Answer
$$\$32719.53$$
Work Step by Step
Since
\begin{align*}
\begin{aligned}
\left(-\frac{e^{-r t}(1+r t)}{r^{2}}+C\right)^{\prime} &=\frac{-1}{r^{2}}\left(-r \cdot e^{-r t}(1+r t)+e^{-r t}(0+r)\right)\\
&=\frac{-1}{r^{2}}\left(-r e^{-r t}-r^{2} t e^{-r t}+r e^{-r t}\right) \\
&=\frac{-1}{r^{2}}\left(-r^{2} t e^{-r t}\right) \\
&=t e^{-r t}
\end{aligned}
\end{align*}
Then
$$\int t e^{-r t} d t=-\frac{e^{-r t}(1+r t)}{r^{2}}+C$$
Since for $$ R(t)=5000+1000 t, \ \ \ \ \ T=5,\ \ \ \ r=0.05$$
Then
\begin{aligned}
P V &=\int_{0}^{T} R(t) e^{-r t} d t \\
&=\int_{0}^{5}(5000+1000 t) e^{-0.05 t} d t \\
&=5000 \int_{0}^{5} e^{-0.05 t} d t+1000 \int_{0}^{5} t e^{-0.05 t} d t\ \ \text{integrate by parts }\\
&= \left.\frac{5000}{-0.05} e^{-0.05 t}\right|_{0} ^{5}+ \left.1000\left(-\frac{e^{-0.05 t}(1+0.05 t)}{0.05^{2}}\right)\right|_{0} ^{5}\\
&= 22119.92+10599.61\\
&=\$32719.53
\end{aligned}
