Question

The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded continuously and find the following: (a) the present value; (b) the accumulated amount of money flow at t=10.
f(t)=0.01t+100

Answer 1

Answer:

a) The present value is 688.64 $

b) The accumulated amount is 1532.60 $

Step-by-step explanation:

a) The preset value equation is given by this formula:

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

where:

• T is the period in years (T = 10 years)
• r is the annual interest rate (r=0.08)

So we have:

$P=\int^{T}_{0}(0.01t+100)e^{-rt}dt$

Now we just need to solve this integral.

$P=\int^{T}_{0}0.01te^{-rt}dt+\int^{T}_{0}100e^{-rt}dt$

$P=e^{-0.08t}(-1.56-0.13t)|^{10}_{0}+1250e^{-0.08t}|^{10}_{0}$

$P=0.30+688.34=688.64$

The present value is 688.64 $

b) The accumulated amount of money flow formula is:

$A=e^{r\tau}\int^{T}_{0}f(t)e^{-rt}dt$

We have the same equation but whit a term that depends of τ, in our case it is 10.

So we have:

$A=e^{r\tau}\int^{T}_{0}(0.01t+100)e^{-rt}dt=e^{0.08\cdot 10}P$

$A=e^{0.08\cdot 10}688.64=1532.60$

The accumulated amount is 1532.60 $

Have a nice day!