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

$\begin{array}{1 1} \large \frac{9}{8}sq.units. \\ \large \frac{7}{8}sq.units. \\ \large \frac{11}{8}sq.units. \\ \large \frac{5}{4}sq.units. \end{array}$

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• If we are given two curves represented by y=f(x),y=g(x),where $f(x)\geq g(x)$ in [a,b],the point of intersection of two curves are given by x=a and x=b by taking common values of y from the equation of the curves.
Clearly $x^2=4y$ 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,$\frac{1}{2})$ 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 -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 fig.
Clearly the curve moves from the point -1 to 0 and from 0 to 2.
The required area is given by
$A=\int_{-1}^0(y_2-y_1)dx+\int_0^2(y_2-y_1)dx.$
$\;\;\;=\begin{bmatrix}\int_{-1}^0\big(\frac{x+2}{4}\big)dx-\int_{-1}^0\big(\frac{x^2}{4}\big)dx\end{bmatrix}+\begin{bmatrix}\int_0^2\big(\frac{x+2}{4}\big)dx-\int_0^2\big(\frac{x^2}{4}\big)dx\end{bmatrix}$
On integrating we get,
$\;\;\;=\begin{bmatrix}\frac{1}{4}\big(\frac{x^2}{2}+2x\big)-\frac{1}{4}\big(\frac{x^3}{3}\big)\end{bmatrix}_{-1}^0+\begin{bmatrix}\frac{1}{4}\big(\frac{x^2}{2}+2x\big)-\frac{1}{4}\big(\frac{x^3}{3}\big)\end{bmatrix}_0^2$
on applying limits we get,
$\;\;\;=\begin{bmatrix}0-\frac{1}{4}\big(\frac{{-1}^2}{2}+2(-1)-\frac{{-1}^3}{3}\big)+\frac{1}{4}\big(\frac{2^2}{2}+2(2)-\frac{2^3}{3}-0\big)\end{bmatrix}$
$\;\;\;=\frac{-1}{4}[\frac{1}{2}-2+\frac{1}{3}]+\frac{1}{4}[\frac{4}{2}+4-\frac{8}{3}]$
$\;\;\;=\frac{-1}{4}[\frac{-3}{2}+\frac{1}{3}]+\frac{1}{4}[6-\frac{8}{3}]$
$\;\;\;=\frac{7}{24}+\frac{10}{12}=\frac{27}{24}=\frac{9}{8}$sq.units.
Hence the required area is $\frac{9}{8}$sq.units.