Answer
The statement is True.
Work Step by Step
The statement is True.
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 $f(t)=1000e^{-0.05t}$ is the rate of continuous money flow at an interest rate $r=0.0045$ for $T=5$ years, then the present value is
$$
\begin{aligned}
P& =\int_{0}^{T} f(x) e^{-rt}dt\\
&=\int_{0}^{5} (1000e^{-0.05t})e^{-0.0045t}dt\\
&=\int_{0}^{5} (1000e^{-0.005t})dt\\
\end{aligned}
$$
Therefore, the statement is True.
