1. **Problem Statement:** Find the area bounded by the curves $y = x$ and $y = x^2$. 2. **Formula and Rules:** The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ We find the points of intersection to determine $a$ and $b$. 3. **Find points of intersection:** Set $x = x^2$: $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ So, $x=0$ or $x=1$. 4. **Determine which function is on top:** For $x$ in $(0,1)$, $x > x^2$ since $x^2$ grows slower. So, $y = x$ is the upper curve and $y = x^2$ is the lower curve. 5. **Set up the integral:** $$\text{Area} = \int_0^1 (x - x^2) \, dx$$ 6. **Calculate the integral:** $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ 7. **Subtract to find area:** $$\text{Area} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$ **Final answer:** The area bounded by the curves $y = x$ and $y = x^2$ is $\boxed{\frac{1}{6}}$.