Calculate present value of a continuous stream of payments with variable force of interest. The continuous rate of payments is $\rho(t)= \begin{cases} 
      0 & t< 1 \\
      12 & 1\leq t\leq 2 \\
      10+t & 2\leq x\leq 5 \\
      15 & 5\leq t\leq 6 \\
      0 & t>6
   \end{cases},
$ and the force of interest is $\delta(t)= \begin{cases} 0.08 & t\leq 2 \\ \frac{1}{10+t} & 2<t\leq 5 \\ 0.06 & t>5 \end{cases} $.
The question is: What is the present value of the payments at time, $t = 0$?
I know that the present value is: $$\int_{0}^{T} \rho(s) e^{ -\int_0^s \delta(u) du} ds$$
However, I cannot figure out how to evaluate this integral using the two piecewise functions. The main problem is understanding which part of $\delta(t)$ function I should use in the exponent and how to integrate from $0$ to $s$ when we don't know what the value of $s$ is.
Thanks for any help.

As the functions are interval functions, and the integral depends on a parameter $T$, then the integrals results will be also interval type functions. So to start with the integral, you should first get the integral of the exponent. This integral($\int_0^T\delta(u)du$) will have also a 3 interval function as a result, being the answer:
$1^{st}$ interval : $s\leq2\rightarrow \int_0^s0.08du=0.08s$
$2^{nd}$ interval: $2<s\leq5\rightarrow\int_0^20.08du+\int_2^s\frac{1}{10+u}du=0.16+\ln{(10+s)}-\ln12=0.16+\ln{\frac{10+s}{12}}$
$3^{rd}$ interval: $s>5\rightarrow\int_0^20.08du+\int_2^5\frac{1}{10+u}du+\int_5^s0.06du=0.16+\ln{\frac{15}{12}}+0.06s - 0.3$ 
So that would be the answer for the integral of the exponent, to complete the integral, you should first get the combined intervals for function $\rho(s) $ and the exponent's integral and then integrate each of the intervals to get the final function dependant on T.
