Payments are made to an account at a continuous rate of [math]100e^{0.5t}[/math] from time t = 1 to time t = 3. The force of interest for this account is [math]\delta = 8\%[/math].

Calculate the value of the account at time t = 4.

• 313
• 432
• 477
• 606
• 657

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

The accumulated value to time 4 is

[[math]] \int_{1}^{3}100e^{0.5t}e^{0.88(4-t)}d t=100e^{0.32}\int_{1}^{3}e^{0.42t}d t=100e^{0.32}\frac{e^{0.42t}}{0.42}\biggl|_{1}^{3}=\frac{100e^{0.32}(e^{1.26}-e^{0.42})}{0.42}=657. [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.