# Choose the correct answer in the area lying between the curves $$y^2 = 4x$$ and $$y = 2x$$ is

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

Toolbox:
• 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.