Answer
The population at the beginning of the year 2015 is 130,004
Work Step by Step
\[
\begin{array}{l}
\because P^{\prime}(t)=(0.12 t+3)^{\frac{3}{2}} \\
\therefore P(t)=\int(0.12 t+3)^{\frac{3}{2}} d t
\end{array}
\]
By using the substitution $u=0.12 t+3 \Rightarrow d u=0.12 d t$
\[
P(t)=\int(0.12 t+3)^{\frac{3}{2}} d t=\frac{1}{0.12} \int u^{\frac{3}{2}} d u
\]
\[
\begin{array}{l}
=\frac{2}{.12 \times 5} u^{\frac{5}{2}}+C \\
P(t)=\frac{1}{0.3}(0.12 t+3)^{\frac{5}{2}}+C
\end{array}
\]
$P(0)=\frac{1}{0.3}(0.12 t(0)+3)^{\frac{5}{2}}+C=100$
\[
\Rightarrow 100=30 \sqrt{3}+C \Rightarrow[C=-30 \sqrt{3}+100 \approx 48.038
\]
So
\[
P(t)=\frac{1}{0.3}(0.12 t+3)^{\frac{5}{2}}+48.038
\]
At the beginning of the year $2015,$ $t=5$ (in years)
$P(5)=48.038+\frac{1}{0.3}(0.12(5)+3)^{\frac{5}{2}}=130$
