1. ## antiderivative

Say you want to write $\int^{x+ct}_{x-ct} g(x) \ dx = G(x+ct) - G(x-ct)$ (where $G$ is an antiderivative of $g$)

for the odd 2-periodic function $g(x)=\left\{\begin{array}{ccc}-10,&\mbox{ if }
-1 \le x<0 \\10, & \mbox{ if } 0 \le x <1 \\g(x+2L) & \text{otherwise}\end{array}\right.$

then would G(x) be

$G(x)=\left\{\begin{array}{ccc}-\int_{-1}^{x} 10 \ dx = -10x-10 \,&\mbox{ if }
-1 \le x<0 \\ -\int_{-1}^{0} 10 \ dx + \int_{0}^{x}10 \ dx = -10 + 10x, & \mbox{ if } 0 \le x <1 \\F(x+2L) & \text{otherwise}\end{array}\right.$
?

You couldn't just write

$G(x)=\left\{\begin{array}{ccc}-10x\,&\mbox{ if }
-1 \le x<0 \\ 10x, & \mbox{ if } 0 \le x <1 \\F(x+2L) & \text{otherwise}\end{array}\right.$

could you?

2. I made my question less abstract in hopes it makes sense to someone.

3. Hello,

If you look at the antidervative of g over a given interval (say -1,0), then you have g(t)=-10, which has -10x as an antiderivative
So over this interval, $G(t)=-10(x)-(-10)(-1)=-10x-10$

So no you can't write the last part you did. It would be like you integrate from 0 to x, which is false !
It's easy to see that on a graph, since the integral is the area under the line at y=-10

I don't know if it's clear... ?