It is projected that t years from now the population of a city will be changing at the rate of $t ^ { 2 } e ^ { 0.01 t }$ thousand people per year. If the current population is 1 million, what will the population be 4 years from now?
A) 1,021,984
B) 1,028,800
C) 1,028,000
D) 1,000,022

Question

Quizplus · Accepted Answer

The population $P$ (in thousands) $t$ years from now is given by the integral of the rate of change:$$P(t) = \int t^2 e^{0.01t} \ dt$$Using the substitution $u = 0.01t$, $du = 0.01 \ dt$, we get$$P(t) = \int 100u^2 e^u \ du = 100 \int u^2 e^u \ du$$Integrating by parts, with $u^2$ as the first function and $e^u$ as the second function, we get$$P(t) = 100 \left[ u^2 e^u - \int 2ue^u \ du \right]$$Integrating by parts again, with $2u$ as the first function and $e^u$ as the second function, we get$$P(t) = 100 \left[ u^2 e^u - 2ue^u + \int 2e^u \ du \right]$$$$P(t) = 100 \left[ u^2 e^u - 2ue^u + 2e^u \right] + C$$Substituting back $u = 0.01t$, we get$$P(t) = 100 \left[ (0.01t)^2 e^{0.01t} - 2(0.01t)e^{0.01t} + 2e^{0.01t} \right] + C$$$$P(t) = 100 \left[ 0.0001t^2 e^{0.01t} - 0.02te^{0.01t} + 2e^{0.01t} \right] + C$$Since the current population is 1 million, we have$$P(0) = 1000 = 100 \left[ 0.0001(0)^2 e^{0.01(0)} - 0.02(0)e^{0.01(0)} + 2e^{0.01(0)} \right] + C$$$$C = 1000$$Therefore, the population $t$ years from now is$$P(t) = 100 \left[ 0.0001t^2 e^{0.01t} - 0.02te^{0.01t} + 2e^{0.01t} \right] + 1000$$4 years from now, the population will be$$P(4) = 100 \left[ 0.0001(4)^2 e^{0.01(4)} - 0.02(4)e^{0.01(4)} + 2e^{0.01(4)} \right] + 1000$$$$P(4) = 100 \left[ 0.0016 e^{0.04} - 0.08e^{0.04} + 2e^{0.04} \right] + 1000$$$$P(4) = 100 \left[ 1.9164 \right] + 1000$$$$P(4) = 1021.984$$Therefore, the population 4 years from now will be 1,021,984.

It Is Projected That T Years from Now the Population t2e0.01tt ^ { 2 } e ^ { 0.01 t }t2e0.01t

It Is Projected That T Years from Now the Population $t ^ { 2 } e ^ { 0.01 t }$