Question

find the average value of $f(x)=e^{2x}$ over the interval $0, ln 5$.
a.) $\frac{25}{2ln 5}$
b.) $\frac{24}{ln 5}$
c.) $12$
d.) $\frac{12}{ln 5}$

Explanation:

Step1: Recall average value formula

The average value of $f(x)$ on $[a,b]$ is $\frac{1}{b-a}\int_{a}^{b}f(x)dx$. Here $a=0$, $b=\ln5$, $f(x)=e^{2x}$.

Step2: Set up the integral

$$\text{Average Value} = \frac{1}{\ln5 - 0}\int_{0}^{\ln5} e^{2x}dx$$

Step3: Compute the integral

The antiderivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$. Evaluate from 0 to $\ln5$:
$$\int_{0}^{\ln5} e^{2x}dx = \frac{1}{2}e^{2\ln5} - \frac{1}{2}e^{0}$$

Step4: Simplify the exponential terms

Use $e^{2\ln5}=e^{\ln5^2}=25$ and $e^0=1$:
$$\frac{1}{2}(25) - \frac{1}{2}(1) = \frac{25-1}{2} = 12$$

Step5: Final average value calculation

$$\text{Average Value} = \frac{1}{\ln5} \times 12 = \frac{12}{\ln5}$$

Answer:

d.) $\frac{12}{\ln5}$