$\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.

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