Chapter 8 - Further Techniques and Applications of Integration - 8.3 Continuous Money Flow - 8.3 Exercises - Page 447: 11

$\left( \mathbf{a} \right)\$688.64,\text{ }\left( \mathbf{b} \right)\$1532.59$

\[\begin{align} & f\left( t \right)=0.01t+100 \\ & \text{Let }T=10\text{ and }r=0.08 \\ & \\ & \left( \mathbf{a} \right)\text{The present Value is: } \\ & P=\int_{0}^{T}{f\left( t \right){{e}^{-rt}}dt} \\ & P=\int_{0}^{10}{\left( 0.01t+100 \right){{e}^{-0.08t}}dt} \\ & \text{Integrate by parts} \\ & \text{Let }u=0.01t+100,\text{ }du=0.01dt \\ & dv={{e}^{-0.08t}},\text{ }v=-12.5{{e}^{-0.08t}} \\ & =-12.5\left( 0.01t+100 \right){{e}^{-0.08t}}+\int{12.5{{e}^{-0.08t}}\left( 0.01 \right)dt} \\ & =-12.5\left( 0.01t+100 \right){{e}^{-0.08t}}+0.125\int{{{e}^{-0.08t}}dt} \\ & =-12.5\left( 0.01t+100 \right){{e}^{-0.08t}}-1.5625{{e}^{-0.08t}}+C \\ & \text{Therefore} \\ & P=\left[ -12.5\left( 0.01t+100 \right){{e}^{-0.08t}}-1.5625{{e}^{-0.08t}} \right]_{0}^{10} \\ & P=\left[ -12.5\left( 0.01\left( 10 \right)+100 \right){{e}^{-0.08\left( 10 \right)}}-1.5625{{e}^{-0.08\left( 10 \right)}} \right] \\ & -\left[ -12.5\left( 0.01\left( 0 \right)+100 \right){{e}^{0}}-1.5625{{e}^{0}} \right] \\ & P=-1251.25{{e}^{-0.8}}-1.5625{{e}^{-0.8}}+1251.5625 \\ & P=-1252.8125{{e}^{-0.8}}+1251.5625 \\ & P\approx \$688.64 \\ & \\ & \left( \mathbf{b} \right)\text{The accumulated amount of money flow at time }T\text{ is:} \\ & A={{e}^{rT}}\int_{0}^{T}{f\left( t \right)}{{e}^{-rt}}dt \\ & A={{e}^{rT}}P \\ & A={{e}^{0.08\left( 10 \right)}}\left( \$688.64 \right) \\ & A\approx \$1532.59 \\ \end{align}\]