Question

an automobile company is ready to introduce a new line of cars through a national sales campaign. after test marketing the line in a carefully selected city, the marketing research department estimates that the sales (in millions of dollars) will increase at the monthly rate of the following function for t months after the campaign has started.

s(t)=15 - 13e^{-0.1t}

0 ≤ t ≤ 20

(a) what will be the total sales, s(t), t months after the beginning of the national sales campaign if we assume no sales at the beginning of the campaign?

s(t)=□

Explanation:

Step1: Integrate the rate function

We know that \(S(t)=\int S^{\prime}(t)dt\). Given \(S^{\prime}(t)=15 - 13e^{-0.1t}\), then \(S(t)=\int(15 - 13e^{-0.1t})dt\).
Using the integral rules \(\int a dt=at + C\) (\(a = 15\)) and \(\int e^{kt}dt=\frac{1}{k}e^{kt}+C\) (\(k=- 0.1\)), we have:
\(S(t)=\int15dt-\int13e^{-0.1t}dt\)
\(S(t)=15t-13\times\frac{e^{-0.1t}}{-0.1}+C\)
\(S(t)=15t + 130e^{-0.1t}+C\)

Step2: Determine the constant \(C\)

Since there are no sales at the beginning (\(t = 0\), \(S(0)=0\)).
Substitute \(t = 0\) and \(S(0)=0\) into \(S(t)=15t + 130e^{-0.1t}+C\):
\(0=15\times0+130e^{0}+C\)
Since \(e^{0}=1\), we get \(0 = 130 + C\), so \(C=-130\)

Step3: Write the final function

Substitute \(C = - 130\) into \(S(t)=15t + 130e^{-0.1t}+C\)
\(S(t)=15t+130e^{-0.1t}-130\)

Answer:

\(S(t)=15t + 130e^{-0.1t}-130\)