Answer
The additional amount of area covered by the moss between 4 and 9 days is given by:
$$
\begin{aligned}
\int_{4}^{9} \sqrt{t} \ln t d t &=[\frac{2}{3} t^{3 / 2} \ln t-\frac{4}{9}t^{3 / 2}]_{4}^{9} \\
&=18 \ln 9-\frac{16}{3} \ln 4-\frac{76}{9} \\
& \approx 23.71 \mathrm{sq} \mathrm{cm}
\end{aligned}
$$
Work Step by Step
The area covered by a patch of moss is growing at a rate of
$$
A^{\prime}(t)=\sqrt {t}\ln t,
$$
the additional amount of area covered by the moss between 4 and 9 days is given by:
$$
\int_{4}^{9}\sqrt {t}\ln t d t
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=\ln t, \quad\quad dv= \sqrt {t}dt} \\ {d u= \frac{dt}{t} , \quad\quad v=\frac{2}{3} t^{\frac{3}{2}} }\end{array}\right] , \text { then }\\
$$
First , we find indefinite integral
$$
\begin{aligned} \int \sqrt{t} \ln t d t &=\frac{2}{3} t^{3 / 2} \ln t-\int\left(\frac{2}{3} t^{3 / 2} \cdot \frac{1}{t}\right) d t \\
&=\frac{2}{3} t^{3 / 2} \ln t-\int \frac{2}{3} t^{1 / 2} d t \\ &=\frac{2}{3} t^{3 / 2} \ln t-\frac{4}{9} t^{3 / 2}+C \end{aligned}
$$
Now, we obtain the define integral
$$
\begin{aligned}
\int_{4}^{9} \sqrt{t} \ln t d t &=[\frac{2}{3} t^{3 / 2} \ln t-\frac{4}{9}t^{3 / 2}]_{4}^{9} \\
&=18 \ln 9-\frac{16}{3} \ln 4-\frac{76}{9} \\
& \approx 23.71 \mathrm{sq} \mathrm{cm}
\end{aligned}
$$
Therefore the total accumulated growth during the first 2 days is $23.71$
