Answer
The change in weight of the population from $t=0$ to $t=3$ is given by:
$$
\begin{aligned}
w(t)&=\int_{0}^{3}(3 t+2)^{1 / 3} d t\\
&=\frac{1}{4}\left(11^{4 / 3}-2^{4 / 3}\right)\\
& \approx 5.486 \mathrm{mg}
\end{aligned}
$$
Work Step by Step
A population of $E$. coli bacteria will grow at a rate given by
$$
w^{\prime}(t)=(3 t+2)^{1 / 3}
$$
where $w$ is the weight (in milligrams) after $t$ hours.
The change in weight of the population from $t=0$ to $t=3$ is given by:
$$
\begin{aligned}
w(t)&=\int_{0}^{3}(3 t+2)^{1 / 3} d t\\
&=\left.\frac{1}{3} \cdot \frac{(3 t+2)^{4 / 3}}{\frac{4}{3}}\right|_{0} ^{3}\\
&=\left. \frac{(3 t+2)^{4 / 3}}{4}\right|_{0} ^{3}\\
&=\frac{1}{4}\left(11^{4 / 3}-2^{4 / 3}\right)\\
& \approx 5.486 \mathrm{mg}
\end{aligned}
$$
