Answer
$$
\left( f(x)=500e^{-0.04t}, \quad r =8, \quad 10 \% per yr \right)
$$
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)=500e^{-0.04t} $ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=8$ years, then the accumulated amount of money flow at time $T=8$ is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt.\\
&=e^{0.1(8)} \int_{0}^{8} 500 e^{-0.04 t} \cdot e^{-0.1 t} d t \\
&=e^{0.8} \int_{0}^{8} 500 e^{-0.14 t} d t \\ &=\left.e^{0.8}\left(\frac{500}{-0.14} e^{-0.14 t}\right)\right|_{0} ^{8} \\
&=e^{0.8}\left[\frac{500}{-0.14}\left(e^{-1.12}-1\right)\right] \\
& \approx 5354.971041 \end{aligned}
$$
So, the accumulated amount of money flow at time $T=8$ is $\approx 5354.97.$
Work Step by Step
$$
\left( f(x)=500e^{-0.04t}, \quad r =8, \quad 10 \% per yr \right)
$$
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)=500e^{-0.04t} $ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=8$ years, then the accumulated amount of money flow at time $T=8$ is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt.\\
&=e^{0.1(8)} \int_{0}^{8} 500 e^{-0.04 t} \cdot e^{-0.1 t} d t \\
&=e^{0.8} \int_{0}^{8} 500 e^{-0.14 t} d t \\ &=\left.e^{0.8}\left(\frac{500}{-0.14} e^{-0.14 t}\right)\right|_{0} ^{8} \\
&=e^{0.8}\left[\frac{500}{-0.14}\left(e^{-1.12}-1\right)\right] \\
& \approx 5354.971041 \end{aligned}
$$
So, the accumulated amount of money flow at time $T=8$ is $\approx 5354.97.$
