Question

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

Answer 1

Given $x^2=4y$ is a parabola with vertex (0,0) and it opens upwards.

Let the interior of the parabola be the required region $R_1$

The line x = 4y- 2 represents a straight line.

By putting x = 0 and y = 0 we obtain $y=\frac{1}{2}$ and $x=-2$ respectively. Hence the straight line meets the x-axis at (-2, 0) and at (0, 12) on the y-axis respectively.

Hence $y=\frac{x+2}{4}$

In order to find the points of intersection. Let us equate the equation of parabola and the equation of the straight line.

$\frac{x+2}{4}=\frac{x^2}{4}\Rightarrow x^2-x-2=0$

$(x-2)(x+1)=0$

Hence x = 2 and x = -1

If x = 2, y = -1 and if $x=-1,y=\frac{1}{4}$

Hence the points of intersection are (2, 1) and (-1, $\frac{1}{4}$).

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,

$A={\int }_{-1}^0(y_2-y_1)dx+{\int }_0^2(y_2-y_1)dx$

$={\int }_{-1}^0\left(\frac{x+2}{4}\right)dx-{\int }_{-1}^0\left(\frac{x^2}{4}\right)dx+{\int }_0^2\left(\frac{x+2}{4}\right)dx-{\int }_0^2\left(\frac{x^2}{4}\right)dx$

On integrating we get,

$={\left[\frac{1}{4}\right(\frac{x^2}{2}+2x)-\frac{1}{4}\frac{x^3}{3}]}_{-1}^0+{\left[\frac{1}{4}\right(\frac{x^2}{2}+2x)-\frac{1}{4}\frac{x^3}{3}]}_0^2$

On applying limits we get,

$=[0-\frac{1}{4}(\frac{(-1{)}^2}{2}+2(-1)-\left(\frac{-1^3}{3}\right))+\frac{1}{4}(\frac{2^2}{2}+2\left(2\right))-\frac{2^3}{3}-0]$

$=-\frac{1}{4}[\frac{-3}{2}+\frac{1}{3}]+\frac{1}{4}[6-\frac{8}{3}]$

$=\frac{7}{24}+\frac{10}{12}=\frac{27}{24}=\frac{9}{8}$ Sq.units

Therefore, the required area is $\frac{9}{8}$ sq.units.