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To solve this problem using the Midpoint Rule, let’s go through it step-by-step.

### Problem Recap
We're given:
- The function \( f(x) = \sin(x) \).
- The interval \( \left[0, \frac{\pi}{2}\right] \).
- \( n = 4 \), meaning we will divide the interval into 4 subintervals.

The Midpoint Rule formula for approximating the area is:
\[
\text{Area} \approx \sum_{i=1}^n f\left(\frac{x_i + x_{i-1}}{2}\right) \Delta x
\]
where \( \Delta x = \frac{b - a}{n} \), \( [a, b] \) is the interval, and \( \frac{x_i + x_{i-1}}{2} \) is the midpoint of each subinterval.

### Step-by-Step Solution

1. **Determine \(\Delta x\):**
   \[
   \Delta x = \frac{\frac{\pi}{2} - 0}{4} = \frac{\pi}{8}
   \]

2. **Identify Midpoints:**
   We’ll divide \([0, \frac{\pi}{2}]\) into 4 subintervals of width \(\frac{\pi}{8}\):
   - The interval \([0, \frac{\pi}{8}]\) has a midpoint of \(\frac{\pi}{16}\).
   - The interval \([\frac{\pi}{8}, \frac{2\pi}{8}]\) has a midpoint of \(\frac{3\pi}{16}\).
   - The interval \([\frac{2\pi}{8}, \frac{3\pi}{8}]\) has a midpoint of \(\frac{5\pi}{16}\).
   - The interval \([\frac{3\pi}{8}, \frac{\pi}{2}]\) has a midpoint of \(\frac{7\pi}{16}\).

3. **Calculate \( f \) at each midpoint:**
   Substitute each midpoint into \( f(x) = \sin(x) \):
   - \( f\left(\frac{\pi}{16}\right) = \sin\left(\frac{\pi}{16}\right) \)
   - \( f\left(\frac{3\pi}{16}\right) = \sin\left(\frac{3\pi}{16}\right) \)
   - \( f\left(\frac{5\pi}{16}\right) = \sin\left(\frac{5\pi}{16}\right) \)
   - \( f\left(\frac{7\pi}{16}\right) = \sin\left(\frac{7\pi}{16}\right) \)

4. **Apply the Midpoint Rule:**
   Now, plug these values into the Midpoint Rule formula:
   \[
   \text{Area} \approx \Delta x \left( f\left(\frac{\pi}{16}\right) + f\left(\frac{3\pi}{16}\right) + f\left(\frac{5\pi}{16}\right) + f\left(\frac{7\pi}{16}\right) \right)
   \]
   Substituting \(\Delta x = \frac{\pi}{8}\):
   \[
   \text{Area} \approx \frac{\pi}{8} \left( \sin\left(\frac{\pi}{16}\right) + \sin\left(\frac{3\pi}{16}\right) + \sin\left(\frac{5\pi}{16}\right) + \sin\left(\frac{7\pi}{16}\right) \right)
   \]

5. **Calculate the Sine Values and Final Area Approximation:**
   Approximating the sine values (or using a calculator):
   - \( \sin\left(\frac{\pi}{16}\right) \approx 0.1951 \)
   - \( \sin\left(\frac{3\pi}{16}\right) \approx 0.5556 \)
   - \( \sin\left(\frac{5\pi}{16}\right) \approx 0.8315 \)
   - \( \sin\left(\frac{7\pi}{16}\right) \approx 0.9808 \)

   Substitute these values:
   \[
   \text{Area} \approx \frac{\pi}{8} \times (0.1951 + 0.5556 + 0.8315 + 0.9808)
   \]
   \[
   \text{Area} \approx \frac{\pi}{8} \times 2.563
   \]
   \[
   \text{Area} \approx 1.006
   \]

### Answer
The approximate area under \( f(x) = \sin(x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) using the Midpoint Rule with \( n = 4 \) is about **1.006**.

Would you like more details on any step?

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### Additional Questions
1. How does the Midpoint Rule compare to other numerical methods like the Trapezoidal Rule?
2. How would the approximation change if we increased \( n \) to a larger value, such as \( n = 8 \)?
3. Why is the Midpoint Rule generally more accurate than the Left or Right Riemann Sum?
4. What is the exact area under \( \sin(x) \) from \( 0 \) to \( \frac{\pi}{2} \), and how does it compare with this approximation?
5. How do we apply the Midpoint Rule if we have a more complex function like \( f(x) = x^2 \sin(x) \)?

### Tip
For improved accuracy in numerical integration, increase the number of subintervals \( n \); this decreases the approximation error in methods like the Midpoint and Trapezoidal Rule.

Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of the function f(x) = sin(x) over the interval [0, π/2].

Learn how to use the Midpoint Rule with n = 4 to approximate the area under the curve f(x) = sin(x) over the interval [0, π/2]. This detailed solution involves step-by-step calculations and applications of numerical integration.

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