Answer
$$
\left( f(x)=1000, \quad r =5, \quad 6 \% \text {per yr.} \right)
$$
If $f(t)=1000 $ is the rate of continuous money flow at an interest rate $r=0.06$ for $T=5 $ years, then the accumulated amount of money flow at time $T=5$ is
$$
\begin{aligned} A &=e^{0.06(5)} \int_{0}^{5} 1000 e^{-0.06 t} d t \\
& \approx 5830.980126 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=1000, \quad r =5, \quad 6 \% per yr \right)
$$
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)=1000 $ is the rate of continuous money flow at an interest rate $r=0.06$ for $T=5 $ years, then the accumulated amount of money flow at time $T=5$ is
$$
\begin{aligned} A &=e^{0.06(5)} \int_{0}^{5} 1000 e^{-0.06 t} d t \\
&=\left.e^{0.3}\left(\frac{1000}{-0.06} e^{-0.06 t}\right)\right|_{0} ^{5} \\
&=e^{0.3}\left[\frac{1000}{-0.06}\left(e^{-0.3}-1\right)\right] \\
& \approx 5830.980126 \end{aligned}
$$
So, the accumulated amount of money flow at time $T=5$ is $\approx 5830.98.$
