Calculating a continuously varying, continuously paid annuity, using different focal dates I am working on this question: http://math.illinoisstate.edu/krzysio/KO-FM-Exercise143.pdf
Payments are made to an account at a continuous rate of $(8k+tk)$, where $0≤t≤10$. Interest is credited at a force of interest $\delta_t=\frac{8}{8+t}$. After 10 years, the account is worth 20,000. Calculate $k$.
I found an answer here Calculating a continuously varying, continuously paid annuity
I understand the method but I have a question. If, instead of calculating the present value of both the 20,000 and the annuity payments, I calculate the future value (after the 10 years), I end up with a different value for $k$:
$$
\int_{0}^{10}(8k+tk) \left( e^{ \int_{0}^{t}\frac{1}{8+s}ds } \right) dt = \int_{0}^{10}(8k+tk) \left( \frac{8+t}{8} \right) dt = k \int_{0}^{10}(8+t) \left( \frac{8+t}{8} \right) dt = \frac{665}{3}k
$$
Thus 
$\frac{665}{3}k = 20000 $ and $k=90.23$
I assumed that I could use any point as the focal point and still obtain the same answer. Can someone explain?
 A: Generally speaking, if $r$ is your rate of payment, it is not true that 
$$\text{FV} \neq \int_{0}^{T}r(t)\exp\left(\int_{0}^{t}\delta_s\text{ d}s \right)\text{ d}t\tag{1}$$
The correct formula is
$$\text{FV} = \exp\left(\int_{0}^{T}\delta_s\text{ d}s \right) \cdot \text{PV} = \exp\left(\int_{0}^{T}\delta_s\text{ d}s \right)\int_{0}^{T}r(t)\exp\left(-\int_{0}^{t}\delta_s\text{ d}s \right)\text{ d}t\text{.}\tag{2}$$
To see why $(2)$ is true: it's just that familiar concept of if you want the future value at time $T$, you take the present value (at time $0$) and "accumulate" it to time $T$.
To see why $(1)$ isn't true: let's consider the integrand 
$$r(t)\exp\left(\int_{0}^{t}\delta_s\text{ d}s \right)$$
intuitively: what you are doing is for each time $t$, you are multiplying by the force of interest which accumulates from time $0$ to time $t$. This doesn't do what you think it should do: what you want it to do is accumulate from time $t$ to $T$. So actually, the correct integrand - we would expect - is
$$r(t)\exp\left(\int_{t}^{T}\delta_s\text{ d}s \right)\text{.}$$
It turns out that this is correct, and is actually equivalent to $(2)$. To see this, notice
$$\begin{align}
\int_{0}^{T}r(t)\exp\left(\int_{t}^{T}\delta_s\text{ d}s \right)\text{ d}t &= \int_{0}^{T}r(t)\exp\left(\int_{0}^{T}\delta_s\text{ d}s-\int_{0}^{t}\delta_s\text{ d}s \right)\text{ d}t \\
&= \int_{0}^{T}r(t)\exp\left(\int_{0}^{T}\delta_s\text{ d}s \right)\exp\left( -\int_{0}^{t}\delta_s\text{ d}s\right)\text{ d}t \\
&= \exp\left(\int_{0}^{T}\delta_s\text{ d}s \right)\int_{0}^{T}r(t)\exp\left( -\int_{0}^{t}\delta_s\text{ d}s\right)\text{ d}t \\
&= (2)\text{.}
\end{align}$$
