Answer
The total accumulated growth of the microbe
population during the first 2 days is given by:
$$
\begin{aligned}
\\ \int_{0}^{2} 27 t e^{3 t} d t \approx 6054. \end{aligned} $$
Work Step by Step
The rate of growth of a microbe population is given by $$ m^{\prime}(t)=27te^{3t}, $$ the total accumulated growth during the first 2 days is given by $$ \int_{0}^{2} 27te^{3t} d t $$ use integration by parts with $$ \quad\quad\quad \left[\begin{array}{c}{u=27t, \quad\quad dv= e^{3t}dt} \\ {d u= 27dt , \quad\quad v=\frac{e^{3t}}{3} }\end{array}\right] , \text { then }\\ $$ First , we find indefinite integral $$ \begin{aligned} \int 27 t e^{3 t} d t &=27 t \cdot \frac{e^{3 t}}{3}-\int \frac{e^{3 t}}{3} \cdot 27 d t \\ &=9 t e^{3 t}-3 e^{3 t} \end{aligned} $$ Now, we obtain the define integral $$ \begin{aligned} \\ \int_{0}^{2} 27 t e^{3 t} d t &=\left.\left(9 t e^{3 t}-3 e^{3 t}\right)\right|_{0} ^{2} \\ &=\left(18 e^{6}-3 e^{6}\right)-(0-3) \\ &=15 e^{6}+3\\ & \approx 6054. \end{aligned} $$ Therefore the total accumulated growth during the first 2 days is $6054$
