Answer
$$
\left( f(x)=15t, \quad r =18 \text { months}, \quad 8 \% \right)
$$
If $f(t)=15t $ is the rate of continuous money flow at an interest rate $r=0.08$ for $T=1.5 $ years, then the present value is
$$
\begin{aligned} P & =15 \int_{0}^{1.5} t e^{-0.08 t} d t\\
& = 15\left(-175 e^{-0.12}+156.25\right)\\
&\approx 15.58385362 \end{aligned}
$$
So, the present value is $\approx 15.584.$
Work Step by Step
$$
\left( f(x)=15t, \quad r =18 \text { months}, \quad 8 \% \right)
$$
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)=15t $ is the rate of continuous money flow at an interest rate $r=0.08$ for $T=1.5 $ years, then the present value is
$$
\int_{0}^{1.5} 15t e^{-0.08 t} d t
$$
First, we evaluate the indefinite integral as follows:
$$
\int 15t e^{-0.08 t} d t
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=t, \quad\quad dv= e^{-0.08 t} d t } \\ {d u= dt, \quad\quad v=\frac{1}{-0.08} e^{-0.08 t} }\end{array}\right] ,
$$
$$
\begin{aligned} \int t e^{-0.08 t} d t &=-12.5 t e^{-0.08 t}-\int\left(-12.5 e^{-0.08 t}\right) d t \\ &=-12.5 t e^{-0.08 t}-156.25 e^{-0.08 t}+C \end{aligned}
$$
Now we will evaluate the definite integral as follows:
$$
\begin{aligned} P & =15 \int_{0}^{1.5} t e^{-0.08 t} d t\\
&=\left.15\left(-12.5 t e^{-0.08 t}-156.25 e^{-0.08 t}\right)\right|_{0} ^{1.5}\\
&=15\left[\left(-18.75 e^{-0.12}-156.25 e^{-0.12}\right)\right.\\
& \quad-(0-156.25)] \\
& = 15\left(-175 e^{-0.12}+156.25\right)\\
&\approx 15.58385362 \end{aligned}
$$
So, the present value is $\approx 15.584.$
