To approximate the integral using the trapezoidal rule, we are given the function $f(x) = \frac{5}{x}$ and the interval from $3$ to $4.5$, divided into 3 equal subintervals.

The Trapezoidal Rule for approximating the integral of a function $f(x)$ over the interval $[a, b]$ using $n$ subintervals is given by:

$T_n = \frac{b - a}{2n} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right)$ Where:

• $a = 3$
• $b = 4.5$
• $n = 3$
• The points $x_i$ are the endpoints of the subintervals.

Step 1: Find the length of each subinterval

The length of each subinterval is:

$\Delta x = \frac{b - a}{n} = \frac{4.5 - 3}{3} = \frac{1.5}{3} = 0.5$

Thus, the subintervals will have a width of 0.5.

Step 2: Compute the values of $f(x)$ at the endpoints and the interior points

The points of evaluation are:

• $x_0 = 3$
• $x_1 = 3.5$
• $x_2 = 4$
• $x_3 = 4.5$

Now, compute the values of $f(x) = \frac{5}{x}$ at these points:

• $f(3) = \frac{5}{3} \approx 1.6667$
• $f(3.5) = \frac{5}{3.5} \approx 1.4286$
• $f(4) = \frac{5}{4} = 1.25$
• $f(4.5) = \frac{5}{4.5} \approx 1.1111$

Step 3: Apply the trapezoidal rule formula

Now, apply these values into the trapezoidal rule formula:

$T_3 = \frac{0.5}{2} \left( f(3) + 2(f(3.5) + f(4)) + f(4.5) \right)$ Substitute the function values:

$T_3 = \frac{0.5}{2} \left( 1.6667 + 2(1.4286 + 1.25) + 1.1111 \right)$

Simplify inside the parentheses:

$T_3 = \frac{0.5}{2} \left( 1.6667 + 2(2.6786) + 1.1111 \right)$ $T_3 = \frac{0.5}{2} \left( 1.6667 + 5.3572 + 1.1111 \right)$ $T_3 = \frac{0.5}{2} \times 8.135$ $T_3 = 0.25 \times 8.135 = 2.03375$

Thus, the value of the trapezoidal sum approximation is:

$T_3 \approx 2.034$

Final Answer:

The trapezoidal sum approximation for the integral is approximately 2.034.

Would you like further details or have any questions?

Additional Questions:

1. How would the result change if we used more subintervals?
2. What is the exact value of the integral for comparison?
3. How does the Trapezoidal Rule compare to Simpson's Rule for this problem?
4. What happens if the subintervals are not of equal length?
5. How does the Trapezoidal Rule approximate the integral when the function is linear?

Tip: Increasing the number of subintervals typically improves the accuracy