Answer
$$
\left( f(x)=20t, \quad T =6, \quad 4\% \text{ per yr. } \right)
$$
the accumulated amount of money flow at time $T$ is
$$
\begin{aligned} A & =20 e^{0.24} \int_{0}^{6} t e^{-0.04 t} d t \\
& \approx 390.614379
\end{aligned}
$$
So, the accumulated amount of money flow at time $T=6$ is $\approx 390.614.$
Work Step by Step
$$
\left( f(x)=20t, \quad T =6, \quad 4\% \text{ per yr. } \right)
$$
If $f(t)$ is the rate of money flow at an interest rate $r$ at time $t$, the accumulated amount of money flow at time $T$ is
$$
A=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt.
$$
If $f(t)=20t $ is the rate of continuous money flow at an interest rate $r=0.4$ for $T=6$ years, then the accumulated amount of money flow at time $T=6$ is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt\\
&=e^{0.04(6)} \int_{0}^{6} 20 t e^{-0.044 t} d t \\
&=20 e^{0.24} \int_{0}^{6} t e^{-0.04 t} d t \end{aligned}
$$
First, we evaluate the indefinite integral as follows:
$$
\int_{0}^{6} t e^{-0.04 t} d t
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=t, \quad\quad dv= e^{-0.04 t} d t } \\ {d u= dt, \quad\quad v=\frac{1}{-0.04} e^{-0.08 t} }\end{array}\right] ,
$$
$$
\begin{aligned} \int t e^{-0.044 t} d t &=-25 t e^{-0.04 t}-\int\left(-25 e^{-0.04 t}\right) d t \\ &=-25 t e^{-0.04 t}-625 e^{-0.04 t}+C \end{aligned}
$$
Now we will evaluate the definite integral as follows:
$$
\begin{aligned} A & =20 e^{0.24} \int_{0}^{6} t e^{-0.04 t} d t \\
&=\left.20 e^{0.24}\left(-25 t e^{-0.04 t}-625 e^{-0.04 t}\right)\right|_{0} ^{6}\\
&=20 e^{0.24}\left[\left(-150 t e^{-0.24}-625 e^{-0.24}\right)-(0-625)\right]\\
&=20\left(625 e^{0.24}-775\right)\\
& \approx 390.614379
\end{aligned}
$$
So, the accumulated amount of money flow at time $T=6$ is $\approx 390.614.$
