Is it always true that if $f$ is odd, then $\displaystyle\int_{-b}^b f(x) dx =0$? Obviously assuming $f$ integrable. The standard proof seems to integrate by substitution(substitute $x$ by $-u$, $dx$ by $-du$,...etc.). Nevertheless, in the proof of the theorem of substitution, we assume(at least in the proof I know) that for
$$\int _{g(a)}^{g(b)} h(x) dx=\int _a^b h(g(t))g´(t)dt$$ to hold, we need h to have a primitive $H$, which I'm not assuming in this problem...
So, there exists some counterexample?
 A: Yes it is true. Let me outline a proof. 
First note that if $f$ is integrable on $[a,b]$ and $c\in [a,b]$ then (by theorem):
$$\int_a^b f(x) ~dx=\int_a^cf(x)~dx+\int_c^b f(x)~ dx$$
Applying this to $[-b,b]$ and $c=0$ we get:
$$\int_{-b}^b f(x)~dx=\int_{-b}^0 f(x) ~dx+\int_0^b f(x)~dx$$
Note that: 
$$\int_0^b f(x)~dx =\lim_{|P|\to 0}\sum_{[u,v]\in P}f(x^{\ast})(v-u)$$
Where $x^{\ast}$ is any point in $[u,v]$. For concreteness pick the midpoint. $P$ is any partition of $[0,b]$ and $|P|$ is the mesh size. If this is not your definition of the integral, then this is a theorem (or substitute your definition). :)
Note that $f(x^{\ast})=-f(-x^{\ast})$. Also note that if $P=\{0,x_1,x_2,...,b\}$ is a partition of $[0,b]$ then $P'=\{-b,...,-x_2,-x_1,0\}$ is a partition of $[-b,0]$. Further, note if $[u,v]\in P$ then $[-v,-u]\in P'$. 
Putting these together:
$$\begin{align*}\int_0^b f(x)~dx &=\lim_{|P|\to 0}\sum_{[u,v]\in P}f(x^{\ast})(v-u)\\
&=\lim_{|P|\to 0}\sum_{[u,v]\in P}-f(-x^{\ast})(v-u)\\
&=\lim_{|P|\to 0}\sum_{[u,v]\in P}-f(-x^{\ast})((-u)-(-v))\\
&=-\lim_{|P|\to 0}\sum_{[u,v]\in P}f(-x^{\ast})((-u)-(-v))
\end{align*}$$
Note that if $x^{\ast}$ is the midpoint of $[u,v]$ then $-x^{\ast}$ is the midpoint of $[-v,-u]$. Call this new midpoint $x'$, define $u'=-u$ and $v'=-v$. So just renaming variables:
$$-\lim_{|P|\to 0}\sum_{[u,v]\in P}f(-x^{\ast})((-u)-(-v))=-\lim_{|P'|\to 0}\sum_{[v',u']\in P'}f(x')(u'-v')$$
This last sum is just $-\int_{-b}^0 f(x)~dx$. Thus $\int_{-b}^bf(x)~dx=\int_{-b}^0f(x)~dx+\int_0^bf(x)~dx=0$
There are small details to deal with, but I hope you can use your imagination to fill them in.
A: Yes, it is always true. (Assuming the integral makes sense). First, notice that 
$$\int_{-b}^b \mathrm f(x)~\mathrm dx = \int_{-b}^0 \mathrm f(x)~\mathrm dx + \int_{0}^b \mathrm f(x)~\mathrm dx$$
If we make the substitution $x = -u$ then $\mathrm dx = -\mathrm du$ and so
\begin{eqnarray*}
\int_{x=-b}^{x=0} \mathrm f(x)~\mathrm dx &=& \int_{u=b}^{u=0} -\mathrm f(-u)~\mathrm du \\ \\
&=& \int_{0}^b \mathrm f(-u)~\mathrm du
\end{eqnarray*}
The choice of variable does not change the value of the integral, so the $u$ in the last line could be an $x$ and we would still have the same value. Hence
\begin{eqnarray*}\int_{-b}^b \mathrm f(x)~\mathrm dx &=& \int_{0}^b \mathrm f(-x)~\mathrm dx + \int_{0}^b \mathrm f(x)~\mathrm dx \\ \\
&=& \int_{0}^b \mathrm f(-x)+\mathrm f(x)~\mathrm dx
\end{eqnarray*}
This is true for all integrable functions $\mathrm f$. If we now say that $\mathrm f$ is an odd function then, by definition, $\mathrm f(-x) \equiv -\mathrm f(x)$, and so $\mathrm f(-x)+\mathrm f(x) \equiv -\mathrm f(x)+\mathrm f(x) \equiv 0$. Hence 
\begin{eqnarray*}\int_{-b}^b \mathrm f(x)~\mathrm dx &=& \int_{0}^b \mathrm f(-x)+\mathrm f(x)~\mathrm dx \\ \\
&=& \int_{0}^b 0~\mathrm dx \\ \\
&=& 0
\end{eqnarray*} 
A: This is always true!
$f$ is odd if $f(x) = -f(-x)$
Then 
$\int_{-b}^{b} f(x) dx = \int_{-b}^{0} f(x) dx + \int_{0}^{b} f(x) dx = \int_{-b}^{0} -f(-x) dx + \int_{0}^{b} f(x) dx = -\int_{0}^{b} f(x) dx + \int_{0}^{b} f(x) dx = 0$
