Given the table for time $$ (\mathrm{t}) $$ and velocity (v), use the trapezoidal rule for $$ \mathrm{n}=4 $$ to find the approximation for the total distance traveled from 0 to 8 seconds. \begin{tabular}{rl|lllllllll}$$ \mathrm{t}(\mathrm{sec}) $$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \ \hline $$ \mathrm{v}(\mathrm{ft} / \mathrm{sec}) $$ & 0 & 6 & 9 & 14 & 17 & 13 & 20 & 22 & 26\end{tabular} $$ \mathrm{~d}=[?] $$

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We are given data for $$ t $$ and $$ v $$ and need to approximate the total distance traveled from 0 to 8 seconds using the trapezoidal rule with $$ n=4 $$ subintervals.

Since the interval is from 0 to 8 seconds, the width of each subinterval is

$$
h = \frac{8-0}{4} = 2.
$$

The trapezoidal rule formula for approximating the integral is

$$
I \approx \frac{h}{2} \left[ f(t_0) + f(t_n) + 2\sum_{k=1}^{n-1} f(t_k) \right],
$$

where $$ f(t) $$ is the velocity at time $$ t $$.

We choose the points corresponding to $$ t = 0, 2, 4, 6, 8 $$ seconds. From the table:

- At $$ t=0 $$: $$ v = 0 $$
- At $$ t=2 $$: $$ v = 9 $$
- At $$ t=4 $$: $$ v = 17 $$
- At $$ t=6 $$: $$ v = 20 $$
- At $$ t=8 $$: $$ v = 26 $$

Substitute these values in the formula:

$$
I \approx \frac{2}{2} \left[ 0 + 26 + 2(9 + 17 + 20) \right].
$$

First, compute the sum inside:

$$
9 + 17 + 20 = 46.
$$

Then multiply by 2:

$$
2 \times 46 = 92.
$$

Add the endpoints:

$$
0 + 26 = 26.
$$

So the approximation becomes

$$
I \approx 1 \times (26 + 92) = 118.
$$

Thus, the approximate total distance traveled is

$$
d \approx 118 \text{ feet}.
$$
The approximate total distance traveled is 118 feet.

Given the table for time and velocity
(v), use the trapezoidal rule for to
find the approximation for the total
distance traveled from 0 to 8 seconds.

0 1 2 3 4 5 6 7 8

0 6 9 14 17 13 20 22 26

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Given the table for time and velocity (v), use the trapezoidal rule for to find the approximation for the total distance traveled from 0 to 8 seconds. 0 1 2 3 4 5 6 7 8 0 6 9 14 17 13 20 22 26