Answer
$$
\left( f(x)=Ce^{kt} \text{ where }C=1000, \quad k =0.05 \right)
$$
If $f(t)=1000 e^{0.05t} $ is the rate of continuous money flow at an interest rate $r=0.11$ for $T=7$ years, then the present value is
$$
\begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt\\
&=\int_{0}^{7} 1000 e^{0.05 t} \cdot e^{-0.11 t} d t \\
&\approx \$ 5715.89 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=Ce^{kt} \text{ where }C=1000, \quad k =0.05, \right)
$$
If $f(t)$ is the rate of continuous money flow at an interest rate $r$ for $T$ years, then the present value is
$$
P=\int_{0}^{T} f(x) e^{-rt}dt.
$$
If $f(t)=1000 e^{0.05t} $ is the rate of continuous money flow at an interest rate $r=0.11$ for $T=7$ years, then the present value is
$$
\begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt\\
&=\int_{0}^{7} 1000 e^{0.05 t} \cdot e^{-0.11 t} d t \\
&=1000 \int_{0}^{7} e^{-0.06 t} d t \\
&=\left.1000\left(\frac{1}{-0.06} e^{-0.06 t}\right)\right|_{0} ^{7} \\ &=\frac{1000}{-0.06}\left(e^{-0.42}-1\right) \\
&\approx \$ 5715.89 \end{aligned}
$$
