To find the area of the region defined by $\{(x, y) : x^2 \le y \le x\}$, we need to calculate the definite integral of the difference between the upper curve ($y=x$) and the lower curve ($y=x^2$) between their points of intersection.

Finding Intersection Points

Set the equations equal to find the intersection points:

$x^2 = x$

$x^2 - x = 0$

$x(x - 1) = 0$

The points of intersection are at $x=0$ and $x=1$.

Setting Up the Integral

In the interval $[0, 1]$, the line $y=x$ is above the parabola $y=x^2$. We can verify this by testing a point, e.g., $x=0.5$: $y=0.5$ for the line and $y=(0.5)^2=0.25$ for the parabola. Since $0.5 > 0.25$, $y=x$ is the upper boundary.

The area $A$ is given by the integral:

$A = \int_{0}^{1} (\text{upper curve} - \text{lower curve}) \, dx$

$A = \int_{0}^{1} (x - x^2) \, dx$

Calculating the Area

Evaluate the definite integral:

$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}$

$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right)$

$A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0)$

$A = \frac{3 - 2}{6}$

$A = \frac{1}{6}$

The area of the region is $\frac{1}{6}$ square units.