x²/4)dx + ∫₀²((x+2)/4)dx - ∫₀²(x²/4)dx
On integrating we get,
= [1/4(x²/2 + 2x) - x³/12]₋₁⁰ + [1/4(x²/2 + 2x) - x³/12]₀²
On applying limits we get,
= [0 - ((-1)²/8 + 2(-1) - (-1)³/12)] + [1/4(2²/2 + 2(2)) - 2³/12 - 0]
= [-(-1/8 -2 + 1/12)] + [1/4(2+4) - 8/12]
= [1/8 + 2 - 1/12] + [6/4 - 2/3]
= 25/24 + 5/6
= 25/24 + 20/24
= 45/24 = 15/8 Sq.units
Therefore, the required area is 15/8 sq.units.