Answer:
Step-by-step explanation:
Consider the function of the rate of flow of money in dollar per year is
[tex]f(t)= 400e^{0.03t}[/tex]
The objective is to find the present value of this income over 10 years period an assume an annual interest rate of 8% compounded continuously.
Given that,
[tex]f(t)= 400e^{0.03t}[/tex]
r = 0.08, t = 10
if f(t) is the rate of continuously money flow at an interest rate r to T year,
the present value is,
[tex]P= \int\limits^T_0 {f}(t)e^{-rt} \, dt[/tex]
Now input the values to get
[tex]p=\int\limits^{10}_0 400e^{0.03t}*e^{-0.08t} dt[/tex]
[tex]=400\int\limits^{10}_0 e^{0.03t}*^{-0.08t} dt[/tex]
[tex]= 400\int\limits^{10}_0 e^{-0.05t} dt[/tex]
[tex]=400(-\frac{e^{-0.05t}}{0.05} )|_0^1^0[/tex]
[tex]=-\frac{400}{0.05} (e^{-0.05*10}-e^{-0.05*0})\\\\=-\frac{400}{0.05} (e^{-0.5}-e^{*0})\\\\=3147.75[/tex]
