Answer
$\left( \mathbf{a} \right)\$3443.1877,\text{ }\left( \mathbf{b} \right)\$7662.95$
Work Step by Step
\[\begin{align}
& f\left( t \right)=0.05t+500 \\
& \text{Let }T=10\text{ and }r=0.08 \\
& \\
& \left( \mathbf{a} \right)\text{The present Value is: } \\
& P=\int_{0}^{T}{f\left( t \right){{e}^{-rt}}dt} \\
& P=\int_{0}^{10}{\left( 0.05t+500 \right){{e}^{-0.08t}}dt} \\
& \text{Integrate by parts} \\
& \text{Let }u=0.05t+500,\text{ }du=0.05dt \\
& dv={{e}^{-0.08t}},\text{ }v=-12.5{{e}^{-0.08t}} \\
& =-12.5\left( 0.05t+500 \right){{e}^{-0.08t}}+\int{12.5{{e}^{-0.08t}}\left( 0.05 \right)dt} \\
& =-0.625t{{e}^{-0.08t}}-6250{{e}^{-0.08t}}+0.625\int{{{e}^{-0.08t}}dt} \\
& =-0.625t{{e}^{-0.08t}}-6250{{e}^{-0.08t}}-7.8125{{e}^{-0.08t}}+C \\
& =-0.625t{{e}^{-0.08t}}-6242.1875{{e}^{-0.08t}}+C \\
& \text{Therefore} \\
& P=\left[ -0.625t{{e}^{-0.08t}}-6242.1875{{e}^{-0.08t}} \right]_{0}^{10} \\
& P=\left[ -0.625\left( 10 \right){{e}^{-0.8}}-6242.1875{{e}^{-0.8t}} \right]-\left[ -6242.1875 \right] \\
& P\approx \$3443.1877 \\
& \\
& \left( \mathbf{b} \right)\text{The accumulated amount of money flow at time }T\text{ is:} \\
& A={{e}^{rT}}\int_{0}^{T}{f\left( t \right)}{{e}^{-rt}}dt \\
& A={{e}^{rT}}P \\
& A={{e}^{0.08\left( 10 \right)}}\left( \$3443.1877 \right) \\
& A\approx \$7662.95 \\
\end{align}\]
