1. **State the problem:** We need to find the area of the region bounded by the curve $y = x^2$, the x-axis, and the vertical lines $x=1$ and $x=3$. 2. **Formula used:** The area under a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the definite integral: $$\text{Area} = \int_a^b f(x) \, dx$$ 3. **Apply the formula:** Here, $f(x) = x^2$, $a=1$, and $b=3$. So, $$\text{Area} = \int_1^3 x^2 \, dx$$ 4. **Calculate the integral:** The integral of $x^2$ is $$\int x^2 \, dx = \frac{x^3}{3} + C$$ 5. **Evaluate the definite integral:** $$\int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$$ 6. **Conclusion:** The area of the region bounded by the curve $y=x^2$, the x-axis, and the lines $x=1$ and $x=3$ is $$\boxed{\frac{26}{3}}$$ square units.