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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by the curve \(y^2 = 4x\) and the line \(x = 3.\)

$\begin{array}{1 1} 8\sqrt 3sq.units. \\\sqrt 3sq.units. \\ 4\sqrt 3sq.units. \\ 16\sqrt 3sq.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 two curves.
Clearly the equation $y^2=4x$ represents parabola with vertex (0,0) at the origin and axis of symmetry along the positive direction of x-axis as shown in the fig.
x= 3 is a line parallel to the y-axis.The region is the shaded portion as shown in the fig.
Since $y^2=4x is symmetrical about x-axis the required area is
A=2(area bounded by the parabola and the line and the x-axis).
Hence the required area is given by
$A=2\int_3^0y\;dx=2\int_3^02\sqrt x\;dx.$
$A=4\int_3^0x^\frac{1}{2}dx.$
on integrating we get,
$A=4\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_0^3$
$\;\;\;=4\times\frac{2}{3}[x^\frac{3}{2}]_0^3$
on applying limits we get,
$A=\frac{8}{3}\begin{bmatrix}(3)^\frac{3}{2}\end{bmatrix}$sq.units.
$\;\;\;=\frac{8}{3}\times 3\sqrt 3$
$\;\;\;=8\sqrt 3sq.units.$
Hence the required area is $8\sqrt 3$sq.units.
answered Dec 20, 2013 by yamini.v
 

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