009B Sample Midterm 1, Problem 2

Otis Taylor plots the price per share of a stock that he owns as a function of time

and finds that it can be approximated by the function

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s(t)=t(25-5t)+18}

where  $t$  is the time (in years) since the stock was purchased.

Find the average price of the stock over the first five years.

Foundations:
The average value of a function  $f(x)$  on an interval  $[a,b]$  is given by
$f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)~dx.$

Solution:

Step 1:
This problem wants us to find the average value of  $s(t)$  over the interval  $[0,5].$
Using the average value formula, we have
$s_{\text{avg}}={\frac {1}{5-0}}\int _{0}^{5}t(25-5t)+18~dt.$

Step 2:
First, we distribute to get
$s_{\text{avg}}={\frac {1}{5}}\int _{0}^{5}25t-5t^{2}+18~dt.$
Then, we integrate to get
$s_{\text{avg}}=\left.{\frac {1}{5}}{\bigg [}{\frac {25t^{2}}{2}}-{\frac {5t^{3}}{3}}+18t{\bigg ]}\right|_{0}^{5}.$

Step 3:
We now evaluate to get
${\begin{array}{rcl}\displaystyle {s_{\text{avg}}}&=&\displaystyle {{\frac {1}{5}}{\bigg [}{\frac {25(5)^{2}}{2}}-{\frac {5(5)^{3}}{3}}+18(5){\bigg ]}-0}\\&&\\&=&\displaystyle {{\frac {233}{6}}.}\end{array}}$

Final Answer:
${\frac {233}{6}}$

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