\[ \begin{array}{1} (A) \;\frac{2}{3} \qquad & (B)\; \frac{1}{3} \qquad &(C)\; \frac{1}{4} \qquad & (D)\; \frac{3}{4} \end{array} \]

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- If we are given two curves represented by y=f(x) and 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 the common values of y from the equation of the two curves.

Let us find the point of intersection by solving the two equations.

$y^2=4x$ and $y=2x$.

$4x^2=4x \Rightarrow x=0,1$

if x=0,y=0 and if x=1,y=2.

Let the area bounded by the parabola and the x-axis be $y_1$.

Area of $y_1=\int_0^1 2\sqrt x dx$.

on integrating we get,

$\qquad\qquad=\begin{bmatrix}2\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_0^1$

$\qquad\qquad=\frac{4}{3}\begin{bmatrix}x^\frac{3}{2}\end{bmatrix}_0^1$

On applying limits we get

Area of $y_1=\frac{4}{3}[1]=\frac{4}{3}$ sq.units.

Let the area bounded by the parabola and the x-axis be $y_2$.

Area of $y_2=\int_0^1 2x dx$.

On integrating we get,

Area of $y_2=\begin{bmatrix}\frac{2x^2}{2}\end{bmatrix}_0^1$.

On applying limits we get,

Area of $y_2$=1.

Now the required area is $y_2-y_1$.

$\qquad\qquad=\mid 1-\frac{4}{3}\mid=\frac{1}{3}$ sq.units.

Hence the correct answer is B.

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