Find the average value of the function on the given interval. Round the answer to two decimal places if necessary.

f left parenthesis x right parenthesis equals e Superscript x divided by 2f(x)=ex/2​; 

​[0,25]

To find the average value of the function \( f(x) = \frac{e^x}{2} \) over the interval \([0, 25]\), we use the formula for the average value of a continuous function over a given interval:

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

For this problem, \( f(x) = \frac{e^x}{2} \), \( a = 0 \), and \( b = 25 \). Therefore, the average value is:

\[
\text{Average value} = \frac{1}{25-0} \int_0^{25} \frac{e^x}{2} \, dx
\]

First, we compute the integral:

\[
\int_0^{25} \frac{e^x}{2} \, dx
\]

To integrate \( \frac{e^x}{2} \), we can factor out the constant \( \frac{1}{2} \):

\[
\frac{1}{2} \int_0^{25} e^x \, dx
\]

The integral of \( e^x \) is simply \( e^x \), so we have:

\[
\frac{1}{2} \left[ e^x \right]_0^{25}
\]

Evaluating this at the bounds \( x = 0 \) and \( x = 25 \):

\[
\frac{1}{2} \left( e^{25} - e^0 \right)
\]

Since \( e^0 = 1 \):

\[
\frac{1}{2} \left( e^{25} - 1 \right)
\]

Next, we divide this result by the length of the interval \( 25 - 0 = 25 \):

\[
\text{Average value} = \frac{1}{25} \cdot \frac{1}{2} \left( e^{25} - 1 \right)
\]

\[
= \frac{1}{50} \left( e^{25} - 1 \right)
\]

To find a numerical value, we need to compute \( e^{25} \). Using a calculator, we find:

\[
e^{25} \approx 7.20048993 \times 10^{10}
\]

Substituting this value back in:

\[
\text{Average value} = \frac{1}{50} \left( 7.20048993 \times 10^{10} - 1 \right)
\]

\[
\approx \frac{1}{50} \left( 7.20048993 \times 10^{10} \right)
\]

\[
\approx 1.44009799 \times 10^9
\]

Rounding to two decimal places:

\[
\text{Average value} \approx 1440097990.00
\]

So, the average value of the function \( f(x) = \frac{e^x}{2} \) on the interval \([0, 25]\) is approximately **1,440,097,990.00**.

Would you like more details or have any questions?

Here are 8 related questions to expand your understanding:

1. How does the exponential function \( e^x \) affect the average value compared to polynomial functions?
2. What are the applications of finding the average value of an exponential function in real-world problems?
3. How would the result change if the interval was different, such as \([0, 10]\)?
4. What is the significance of the constant \( \frac{1}{2} \) in the function \( f(x) = \frac{e^x}{2} \)?
5. How do you interpret the average value of an exponentially growing function over a large interval?
6. What are some common methods for approximating integrals if an exact solution is difficult to compute?
7. How does the value of \( e^{25} \) compare to other large exponential values?
8. What other types of functions might exhibit similar rapid growth over large intervals?

**Tip:** When dealing with functions involving \( e^x \), be aware of the rapid growth rate of the exponential function, especially over large intervals, as it can lead to very large values.

Explore how to find the average value of the function f(x) = e^x/2 over the interval [0, 25]. Learn the integral calculus techniques involved and the application of exponential functions in mathematics. Suitable for advanced undergraduate students.

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