question a company models its income, in thousands of dollars, using the continuous stream…

question a company models its income, in thousands of dollars, using the continuous stream f(t)=20e^(-0.025t)+(t/12), where t is measured in months. what is the total income produced in the second year? give your answer in thousands of dollars to the nearest thousand dollars. when giving your answer, use numbers only. do not include the dollar symbol, commas, or any symbols to denote thousands. provide your answer below: $□ thousand feedback more instruction submit

question a company models its income, in thousands of dollars, using the continuous stream f(t)=20e^(-0.025t)+(t/12), where t is measured in months. what is the total income produced in the second year? give your answer in thousands of dollars to the nearest thousand dollars. when giving your answer, use numbers only. do not include the dollar symbol, commas, or any symbols to denote thousands. provide your answer below: $□ thousand feedback more instruction submit

Answer

Explanation:

Step1: Determine the limits of integration

The second - year corresponds to (t) values from (t = 12) to (t=24) months. The total income (I) from a continuous income stream (f(t)) over the interval ([a,b]) is given by (I=\int_{a}^{b}f(t)dt). Here, (a = 12), (b = 24), and (f(t)=20e^{- 0.025t}+\frac{t}{12}).

Step2: Split the integral

We use the property (\int_{a}^{b}(g(t)+h(t))dt=\int_{a}^{b}g(t)dt+\int_{a}^{b}h(t)dt). So, (\int_{12}^{24}(20e^{-0.025t}+\frac{t}{12})dt=\int_{12}^{24}20e^{-0.025t}dt+\int_{12}^{24}\frac{t}{12}dt).

Step3: Integrate the first integral

For (\int_{12}^{24}20e^{-0.025t}dt), let (u=-0.025t), then (du=-0.025dt) and (dt =-\frac{1}{0.025}du). When (t = 12), (u=-0.025\times12=-0.3); when (t = 24), (u=-0.025\times24 = - 0.6). (\int_{12}^{24}20e^{-0.025t}dt=20\int_{-0.3}^{-0.6}e^{u}\left(-\frac{1}{0.025}\right)du=- \frac{20}{0.025}\int_{-0.3}^{-0.6}e^{u}du=-800\left[e^{u}\right]_{-0.3}^{-0.6}=-800(e^{-0.6}-e^{-0.3})).

Step4: Integrate the second integral

For (\int_{12}^{24}\frac{t}{12}dt=\frac{1}{12}\int_{12}^{24}t\ dt). Using the power - rule (\int t^n dt=\frac{t^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (\frac{1}{12}\left[\frac{t^{2}}{2}\right]{12}^{24}=\frac{1}{24}(t^{2})\big|{12}^{24}=\frac{1}{24}(24^{2}-12^{2})=\frac{1}{24}(576 - 144)=\frac{432}{24}=18).

Step5: Calculate the total value

(-800(e^{-0.6}-e^{-0.3})+18). (e^{-0.6}\approx0.5488), (e^{-0.3}\approx0.7408). (-800(0.5488 - 0.7408)+18=-800\times(-0.192)+18 = 153.6+18=171.6\approx172).

Answer:

172