Determine the average value of the function $$ f(x) = e^{x} $$ from $$ x = 0 $$ to $$ x = 2 $$.

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To find the **average value** of the function $$ f(x) = e^{x} $$ over the interval from $$ x = 0 $$ to $$ x = 2 $$, we use the formula for the average value of a function over an interval $$[a, b]$$:

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

**Applying the formula:**

1. **Identify the bounds and the function:**
   - $$ a = 0 $$
   - $$ b = 2 $$
   - $$ f(x) = e^{x} $$

2. **Set up the integral:**
   $$
   \text{Average} = \frac{1}{2 - 0} \int_{0}^{2} e^{x} \, dx = \frac{1}{2} \int_{0}^{2} e^{x} \, dx
   $$

3. **Compute the integral:**
   $$
   \int e^{x} \, dx = e^{x} + C
   $$
   
   Evaluating from 0 to 2:
   $$
   \int_{0}^{2} e^{x} \, dx = e^{2} - e^{0} = e^{2} - 1
   $$

4. **Calculate the average value:**
   $$
   \text{Average} = \frac{e^{2} - 1}{2}
   $$

**Final Answer:**
$$
\boxed{\dfrac{e^{2} - 1}{2}}
$$

If you prefer a numerical approximation:
- $$ e^{2} \approx 7.389 $$
- Therefore, $$ \frac{7.389 - 1}{2} \approx \frac{6.389}{2} \approx 3.1945 $$

So, the average value is approximately **3.1945**.
The average value of $$ f(x) = e^{x} $$ from $$ x = 0 $$ to $$ x = 2 $$ is $$ \frac{e^{2} - 1}{2} $$, which is approximately 3.1945.

Determine the average value of the function from to .

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