find the average value of $g(t)=4^{t}+5$ over the interval $0,6$. round your answer to the nearest…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = g(t)$ over the interval $[a,b]$ is given by $\bar{g}=\frac{1}{b - a}\int_{a}^{b}g(t)dt$. Here, $a = 0$, $b = 6$, and $g(t)=4^{t}+5$. So, $\bar{g}=\frac{1}{6 - 0}\int_{0}^{6}(4^{t}+5)dt=\frac{1}{6}\int_{0}^{6}(4^{t}+5)dt$.

Step2: Integrate term - by - term

We know that $\int 4^{t}dt=\frac{4^{t}}{\ln(4)}+C$ and $\int 5dt = 5t + C$. Then $\int_{0}^{6}(4^{t}+5)dt=\left[\frac{4^{t}}{\ln(4)}+5t\right]_{0}^{6}$.

Step3: Evaluate the definite integral

First, substitute $t = 6$: $\frac{4^{6}}{\ln(4)}+5\times6=\frac{4096}{\ln(4)} + 30$. Then substitute $t = 0$: $\frac{4^{0}}{\ln(4)}+5\times0=\frac{1}{\ln(4)}$. Subtract the two results: $\left(\frac{4096}{\ln(4)}+30\right)-\frac{1}{\ln(4)}=\frac{4096 - 1}{\ln(4)}+30=\frac{4095}{\ln(4)}+30$.

Step4: Calculate the average value

$\bar{g}=\frac{1}{6}\left(\frac{4095}{\ln(4)}+30\right)$. $\frac{4095}{\ln(4)}\approx\frac{4095}{1.3863}\approx2954.05$. $\frac{1}{6}\left(2954.05 + 30\right)=\frac{1}{6}\times2984.05\approx497.34$.

Answer:

$497.34$