The area of the region bounded by parabola $y^2=x$ and the straight line $2y=x$ is \begin{array}{1 1}(A)\;\frac{4}{3} sq.units & (B)\;1 sq.units\\(C)\;\frac{2}{3} sq.units & (D)\;\frac{1}{3}sq.units \end{array}

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$+c.
Step 1:
Given curve $y^2=x$ and the line $2y=x$.
Required area is the shaded portion shown in the fig.
Step 2:
Next let us find the point of intersection.
Clearly one point is (0,0)
$y^2=x$ and $4y^2=x^2\Rightarrow y^2=\large\frac{x^2}{4}$
$x=\large\frac{x^2}{4}$$\Rightarrow 4x=x^2 \Rightarrow x^2-4x=0 x(x-4)=0\Rightarrow x=0 or x=4 Hence Area of the shaded portion is A=\int_0^4(y_2-y_1)dx. Where y_2=\sqrt x and y_1=\large\frac{x}{2} A=\int_0^4(\sqrt x-\large\frac{x}{2}) Step 3: On integrating we get, A=\int_0^4\sqrt xdx-\large\frac{1}{2}$$\int_0^4 xdx$
$\;\;\;=\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_0^4-\large\frac{1}{2}\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_0^4$
$\;\;\;=\large\frac{2}{3}\begin{bmatrix}x^{\Large\frac{3}{2}}\end{bmatrix}_0^4-\large\frac{1}{4}[x^2]_0^4$
On applying limits we get,
$A=\large\frac{2}{3}$$[4^{\Large\frac{3}{2}}-0]-\large\frac{1}{4}$$[4^2-0]$
$\;\;\;=\large\frac{2}{3}$$[8]-\large\frac{1}{4}$$\times 16$
$\;\;\;=\large\frac{16}{3}$$-4=\large\frac{4}{3}$sq.units.
Hence A is the correct option.