Question

The function f(x) = 900 represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded continuously. Find (A) the present value, and (B) the accumulated amount of money flow at t = 10. (A) The present value is $ (Do not round until the final answer. Then found to the nearest cent as needed) (B) The accumulated amount of money flow at t = 1053 (Do not round until the final answer. Then round to the nearest cent as needed.)

Answer 1

Given f(x)=900, r=0.08, T=10

a) present value at time t is:

$P=\int_{0}^{T}f(t)e^{-rt}dt$

  $=\int_{0}^{10}900e^{-0.08x}dx$

  $=900\int_{0}^{10}e^{-0.08x}dx$

  $=(-90000/8)[e^{-0.08x}]_0^{10}$

$=-11250[e^{-0.8}-1]$

$\approx 6195.05$

(b) Accumulated value $A=e^{rt}(p)$

At t=10,

$A=e^{0.08*10}(6195.05)=e^{0.8}(6195.05)$

A=13787.34