Answer
$$
\left( f(x)=1000+200t, \quad T =10, \quad 9\% \text{ per yr. } \right)
$$
If $f(t)=1000+200t $ is the rate of continuous money flow at an interest rate $r=0.09$ for $T=10$ years, then the accumulated amount of money flow at time $T=10$ is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt\\
&=e^{(0.09)(10)} \int_{0}^{10} (1000+200 t) e^{-0.09 t} d t \\
&\approx \$ 30,035.17
\end{aligned}
$$
Work Step by Step
$$
\left( f(x)=1000+200t, \quad T =10, \quad 9\% \text{ per yr. } \right)
$$
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)=1000+200t $ is the rate of continuous money flow at an interest rate $r=0.09$ for $T=10$ years, then the accumulated amount of money flow at time $T=10$ is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(x) e^{-rt}dt\\
&=e^{(0.09)(10)} \int_{0}^{10} (1000+200 t) e^{-0.09 t} d t \\
&=\left.e^{0.9}\left[\frac{1000}{-0.09} e^{0.09 t}+\frac{200}{(0.09)^{2}}(-0.09 t-1) e^{-0.09 t}\right]\right|_{0} ^{10} \\
&=e^{0.9}\left[\frac{1000}{-0.09}\left(e^{-0.9}-1\right)+\frac{200}{(0.09)^{2}}\left(-1.9 e^{-0.9}+1\right)\right]\\
&\approx \$ 30,035.17
\end{aligned}
$$
So, the accumulated amount of money flow at time $T=10$ is $\approx 30,035.17$
