To find the average value of the function $f(x) = 8e^x$ over the interval $[-4, 4]$, we use the formula for the average value of a continuous function $f(x)$ over the interval $[a, b]$:

$\text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx$

For this problem:

• $f(x) = 8e^x$
• $a = -4$
• $b = 4$

So, we need to calculate:

$\text{Average value} = \frac{1}{4 - (-4)} \int_{-4}^{4} 8e^x \, dx$

Step 1: Simplify the Expression

Since $b - a = 4 - (-4) = 8$, the formula becomes:

$\text{Average value} = \frac{1}{8} \int_{-4}^{4} 8e^x \, dx$

We can factor out the 8 from the integral:

$= \frac{8}{8} \int_{-4}^{4} e^x \, dx = \int_{-4}^{4} e^x \, dx$

Step 2: Integrate $e^x$

The integral of $e^x$ is simply $e^x$, so:

$\int_{-4}^{4} e^x \, dx = e^x \Big|_{-4}^{4} = e^4 - e^{-4}$

Step 3: Calculate the Average Value

Thus, the average value is:

$e^4 - e^{-4}$

Now, we substitute approximate values for $e^4 \approx 54.598$ and $e^{-4} \approx 0.0183$:

$e^4 - e^{-4} \approx 54.598 - 0.0183 = 54.580$

Answer (rounded to three decimal places):

$\boxed{54.580}$

Would you like more details on the integration steps, or have any questions?

Here are some related questions:

1. How would the average value change if the interval were different?
2. What happens to the average value if $f(x)$ is modified to $e^{2x}$ instead?
3. Can we apply this method to functions with more complex expressions, like $f(x) = e^{x^2}$?
4. How does the exponential function $e^x$ behave over different intervals?
5. What are other methods to approximate the integral if it were too complex to solve directly?

Tip: Always recheck the interval boundaries when computing an average value to ensure correct setup in the integral formula.