To find the average value of a function $f(t)$ over an interval $[a, b]$, we use the formula:

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

In this case, the function is $f(t) = e^{0.05t}$ and the interval is $[0, 10]$. So, we have:

$\text{Average value} = \frac{1}{10-0} \int_0^{10} e^{0.05t} \, dt$

Step 1: Compute the integral

The integral of $e^{0.05t}$ is:

$\int e^{0.05t} \, dt = \frac{1}{0.05} e^{0.05t} = 20 e^{0.05t}$

Step 2: Evaluate the definite integral

Now, evaluate the integral from 0 to 10:

$\int_0^{10} e^{0.05t} \, dt = 20 \left[ e^{0.05t} \right]_0^{10} = 20 \left( e^{0.5} - e^{0} \right)$

Since $e^0 = 1$, we have:

$\int_0^{10} e^{0.05t} \, dt = 20 \left( e^{0.5} - 1 \right)$

Using the approximation $e^{0.5} \approx 1.6487$:

$\int_0^{10} e^{0.05t} \, dt \approx 20 (1.6487 - 1) = 20 \times 0.6487 = 12.974$

Step 3: Divide by the length of the interval

The length of the interval is $10 - 0 = 10$, so the average value is:

$\text{Average value} \approx \frac{12.974}{10} = 1.2974$

Thus, the average value of the function $f(t) = e^{0.05t}$ over the interval $[0, 10]$ is approximately $1.2974$.

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Would you like further clarification or a more detailed explanation of any step? Here are some related questions:

1. What is the average value formula for trigonometric functions?
2. How would the answer change if the interval were different?
3. How do you compute the integral of exponential functions with a negative exponent?
4. What is the geometric interpretation of the average value of a function?
5. How can we use numerical integration methods for more complicated functions?

Tip: When dealing with exponential functions, the constant factor in the exponent significantly affects the rate of growth or decay.