Prove that this is a metric space? I'm supposed to show that If X is the set of all functions on the interval $[a,b]$ and $\displaystyle d(f,g)= \int^{b}_{a}|f(x)-g(x)|dx\,$, then $(X, d)$ is a metric space.


But I don't think it is. The problem I'm having is that in order for $(X,d)$ to be a metric space, it must be true that $d(f,g) = 0$ iff $f=g$. But d could also be $0$ if, for instance, 

$f(x) - g(x)$ is odd and is symmetrical across $x=(a+b)/2$

Am I right about this or am I missing something?

Sorry, about the format by the way. I don't know how to type the actual symbols on here!
No: you are taking the integral of the absolute value of the difference  $$\displaystyle d(f,g)= \int_a^b \left|f(x)−g(x)\right|dx$$ while your counterexample would apply if you were taking the absolute value of the integral of the difference $$\displaystyle \left|\int_a^b (f(x)−g(x))\,dx\right|.$$

You may need to watch out for the possibility that $f(x)$ and $g(x)$ differ on a set of measure zero, but apart from that you will have $\left|f(x)−g(x)\right|$ as non-negative by definition, and positive if they differ, making the integral positive if they differ on a large enough subset of $[a,b]$.
I think you are confusing that the limits of integration is arbitrary. The $a$ and $b$ are to be any number, you don't determine them, they are given to you.