The area of the region bounded by parabola y² = x and the straight line 2y = x is

The correct option is 4/3 sq. unit.

To find the area of the region bounded by the parabola $y^2=x$ and the straight line $2y=x$, we follow a step-by-step mathematical approach.

Step 1: Find the points of intersection of the two curves.
We have the equations:
1) $y^2=x$
2) $x=2y$

Substitute the value of $x$ from equation (2) into equation (1):
$y^2=2y$
$y^2-2y=0$
$y(y-2)=0$

This gives two values for $y$:
$y=0$ and $y=2$

Corresponding values of $x$ are:
When $y=0$, $x=2\left(0\right)=0$
When $y=2$, $x=2\left(2\right)=4$

Thus, the curves intersect at the points $(0,0)$ and $(4,2)$.

Step 2: Set up the integral for the area.
We integrate with respect to $y$ from $y=0$ to $y=2$. In this interval, the straight line $x=2y$ lies to the right of the parabola $x=y^2$ (i.e., $2y>y^2$ for $0<y<2$).

The area $A$ is given by the formula:
$$A={\int }_0^2\left(x_{\text{line}}-x_{\text{parabola}}\right)dy$$

Substituting the expressions for $x$:
$$A={\int }_0^2\left(2y-y^2\right)dy$$

Step 3: Evaluate the definite integral.
Find the antiderivative:
$$\int \left(2y-y^2\right)dy=y^2-\frac{y^3}{3}$$

Now, apply the integration limits from 0 to 2:
$$A={\left[y^2-\frac{y^3}{3}\right]}_0^2$$

$$A=\left(2^2-\frac{2^3}{3}\right)-\left(0^2-\frac{0^3}{3}\right)$$

$$A=\left(4-\frac{8}{3}\right)-0$$

$$A=\frac{12-8}{3}=\frac{4}{3}$$

Thus, the area of the bounded region is 4/3 sq. unit.