Answer
a. $2065;
b. $4595.8
Work Step by Step
a. $P=\int ^{10}_{0} 300e^{-0.08t}=300\frac{e^{-0.08t}}{-0.08}|^{10}_{0}=-3750(e^{-0.8}-1)=2065.0164 \approx 2065$
b. $A=e^{(0.08)10}\int^{10}_{0}300e^{-0.08t}dt = e^{0.8}.300\int^{10}_{0}e^{-0.08t}dt=e^{0.8}.300.\frac{1}{-0.08}(e^{-0.08t}|^{10}_{0}) \approx 4595.8$
