# Average value, integrals, compound interest

An amount of money $M$ compounded continuously at interest rate $r$ increases according to the law $M(t)=Ce^{rt}$ ($t$ = time in years, $C$ = the initial amount).

What is the average amount of money A in the bank over from 0 to T years?

I thought I would do something like:

$$A = {C \over T} \int_{0}^{T} e^{rt}$$

Evaluating with respect to $t$ I think? Would give me:

$$Ce^{rT} \over Tr$$

But I'm getting the incorrect answer. How do I go about doing this?

• Close, but the integral is $\frac{1}{r}(e^{rT}-1)$. You did not evaluate at the lower limit. Apr 20, 2016 at 20:47
• oh i see! I saw the zero and thought, like many others that I've done that the lower limit would evaluate to zero but it is indeed $e^{0} = 1$ Apr 20, 2016 at 21:41

$$A=\frac{1}{T}\int_0^TC \mathrm e^{rt}\mathrm d t=\frac{C}{T}\left[\frac{\mathrm e^{rt}}{r}\right]_0^T=\frac{C}{T}\left[\frac{\mathrm e^{rT}}{r}-\frac{\mathrm e^{r\cdot 0}}{r}\right]=\frac{C}{Tr}\left(\mathrm e^{rT}-1\right)$$