The composite trapezoidal rule for numerical integration is given by:

$\int_a^b f(x) \, dx \approx \frac{b-a}{n} \left[ \frac{f(x_0) + f(x_n)}{2} + \sum_{k=1}^{n-1} f(x_k) \right],$

where $x_k = a + k \cdot h$ and $h = \frac{b-a}{n}$.

To use five function evaluations, we set $n = 4$, resulting in five equally spaced points (including the endpoints $a$ and $b$).

Step-by-step solution:

1. Given function and interval: [ f(x) = e^{-x} \cos(x), \quad [a, b] = [0, \frac{\pi}{2}] ]

2. Number of intervals and step size: $n = 4, \quad h = \frac{b-a}{n} = \frac{\frac{\pi}{2} - 0}{4} = \frac{\pi}{8}.$

3. Equally spaced points: $x_0 = 0, \quad x_1 = \frac{\pi}{8}, \quad x_2 = \frac{\pi}{4}, \quad x_3 = \frac{3\pi}{8}, \quad x_4 = \frac{\pi}{2}.$

4. Evaluate $f(x)$ at these points: $f(x_0) = e^{-0} \cos(0) = 1,$ $f(x_1) = e^{-\frac{\pi}{8}} \cos\left(\frac{\pi}{8}\right),$ $f(x_2) = e^{-\frac{\pi}{4}} \cos\left(\frac{\pi}{4}\right),$ $f(x_3) = e^{-\frac{3\pi}{8}} \cos\left(\frac{3\pi}{8}\right),$ $f(x_4) = e^{-\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) = 0.$

5. Substitute into the trapezoidal rule: $\int_0^{\frac{\pi}{2}} e^{-x} \cos(x) \, dx \approx \frac{\pi}{8} \left[ \frac{f(x_0) + f(x_4)}{2} + f(x_1) + f(x_2) + f(x_3) \right].$

   Simplifying: $\int_0^{\frac{\pi}{2}} e^{-x} \cos(x) \, dx \approx \frac{\pi}{8} \left[ \frac{1 + 0}{2} + f(x_1) + f(x_2) + f(x_3) \right].$

6. Numerical computation: Evaluate $f(x_1), f(x_2),$ and $f(x_3)$. numpy as np

Define the function

def f(x): return np.exp(-x) * np.cos(x)

Define interval and step size

a = 0 b = np.pi / 2 n = 4 h = (b - a) / n

Calculate the x values

x_values = np.linspace(a, b, n + 1)

Evaluate the function at these points

f_values = f(x_values)

Apply the composite trapezoidal rule

integral = (h / 2) * (f_values[0] + 2 * np.sum(f_values[1:-1]) + f_values[-1]) integralThe value of the integral of $f(x) = e^{-x} \cos(x)$ over $[0, \pi/2]$ using the composite trapezoidal rule with five function evaluations is approximately:

$\int_0^{\pi/2} e^{-x} \cos(x) \, dx \approx 0.6142$

Would you like further details about the calculations, or assistance with other related problems?

Here are five follow-up questions to consider:

1. What is the error bound for the trapezoidal rule in this case?
2. How does this compare to the true value of the integral computed analytically?
3. What would the result be if a higher number of intervals were used?
4. Can this integral be solved using Simpson's rule for improved accuracy?
5. How does the behavior of $e^{-x} \cos(x)$ affect the accuracy of numerical methods?

Tip: The accuracy of the trapezoidal rule improves with more intervals, especially when the function being integrated is smooth.