Answer
$$
\left( f(x)=150,e^{0.04t}, \quad r =5, \quad 6 \% \right)
$$
If $f(t)=150e^{0.04t} $ is the rate of continuous money flow at an interest rate $r=0.06$ for $T=5 $ years, then the present value is
$$
\begin{aligned} P & =\int_{0}^{T} f(x) e^{-rt}dt\\
&=\int_{0}^{5} 150 e^{0.04 t} e^{-0.06 t} d t \\
& \approx 713.7193647 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=150,e^{0.04t}, \quad r =5, \quad 6 \% \right)
$$
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)=150e^{0.04t} $ is the rate of continuous money flow at an interest rate $r=0.06$ for $T=5 $ years, then the present value is
$$
\begin{aligned} P & =\int_{0}^{T} f(x) e^{-rt}dt\\
&=\int_{0}^{5} 150 e^{0.04 t} e^{-0.06 t} d t \\
&=\int_{0}^{5} 150 e^{-0.02 t} d t \\
&=\left.\frac{150}{-0.02} e^{-0.02 t}\right|_{0} ^{5} \\ &=-7500\left(e^{-0.1}-e^{0}\right) \\ &=-7500\left(e^{-0.1}-1\right) \\
& \approx 713.7193647 \end{aligned}
$$
