The population of a town grows at a rate of 2e^(0.2 t)-t people per year (where t is the number of years). Approximately by how many people did the population grow between t=0 and t=5 ? Choose 1 answer: (A) 5 (B) 10 (C) 15 (D) 20

Question

The population of a town grows at a rate of  2e^(0.2 t)-t people per year (where  t is the number of years). Approximately by how many people did the population grow between  t=0 and  t=5 ? Choose 1 answer: (A) 5 (B) 10 (C) 15 (D) 20

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Understand function meaning: Understand the given function and what it represents. The function given is 2e^(0.2t) - t, which represents the rate of population growth per year. To find the total growth over a certain period, we need to integrate this function with respect to time from the start time to the end time.Set up integral calculation: Set up the integral to calculate the total population growth from t=0 to t=5. We need to integrate the function 2e^(0.2t) - t from t=0 to t=5. This will give us the total change in population over the 5-year period.Calculate integral from t=0 to t=5: Calculate the integral of the function 2e^(0.2t) - t from t=0 to t=5. The integral of 2e^(0.2t) is (10e^(0.2t)), and the integral of t is (t^2)/2. So, we need to evaluate the integral (10e^(0.2t) - (t^2)/2) from t=0 to t=5.Evaluate integral at t=5: Evaluate the integral at the upper limit t=5. Substitute t=5 into the integrated function to get (10e^(0.2*5) - (5^2)/2) which simplifies to (10e^1 - 12.5).Evaluate integral at t=0: Evaluate the integral at the lower limit t=0. Substitute t=0 into the integrated function to get (10e^(0.2*0) - (0^2)/2) which simplifies to (10*1 - 0), or just 10.Subtract values for total growth: Subtract the value of the integral at t=0 from the value at t=5 to find the total population growth. Subtracting the lower limit from the upper limit gives us (10e^1 - 12.5) - 10. We calculate 10e^1 as approximately 10*2.71828, which is about 27.1828. So, (10e^1 - 12.5) - 10 is approximately (27.1828 - 12.5) - 10, which equals 14.6828 - 10, or approximately 4.6828.

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