1. **Problem Statement:** Find the area of the region bounded by the parabola $y = x^{2}$, the vertical line $x = 2$, and the horizontal line $y = 0$. 2. **Formula and Explanation:** The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral: $$\text{Area} = \int_{a}^{b} f(x) \, dx$$ Here, the curve is $y = x^{2}$, the vertical boundary is $x = 2$, and the horizontal boundary is $y = 0$ (the x-axis). Since the parabola opens upwards and is above the x-axis between $x=0$ and $x=2$, the limits of integration are from $0$ to $2$. 3. **Set up the integral:** $$\text{Area} = \int_{0}^{2} x^{2} \, dx$$ 4. **Calculate the integral:** $$\int x^{2} \, dx = \frac{x^{3}}{3} + C$$ 5. **Evaluate definite integral:** $$\text{Area} = \left[ \frac{x^{3}}{3} \right]_{0}^{2} = \frac{2^{3}}{3} - \frac{0^{3}}{3} = \frac{8}{3} - 0 = \frac{8}{3}$$ 6. **Interpretation:** The area of the region bounded by the parabola $y = x^{2}$, the vertical line $x = 2$, and the horizontal line $y = 0$ is $\frac{8}{3}$ square units.