the birth rate of a population is b(t)=2100e^{0.023t} people per year and the death rate is…

Explanation:

Step1: Recall the concept of the area between curves

The area (A) between two curves (y = f(t)) and (y = g(t)) from (t=a) to (t = b) is given by (A=\int_{a}^{b}|f(t)-g(t)|dt). Here, (f(t)=2100e^{0.023t}) (birth - rate function) and (g(t)=1450e^{0.018t}) (death - rate function), (a = 0), (b = 10). Since (2100e^{0.023t}\geq1450e^{0.018t}) for (t\geq0), the area (A=\int_{0}^{10}(2100e^{0.023t}-1450e^{0.018t})dt).

Step2: Integrate term - by - term

We know that (\int e^{kt}dt=\frac{1}{k}e^{kt}+C) ((k\neq0)). (\int_{0}^{10}(2100e^{0.023t}-1450e^{0.018t})dt=2100\int_{0}^{10}e^{0.023t}dt - 1450\int_{0}^{10}e^{0.018t}dt). For (\int_{0}^{10}e^{0.023t}dt=\left[\frac{1}{0.023}e^{0.023t}\right]{0}^{10}=\frac{1}{0.023}(e^{0.023\times10}-e^{0})=\frac{1}{0.023}(e^{0.23}-1)). For (\int{0}^{10}e^{0.018t}dt=\left[\frac{1}{0.018}e^{0.018t}\right]_{0}^{10}=\frac{1}{0.018}(e^{0.018\times10}-e^{0})=\frac{1}{0.018}(e^{0.18}-1)).

Step3: Calculate the values

(2100\times\frac{1}{0.023}(e^{0.23}-1)-1450\times\frac{1}{0.018}(e^{0.18}-1)). (e^{0.23}\approx1.2586), (e^{0.18}\approx1.1972). (2100\times\frac{1}{0.023}(1.2586 - 1)-1450\times\frac{1}{0.018}(1.1972 - 1)). (2100\times\frac{1}{0.023}\times0.2586-1450\times\frac{1}{0.018}\times0.1972). (2100\times11.2435 - 1450\times10.9556). (23611.35-15885.62). (\approx7726).

The area between the curves represents the increase in population over a 10 - year period.

Answer:

The area represents the increase in population over a 10 - year period. The value of the area (increase in population) is approximately (7726).