How Does Symmetry Affect Integrals in Wave Equations?

[stunner5000pt]

Of the partial kind

We have just finished solving the one dimensional wave equation for a vibrating string of length L using d'Alembert's solution which is

$$u(x,t) = \frac{1}{2} \displaystyle{[} f(x+ct) + f(x-ct) \displaystyle{]} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds$$

the question is
If g(x) satisifes the conditions taht g(x) = -g(-x) and g(x) = -g(2l -x) then
$$\int_{x}^{x+2l} g(s) ds = 0$$
Well according to a theorem when a periodic function is integrated over a symmetric interval is zero. Also since the velocity of the wave is ct then ct = 2L

$$\int_{x-ct}^{x+ct} g(s) ds = 0$$
$$\int_{x-ct}^{x} g(s) ds + \int_{x}^{x+ct} g(s) ds = 0$$
$$\int_{x-2L}^{x} g(s) ds = -\int_{x}^{x+2L} g(s) ds$$
but here is where i am stuck. How do i change the limit of integration over the left hand side

ALso I am not even sure is if this is the correct way to go? Is it the right way? Your help and advice would be greatly appreciated!
Thank you!

 

[Physics Monkey]

To convert the integral on the left hand side of your final equaiton into something more useful, try the change of variable $$s = s' - 2L$$.