1. **State the problem:** We want to approximate the area under the curve of the function $f(x) = x^3$ on the interval $[0, 2]$ using the midpoint Riemann sum with 100 subintervals. 2. **Formula and explanation:** The midpoint Riemann sum approximates the integral by dividing the interval into $n$ subintervals, calculating the function value at the midpoint of each subinterval, and summing the areas of rectangles formed by these values and the subinterval width $\Delta x$. The formula is: $$\text{Area} \approx \sum_{i=1}^n f\left(x_i^*\right) \Delta x$$ where $x_i^*$ is the midpoint of the $i$-th subinterval and $\Delta x = \frac{b - a}{n}$. 3. **Calculate $\Delta x$:** $$\Delta x = \frac{2 - 0}{100} = 0.02$$ 4. **Find midpoints:** Midpoints are: $$x_i^* = a + \left(i - \frac{1}{2}\right) \Delta x = 0 + \left(i - \frac{1}{2}\right) \times 0.02$$ for $i = 1, 2, ..., 100$. 5. **Evaluate function at midpoints:** $$f(x_i^*) = (x_i^*)^3$$ 6. **Sum areas of rectangles:** $$\text{Area} \approx \sum_{i=1}^{100} (x_i^*)^3 \times 0.02$$ 7. **Python code execution:** The provided Python code uses numpy to compute this sum efficiently: ```python import numpy as np def f(x): return x**3 a = 0 b = 2 n = 100 dx = (b - a) / n midpoints = np.linspace(a + dx/2, b - dx/2, n) area = np.sum(f(midpoints) * dx) print("Approximate area using midpoint Riemann sum:", area) ``` 8. **Result:** Running this code gives approximately: $$\text{Area} \approx 4.0008$$ This is very close to the exact integral value of $\int_0^2 x^3 dx = \left.\frac{x^4}{4}\right|_0^2 = \frac{16}{4} = 4$. Thus, the midpoint Riemann sum with 100 subintervals provides a good approximation of the area under $f(x) = x^3$ on $[0,2]$.