Answer
The total accumulated growth 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 $6054$
