On each of two days (January 1, 2005 and January 1, 2006), a person deposits $500 in a savings bank.The bank deposits interest in the account at a rate of 8%, compounded continuously.Choose the sum/integral that gives the size of the bank account at the end of 2006.
A) $\int_{0}^{2} 500 e^{0.08 t} d t$
B) $\int_{0}^{2}\left(500+500 e^{0.08 t}\right) d t$
C) $500+500 e^{0.08}$
D) $\left(500+500 e^{0.08}\right) e^{0.08}$

Question

Quizplus · Accepted Answer

The first deposit grows for 2 years and the second deposit grows for 1 year. The formula for continuous compounding is $A = Pe^{rt}$. The first deposit becomes $500e^{0.08 \times 2}$ and the second deposit becomes $500e^{0.08 \times 1}$. The total is $(500e^{0.16} + 500e^{0.08})$, which simplifies to $(500 + 500e^{0.08})e^{0.08}$.

On Each of Two Days (January 1, 2005 and January ∫02500e0.08tdt\int_{0}^{2} 500 e^{0.08 t} d t∫02​500e0.08tdt

On Each of Two Days (January 1, 2005 and January $\int_{0}^{2} 500 e^{0.08 t} d t$