Application of Integration on Investments

Question

A small business expects an income stream of $\$300$ per month for a period of $9$ years. The income will be invested at an annual interest rate of $17\%$, compounded continuously. How much interest was earned?

I already have the answer, i.e. $\$2610$ however, I don't know how to get it. I tried using the formula for compounded interest i.e., $$A=Pe^{rt}$$ but doesn't give the same answer. I believe my formula is not right and I don't know which one to use. This is a calculus problem on integration but I don't know how to start.

Answer 1

Answer:

Continously compounded rate $r_c = ln(1+\frac{.017}{12}) = 0.00141566 $

$$I + P = 300(1+e^{r_c*1} + e^{r_c*2} + e^{r_c*3}+\cdots+e^{r_c*107})$$

$$ e^{r_c} = 1.00141667 = a$$

$$I+P = 300(1+a+a^2+a3+\cdots+a^{107}) = \frac{300(a^{108}-1)}{a-1} = \frac{300\times 0.165198812827715}{0.00141667} = 35008.05198585$$

$$P\text{_only} = 300*108 = 32400$$

$$I = 35008-32400 = 2608.05$$

Answer 2

Assuming that income arrives and can be invested continuously, as is clear from the problem, the total expected value at the end of nine years will be $$\int_0^9 300\cdot 12 \cdot {\rm e}^{0.017(9-t)}\,{\rm d}t = -{3600\over 0.017}{\rm e}^{0.017(9-t)}\Bigm\vert_0^9={3600\over 0.017}\Bigl({\rm e}^{0.017\cdot 9}-1\Bigr)=35\,010$$ With no interest, the amount would have been $300\cdot 12\cdot 9=32\,400$, so the interest added is $$ 35\,010 - 32\,400 = 2\,610$$