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?

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.