Is there any case where a function is 1-1 where the integral=0 and the function is not odd? I know that the integral of an odd function from -a to a is always equal to 0. Is this only true for odd 1-1 functions? If I know a function is 1-1 and its integral from -a to a is 0, does that imply that its odd?
 A: Your question is not very specific. I will give two answers depending on possible interpretations of the question.
Interpretation I. Let $a\in \mathbb{R}$ and suppose $f$ is integrable and 1-1 on $[-a,a]$, and that $\int_{-a}^af(x)dx=0$. Does it follow that $f$ is odd on the interval $[-a,a]$?
Answer. Definitely not. For a simple counterexample, take $a=1$, and 
$f(x) = \begin{cases}
       \text{$4x^3$,} &\quad\text{if $x \leq 0$}\\
       \text{$2x$,} &\quad\text{if $x>0$}
     \end{cases}$.
Interpretation II. Suppose $f$ is a continuous, real-valued function such that for every $a>0$, we have $\int_{-a}^af(x)dx=0$. Does it follow that $f$ is an odd function?
(You don't need injectivity, so I'm leaving it out of the interpretation.)
Answer. Yes. Set $g(x)=\int_0^xf(t)dt$ and note that $g$ is even and $g'(x)=f(x)$ for every $x\in \mathbb{R}$. Suppose $a>0$. Then for $|h|<a$, we have
$\frac{g(-a+h)-g(-a)}{h}=\frac{g(a-h)-g(a)}{h}=-\frac{g(a-h)-g(a)}{-h}$. 
Letting $h \to 0$ on both sides of the equation gives $f(-a)=g'(-a)=-g'(a)=-f(a)$, so that $f$ is odd (note also that $f(0)=0$ now follows from a simple continuity argument).
