suppose that the value of a yacht in dollars after t years of use is $v(t)=325000e^{-0.19t}$. what is the…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ over the interval $[a,b]$ is given by $V_{ave}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 0$, $b = 19$, and $V(t)=325000e^{-0.19t}$, so $V_{ave}=\frac{1}{19-0}\int_{0}^{19}325000e^{-0.19t}dt$.

Step2: Calculate the integral

First, find the antiderivative of $325000e^{-0.19t}$. The antiderivative of $e^{-0.19t}$ is $-\frac{1}{0.19}e^{-0.19t}$, so the antiderivative of $325000e^{-0.19t}$ is $325000\times(-\frac{1}{0.19})e^{-0.19t}=-\frac{325000}{0.19}e^{-0.19t}$. Then, evaluate the definite - integral: [ \begin{align*} \frac{1}{19}\int_{0}^{19}325000e^{-0.19t}dt&=\frac{1}{19}\left[-\frac{325000}{0.19}e^{-0.19t}\right]{0}^{19}\ &=\frac{1}{19}\times\left(-\frac{325000}{0.19}\right)\left(e^{-0.19\times19}-e^{-0.19\times0}\right)\ &=-\frac{325000}{19\times0.19}(e^{-3.61}-1) \end{align*} ] Since $e^{-3.61}\approx0.0273$ and $19\times0.19 = 3.61$, we have: [ \begin{align*} V{ave}&=-\frac{325000}{3.61}(0.0273 - 1)\ &=\frac{325000}{3.61}(1 - 0.0273)\ &=\frac{325000}{3.61}\times0.9727\ &\approx\frac{325000\times0.9727}{3.61}\ &\approx\frac{316127.5}{3.61}\ &\approx87570 \end{align*} ]

Answer:

$87570$