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# Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$.

$\begin{array}{1 1} 8\sqrt 3sq.units. \\\sqrt 3sq.units. \\ 4\sqrt 3sq.units. \\ 16\sqrt 3sq.units.\end{array}$

• 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.
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.