Homework question about future value of a given income stream. I'm trying to find the future value of a given income stream.
$R(t) = 28000$, $0 < t < 10$, at $4\%$
I think the formula is something like:
FV = integral(28,000)((e^.4-.04t)/365)dt with a domain of [0,10].
I think the answer is 280,153 but when I submit my answer my homework page says it's incorrect. 
 A: I take it the income stream is at $28000$ a year for $10$ years. The future value, with rate $r$, is given by
$$\int_0^{10} 28000 e^{r(10-t)}\,dt.$$
We can rewrite this as 
$$28000e^{10r}\int_0^{10}e^{-rt}\,dt.$$
Integrate. An antiderivative is $-\dfrac{1}{r}e^{-rt}$. Substitute the end points in the usual way. We get
$$28000e^{10r}\left(\frac{1}{r}\right)(1-e^{-10r}).$$ 
Now the only remaining issue is "What is $r$ ?"  The wording is not clear on this. 
(i) It could be $0.04$. Then the effective annual rate would be $e^{0.04}-1$, about $4.08\%$. Not a big difference, but in the long run it can be significant.  
(ii) Or else it could be the number $r$ such that $e^r=1.04$, that is, $\ln(1.04)$. This is about $0.0392207$. That would make $0.04$ the effective annual rate.
You may know how interest rate in continuous situations is specified in your course. And you can make the calculation in both cases, see which one corresponds to the given answer. 
We first calculate the answer for possibility (i), $r=0.04$. That gives answer $344277.29$.
We next calculate  the answer for possibility (ii), that $0.04$ is the effective annual interest rate. The result (but I am error-prone with a calculator) is about $342850.47$.
The answer under (ii) is lower, because the effective interest rate has gone from about $0.0408$ to $0.04$. 
