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

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**.