Let's break this problem down to solve it step-by-step. Here's what we need to calculate:

Problem

We are tasked to approximate the integral
$\int_0^6 3\sqrt{x} \, dx$
using the trapezoidal rule with 4 subintervals of equal length.

Step 1: Determine the interval width

The interval is $[0, 6]$, and there are $n = 4$ subintervals. The width of each subinterval ($h$) is calculated as: $h = \frac{b-a}{n} = \frac{6-0}{4} = 1.5$

Step 2: Define the $x_i$ values

The endpoints of the intervals are determined by the width $h$: $x_0 = 0, \, x_1 = 1.5, \, x_2 = 3, \, x_3 = 4.5, \, x_4 = 6$

Step 3: Evaluate $f(x)$ at each $x_i$

The function is $f(x) = 3\sqrt{x}$. Compute $f(x)$ at each $x_i$: $f(x_0) = 3\sqrt{0} = 0$ $f(x_1) = 3\sqrt{1.5} = 3 \times 1.2247 \approx 3.6741$ $f(x_2) = 3\sqrt{3} = 3 \times 1.7321 \approx 5.1962$ $f(x_3) = 3\sqrt{4.5} = 3 \times 2.1213 \approx 6.3639$ $f(x_4) = 3\sqrt{6} = 3 \times 2.4495 \approx 7.3485$

Step 4: Apply the trapezoidal rule

The formula for the trapezoidal rule is: $T = \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right]$

Substitute the values: $T = \frac{1.5}{2} \left[ 0 + 2(3.6741) + 2(5.1962) + 2(6.3639) + 7.3485 \right]$ $T = 0.75 \left[ 0 + 7.3482 + 10.3924 + 12.7278 + 7.3485 \right]$ $T = 0.75 \times 37.8169 \approx 28.3627$

Final Answer:

The trapezoidal sum approximation is approximately: $\boxed{28.363} \, \text{(rounded to the nearest thousandth)}$

Would you like a more detailed breakdown of any step, or do you have additional questions?

Here are 5 related questions to expand on this topic:

1. How does the trapezoidal rule compare to Simpson's rule for approximating integrals?
2. What happens if the number of subintervals ($n$) increases—how does it affect the accuracy?
3. How can you derive the error bound for the trapezoidal rule?
4. What are the advantages of numerical integration over symbolic methods?
5. How does the trapezoidal rule extend to functions of two variables?

Tip: The accuracy of the trapezoidal rule improves as the number of subintervals increases.