To approximate the integral $\int_0^{\pi/2} 2 \cos^4(x) \, dx$ using the Midpoint Rule with $n = 4$, we can proceed with the following steps:

1. Determine the interval width $\Delta x$: $\Delta x = \frac{b - a}{n} = \frac{\pi/2 - 0}{4} = \frac{\pi}{8}$

2. Identify the midpoints of each subinterval. The midpoints $x_i$ for each interval are calculated as:

   • $x_1 = a + \Delta x/2 = 0 + \frac{\pi}{16} = \frac{\pi}{16}$
   • $x_2 = a + 3\Delta x/2 = 0 + \frac{3\pi}{16} = \frac{3\pi}{16}$
   • $x_3 = a + 5\Delta x/2 = 0 + \frac{5\pi}{16} = \frac{5\pi}{16}$
   • $x_4 = a + 7\Delta x/2 = 0 + \frac{7\pi}{16} = \frac{7\pi}{16}$

3. Apply the Midpoint Rule formula: $M_4 = \Delta x \sum_{i=1}^4 f(x_i)$ where $f(x) = 2 \cos^4(x)$.

4. Calculate each term in the summation: $M_4 = \frac{\pi}{8} \left[ 2 \cos^4\left(\frac{\pi}{16}\right) + 2 \cos^4\left(\frac{3\pi}{16}\right) + 2 \cos^4\left(\frac{5\pi}{16}\right) + 2 \cos^4\left(\frac{7\pi}{16}\right) \right]$

5. Evaluate each cosine term and substitute into the formula. I'll calculate this numerically to four decimal places.The approximate value of the integral using the Midpoint Rule with $n = 4$ is $M_4 \approx 1.1781$, rounded to four decimal places.

Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. How does the Midpoint Rule compare to other numerical integration methods, such as the Trapezoidal Rule?
2. What are some situations where the Midpoint Rule is particularly effective or ineffective?
3. How does increasing the value of $n$ affect the accuracy of the Midpoint Rule?
4. Can you apply the Midpoint Rule to integrals with varying limits, such as $\int_a^{b} f(x) \, dx$ where $a \neq 0$?
5. What is the theoretical error bound for the Midpoint Rule, and how can it be calculated?

Tip: Increasing the number of subintervals $n$ generally improves the accuracy of numerical integration methods, as it better captures the behavior of the function over the interval.