We are tasked with approximating the integral of $f(x)$ from 0 to 8 using the trapezoidal sum method and the provided table.

Step-by-Step Solution:

1. Formula for the Trapezoidal Rule

For $n$ subintervals, the trapezoidal approximation is:

$\int_a^b f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n) \right].$

Where:

• $\Delta x = \frac{b - a}{n}$ is the width of each subinterval.
• $x_0, x_1, \dots, x_n$ are the endpoints of the subintervals.

2. Extract Values from the Table

From the table:

• $x_0 = 0, x_1 = 4, x_2 = 6, x_3 = 8$,
• $f(0) = 9, f(4) = 25, f(6) = 27, f(8) = 26$.

3. Compute $\Delta x$

The interval is $[0, 8]$, and there are $n = 3$ subintervals. Thus: $\Delta x = \frac{8 - 0}{3} = \frac{8}{3}.$

4. Apply the Trapezoidal Rule

The approximation becomes: $\int_0^8 f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(0) + 2f(4) + 2f(6) + f(8) \right].$

Substitute the known values: $\int_0^8 f(x) \, dx \approx \frac{\frac{8}{3}}{2} \left[ 9 + 2(25) + 2(27) + 26 \right].$

Simplify step-by-step:

1. Compute the weights: $9 + 2(25) + 2(27) + 26 = 9 + 50 + 54 + 26 = 139$.
2. Compute the coefficient: $\frac{\frac{8}{3}}{2} = \frac{4}{3}$.
3. Final calculation: $\int_0^8 f(x) \, dx \approx \frac{4}{3} \times 139 = \frac{556}{3} \approx 185.33.$

Final Answer:

The trapezoidal sum approximation is: $\boxed{185.33 \text{ (approximately)}}.$

Would you like more details or have further questions?

Related Questions:

1. How does the trapezoidal rule compare to other numerical methods like Simpson’s Rule?
2. What happens to the accuracy of the trapezoidal rule as the number of subintervals increases?
3. Why is $\Delta x/2$ a key factor in the formula?
4. Can the trapezoidal rule be used for functions with unequal intervals?
5. How does the trapezoidal rule perform for highly nonlinear functions?

Tip:

The accuracy of the trapezoidal rule improves significantly when the function $f(x)$ is approximately linear over each subinterval.