1. **State the problem:** Find the area bounded by the curves $y = x^2$ and $y = x$. 2. **Identify the points of intersection:** Set $x^2 = x$ to find where the curves intersect. $$x^2 = x$$ $$x^2 - x = 0$$ $$x(x - 1) = 0$$ So, $x = 0$ or $x = 1$. 3. **Determine which curve is on top between $x=0$ and $x=1$:** For $x$ in $(0,1)$, $x > x^2$, so $y = x$ is above $y = x^2$. 4. **Set up the integral for the area between the curves:** $$\text{Area} = \int_0^1 (x - x^2) \, dx$$ 5. **Calculate the integral:** $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ 6. **Subtract the integrals:** $$\text{Area} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$ **Final answer:** The area bounded by the curves $y = x^2$ and $y = x$ is $\boxed{\frac{1}{6}}$.