Question

(1 point) find the average value of the function ( f(x) = 3e^x ) on the interval (-3, ln 6). ( f_{\text{ave}} = ) preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Recall the average value formula

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by \( f_{\text{ave}}=\frac{1}{b - a}\int_{a}^{b}f(x)dx \). Here, \( a=-3 \), \( b = \ln 6 \), and \( f(x)=3e^{x} \).

Step2: Compute the integral of \( f(x) \)

First, find the antiderivative of \( 3e^{x} \). The antiderivative of \( e^{x} \) is \( e^{x} \), so the antiderivative of \( 3e^{x} \) is \( 3e^{x} \). Then, by the Fundamental Theorem of Calculus, \( \int_{-3}^{\ln 6}3e^{x}dx=3e^{x}\big|_{-3}^{\ln 6}=3e^{\ln 6}-3e^{-3} \). Since \( e^{\ln 6}=6 \), this simplifies to \( 3\times6 - 3e^{-3}=18 - \frac{3}{e^{3}} \).

Step3: Compute the length of the interval

The length of the interval \([-3,\ln 6]\) is \( b - a=\ln 6-(-3)=\ln 6 + 3 \).

Step4: Compute the average value

Now, use the average value formula: \( f_{\text{ave}}=\frac{1}{\ln 6 + 3}(18 - \frac{3}{e^{3}})=\frac{3(6-\frac{1}{e^{3}})}{3 + \ln 6}=\frac{18-\frac{3}{e^{3}}}{3+\ln 6} \). We can also simplify the numerator and denominator:
\[

$$\begin{align*} f_{\text{ave}}&=\frac{3(6 - e^{-3})}{3+\ln 6}\\ \end{align*}$$

\]
If we want a numerical approximation, \( e^{3}\approx20.0855 \), so \( \frac{3}{e^{3}}\approx0.1494 \), \( 18 - 0.1494 = 17.8506 \), \( \ln 6\approx1.7918 \), \( 3 + 1.7918 = 4.7918 \), then \( f_{\text{ave}}\approx\frac{17.8506}{4.7918}\approx3.725 \). But the exact form is \( \frac{18 - 3e^{-3}}{3+\ln 6} \) or simplified as \( \frac{3(6 - e^{-3})}{3+\ln 6} \).

Answer:

\(\frac{18 - \frac{3}{e^{3}}}{3+\ln 6}\) (or the simplified form \(\frac{3(6 - e^{-3})}{3+\ln 6}\) or the approximate value \(\approx3.725\))