To find the area between the line $y = x$ and the curve $y = x^2$, we need to calculate the definite integral of the difference between the two functions. First, let's determine the points of intersection between the line and the curve by setting $y = x$ equal to $y = x^2$.

$x = x^2$

Rearranging this equation:

$x^2 - x = 0$ $x(x - 1) = 0$

This gives $x = 0$ and $x = 1$ as the points of intersection.

Now, the area between the curves is given by the integral of the difference between the line $y = x$ and the parabola $y = x^2$ over the interval $[0, 1]$:

$\text{Area} = \int_0^1 (x - x^2) \, dx$

Let's compute the integral:

$\int_0^1 (x - x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx$

The first integral is:

$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$

The second integral is:

$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

Now subtract the two results:

$\text{Area} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$

Thus, the area contained between the line $y = x$ and the curve $y = x^2$ is $\frac{1}{6}$ square units.

Would you like further details on any step?

Here are 5 related questions to explore further:

1. What is the geometrical significance of the area between curves?
2. How would the area change if the functions were different?
3. Can the same process be applied to curves that intersect multiple times?
4. What if the region to find the area extends beyond the interval $[0,1]$?
5. How does this problem relate to real-world applications, like physics or economics?

Tip: Always sketch the graphs of the functions when finding the area between curves to better visualize the region.