Answer
$$
\left( f(x)=25,000, \quad r =12, \quad 10 \% \right)
$$
If $f(t)=25,000$ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=12$ years, then the present value is
$$
\begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\
&=\int_{0}^{12} 25000 e^{-0.1 t} d t \\
& \approx \$ 174701.45 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=25,000, \quad r =12, \quad 10 \% \right)
$$
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 $f(t)=25,000$ is the rate of continuous money flow at an interest rate $r=0.1$ for $T=12$ years, then the present value is
$$
\begin{aligned} P &=\int_{0}^{T} f(x) e^{-rt}dt \\
&=\int_{0}^{12} 25000 e^{-0.1 t} d t \\
&=\left.\frac{25,000}{-0.1} e^{-0.1 t}\right|_{0} ^{12} \\
&=250000(-e^{-01.2}+1) \\
&=250000(-0.3012+1) \\
& \approx \$ 174701.45 \end{aligned}
$$