To find the average value of the function $A(t) = \frac{3}{t}$ between $t = 2$ and $t = 8$, we can use the formula for the average value of a function over an interval $[a, b]$:

$\text{Average value of } A(t) = \frac{1}{b - a} \int_{a}^{b} A(t) \, dt$

Here, $A(t) = \frac{3}{t}$, $a = 2$, and $b = 8$. So, the formula becomes:

$\text{Average value} = \frac{1}{8 - 2} \int_{2}^{8} \frac{3}{t} \, dt$

Step 1: Set up the integral

$\text{Average value} = \frac{1}{6} \int_{2}^{8} \frac{3}{t} \, dt$

Step 2: Solve the integral

The integral of $\frac{1}{t}$ is $\ln|t|$. So, we can solve the integral as follows:

$\int_{2}^{8} \frac{3}{t} \, dt = 3 \int_{2}^{8} \frac{1}{t} \, dt = 3 [\ln|t|]_{2}^{8}$

Evaluating the definite integral:

$3 [\ln(8) - \ln(2)] = 3 \ln \left(\frac{8}{2}\right) = 3 \ln(4)$

Since $\ln(4) = 2 \ln(2)$, we get:

$3 \ln(4) = 6 \ln(2)$

Step 3: Calculate the average value

Now substitute the result of the integral back into the formula for the average value:

$\text{Average value} = \frac{1}{6} \times 6 \ln(2) = \ln(2)$

Thus, the average value of $A(t) = \frac{3}{t}$ between $t = 2$ and $t = 8$ is:

$\boxed{\ln(2)}$

Would you like more details, or do you have any questions?

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Here are 5 related questions to explore further:

1. How do you find the average value of a quadratic function over an interval?
2. Can you calculate the average value of $A(t) = \frac{5}{t^2}$ over the same interval?
3. What is the physical meaning of the average value of a function in an applied problem?
4. How does the result change if we compute the average of $A(t) = \frac{3}{t}$ over a different interval?
5. What if we replace $\frac{3}{t}$ with a trigonometric function like $\sin(t)$? How do we compute the average then?

Tip: When solving integrals involving $\frac{1}{t}$, remember that the result is always related to the natural logarithm function $\ln(t)$.