What is the difference among three kinds of continuous income stream? In the chapter of our book  , we discuss

*

*Total value of continuous income stream:$\int_a^bR(t)dt$

*Future value of continuous income stream:$\int_a^bR(t)e^{r(b-t)}dt$

*Present value of continuous income stream:$\int_a^bR(t)e^{r(a-t)}dt$
But I can't understand the difference between them. Here is an example question:

You being saving for your retirement by investing $700 per month in an annuity with a guaranteed interest rate of 6% per year. You increase the amount you invest at the rate of 3% per year. With continuous investment and compounding, how much will you have accumulated in the annuity by the time you retire in 45 years?

 A: Let us say he is in this 35 and earning. The way you could solve the problem that you have stated is: $A_0$ is 700, payment growth rate(i) is 3%, monthly interest rate (r) is $\frac{6\text{%}}{12}  = 0.5\text{%}$ and number of years (T) is 10 years.
First Annuity $= A_0 \frac{e^{12\times rT} - 1}{e^r - 1}$  this is the formula for an annuity with cashflow of 700 going out to T years.  The additional increase in cashflow whcih is $i\times A_0$ forms the second annuity. In the third year, $i(i+1)A_0$ becomes the third annuity and so on with (T-1)th annuity with $i(i+1)^{T-2}A_o$.  Thus
First Annuity $$FV_1= A_0 \frac{e^{12\times rT} - 1}{e^r - 1}$$
Second Annuity $$FV_2=iA_0\frac{e^{(12\times r(T-1))} - 1}{e^r - 1}$$
Third Annuity  $$FV_3=i(i+1)A_0\frac{e^{(12\times r(T-2))} - 1}{e^r - 1}$$
...
Last Annuity $$FV_{(T-1)}=i(i+1)^{T-2} \frac{e^{(12\times r(T-(T-1))} - 1}{e^r - 1}$$
$$==i(i+1)^{T-2} \frac{e^{(12\times r)} - 1}{e^r - 1}$$
Sum of all FVs will give the future value of your investment compounded continuously for 10 years and get the amount when you retire at 45.  The investment would have grown to 129 thousand and seven hundred approximately.
I have written a program to do and the  seen as an image below.  If you want the EXCEL Sheeet , let me know I shall send it to you.  Goodluck

