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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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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.
$\frac{x+2}{4}=\frac{x^2}{4}\Rightarrow x^2-x-2=0$.
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
On integrating we get,
on applying limits we get,
Hence the required area is $\frac{9}{8}$sq.units.
answered Dec 20, 2013 by yamini.v

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