Answer
$$\$71460.53 $$
Work Step by Step
Since $$ R(t)=(5000+1000 t) e^{0.02 t}, \ \ \ T=10,\ \ r=0.08$$
Then
\begin{aligned}
P V &=\int_{0}^{T} R(t) e^{-r t} d t \\
&=\int_{0}^{10}(5000+1000 t) e^{-0.06 t} d t \\
&=5000 \int_{0}^{10} e^{-0.06 t} d t+1000 \int_{0}^{10} t e^{-0.06 t} d t\ \ \text{integrate by parts }\\
&= \left.\frac{5000}{-0.06} e^{-0.06 t}\right|_{0} ^{10}+ \left.1000\left(-\frac{e^{-0.06 t}(1+0.06 t)}{0.06^{2}}\right)\right|_{0} ^{10}\\
&= 37599.03+33861.5\\
&=\$71460.53
\end{aligned}
