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

The area of the region bounded by the parabola $y^2 = x$ and the straight line $2y = x$.

To find where the curves intersect, from the line $x = 2y$.

Substituting into the parabola equation:

$y^2 = 2y$

$y^2 - 2y = 0$

$y(y - 2) = 0$

So $y = 0$ or $y = 2$

When $y = 0$: $x = 2(0) = 0$ giving point $(0, 0)$

When $y = 2$: $x = 2(2) = 4$ giving point $(4, 2)$

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To determine which curve is on the right, check at $y = 1$:

Parabola: $x = y^2 = 1$

Line: $x = 2y = 2$

Since $2 > 1$, the line is to the right of the parabola between $y = 0$ and $y = 2$.

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Since both equations are in terms of $x$ as a function of $y$, integrate with respect to $y$:

$\text{Area} = \int_0^2 (2y - y^2) \, dy$

$= \left[y^2 - \frac{y^3}{3}\right]_0^2$

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

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

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

$= \frac{4}{3}$

The area of the region is $\frac{4}{3}$ square units.