Questions: A stock analyst plots the price per share of a certain common stock as a function of time and finds that it can be approximated by the function: S(t) = 9^t + 15e^(-0.5t), where t is the time (in years) since the stock was purchased. The question asks to round the average price of the stock to the nearest cent.

Solution Steps

The given function for the stock price is \( S(t) = 37 + 15e^{-0.07t} \), where \( t \) is the time in years since the stock was purchased. We need to find the average price of the stock over the first 10 years.

The average value of a continuous function \( f(t) \) over the interval \([a, b]\) is given by: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(t) \, dt \] In this case, \( f(t) = S(t) \), \( a = 0 \), and \( b = 10 \).

We need to compute: \[ \int_0^{10} (37 + 15e^{-0.07t}) \, dt \]

First, split the integral: \[ \int_0^{10} 37 \, dt + \int_0^{10} 15e^{-0.07t} \, dt \]

Compute each part separately:

1. \[ \int_0^{10} 37 \, dt = 37t \Big|_0^{10} = 37(10) - 37(0) = 370 \]
2. For the second part, use the substitution \( u = -0.07t \), hence \( du = -0.07 \, dt \): \[ \int_0^{10} 15e^{-0.07t} \, dt = 15 \int_0^{10} e^{-0.07t} \, dt \] \[ = 15 \left[ \frac{e^{-0.07t}}{-0.07} \right]_0^{10} = 15 \left[ \frac{e^{-0.07 \cdot 10} - e^{-0.07 \cdot 0}}{-0.07} \right] \] \[ = 15 \left[ \frac{e^{-0.7} - 1}{-0.07} \right] \] \[ = 15 \left[ \frac{1 - e^{-0.7}}{0.07} \right] \]

Using \( e^{-0.7} \approx 0.4966 \): \[ = 15 \left[ \frac{1 - 0.4966}{0.07} \right] \] \[ = 15 \left[ \frac{0.5034}{0.07} \right] \] \[ = 15 \left[ 7.1914 \right] \] \[ \approx 107.871 \]

\[ \int_0^{10} (37 + 15e^{-0.07t}) \, dt = 370 + 107.871 = 477.871 \]

\[ \text{Average value} = \frac{1}{10 - 0} \int_0^{10} S(t) \, dt = \frac{477.871}{10} = 47.7871 \]

Final Answer