I have (hopefully) an easy question.
Suppose I have the following system (derived from an integration)
$I'(t)= f(t), \, t \in [0,\pi]\\ I(0)=0$
My goal is to find $I(\pi)= \int_{0}^{\pi}f(t) \text{d}t$ using a RK method on the differential setting.
Since $f$ does not depend on $I$, a RK method of size $m \geq1$ can be described only through the vectors $b,c \in \mathbb{R}^{m}$.
Suppose we divide $[0,\pi]$ into a collection of equidistant $N_{t}>1$ points. We then have $\Delta t = \frac{\pi}{N_{t}-1}$ This means $t_{k}=k \Delta t $ for $k \in \lbrace 0, \ldots, N_{t}-1\rbrace$.
A single step of Runge Kutta reads: \begin{align} I^{k+1}-I^{k}= \Delta t \sum_{i=1}^{m} b_{i} f(t_{k}+c_{i} \Delta t) \end{align} summing over all $N_{t}$ points we get \begin{align} I^{N_{t}}-I^{0} = \sum_{k=0}^{N_{t}-1}(I^{k+1}-I^{k}) =\sum_{k=0}^{N_{t}-1} \Delta t \sum_{i=1}^{m} b_{i} f(t_{k}+c_{i} \Delta t) \end{align}
By assumption $I(0)=0$ so that we obtain \begin{align} I^{N_{t}} = \Delta t \sum_{k=0}^{N_{t}-1} \sum_{i=1}^{m} b_{i} f(t_{k}+c_{i} \Delta t) \end{align}
Up to here everything is reasonable in my mind. But then I encounter problems when a $c$ contains an entry bigger or equal than one. For example here's Heun's method $c =(0, \, 1); b = (\frac{1}{2}, \, \frac{1}{2})$.
We then have for calculating $I^{N_{t}}$ \begin{align} I^{N_{t}} = \frac{\Delta t}{2} \sum_{k=0}^{N_{t}-1} [ f(t_{k})+f(t_{k+1})] \end{align} consider now only the last summand that is $$f(t_{N_{t}-1})+f(t_{N_{t}}).$$
Here's my problem: While $t_{N_{t}-1}$ is a well defined point in our $t-$grid, $t_{N_{t}}$ is not. Now since $f$ is given and can be computed even on nodes outside of those considered, this should not arise any questions. Anyway I was wondering if having nodes outside of the "chosen" ones is a problem or not.
I apologize for any language related problems, unfortunately English is not my mother tongue.
Thanks a lot in advance for any help!