How can I prove this $\int_{\gamma}f=-\int_{-\gamma}f$ I'm reading Conway's complex analysis book and on page 64 he made the following definition:

Definition. Let $\sigma:[c,d]\to \mathbb C$ and $\gamma:[a,b]\to \mathbb C$ be rectifiable paths. The path $\sigma$ is equivalent to
  $\gamma$ if there is a function $f:[c,d]\to [a,b]$ which is
  continuous, strictly increasing, and with $f(c)=a,\ f(d)=b$; such that
  $\sigma=\gamma\circ f$.

This is a equivalence relation and after this he introduce another definition:

Definition. If $\gamma$ is a rectifiable curve for $a\le t\le b$, then denote by $-\gamma$ the curve defined by$(-\gamma)(t)=\gamma(-t)$ for $-b\le
 t\le -a$.

So based on these definitions how can I prove this
$$\int_{\gamma}f=-\int_{-\gamma}f$$
This is an one line proof, I've already some demonstrations of this fact, but I always get confused with the choice of the parametrization. If someone could make this proof with details I would be very grateful.
 A: Let $(t_0 =a,t_1,...,t_n = b)$ be a partition of $[a,b]$. Then
$(-t_n ,...,-t_0)$ is a partition of $[-b,-a]$.
Assuming $f$ is continuous, it is uniformly continuous on $[a,b]$.
Suppose $\epsilon>0$, then  we
may assume that the mesh size is small enough so that
$|f(t_k) - f(t_{k-1})| < \epsilon$ for all $k$.
Note that $(-\gamma)(t) = \gamma(-t)$. To avoid confusion $-\gamma$ with
negation, let $\eta = -\gamma$, then $\eta(-t) = \gamma(t)$.
Then
\begin{eqnarray}
\sum_k f(\gamma(t_k)) (\gamma(t_k) - \gamma(t_{k-1})) &=& \sum_k f(\eta(-t_k)) (\eta(-t_k) - \eta(-t_{k-1})) \\
&=& -\sum_k f(\eta(-t_k)) ( \eta(-t_{k-1}) - \eta(-t_k)) \\
&=& -\sum_k f(\eta(-t_{k-1})) ( \eta(-t_{k-1}) - \eta(-t_k)) + \sum_k (f(\eta(-t_{k-1}))- f(\eta(-t_{k}))) ( \eta(-t_{k-1}) - \eta(-t_k))
\end{eqnarray}
and so we have
$|\sum_k f(\gamma(t_k)) (\gamma(t_k) - \gamma(t_{k-1})) - (-\sum_k f(\eta(-t_{k-1})) ( \eta(-t_{k-1}) - \eta(-t_k)))| < \epsilon(b-a)$.
Now take limits as the mesh size of the partition goes to zero and we get
$|\int_\gamma f dz - (-\int_{-\gamma} f dz)| \le \epsilon(b-a)$.
Since $\epsilon>0$ was arbitrary, we have the desired result.
A: DISCLAIMER: I'll do the job for path integrals. You can visualize complex integrals as particular cases of this (for instance, the real part and imaginary part can be separated and be seen as path integrals of $(f_1,-f_2)$ and $(f_2,f_1)$), or simply adapt the arguments. In any case, I'll leave the job of transitioning the result for you. If you have difficulties, tell me.
I'll change his notation of $-\gamma$ to $\Gamma$, and call the function $x \mapsto -x$ as $g$. Note that $\Gamma'=-\gamma ' \circ g$. We have that
$$\int_{\Gamma} f=\int_{-b}^{-a} f\circ \Gamma \cdot \Gamma'=\int_{-b}^{-a} f \circ (\gamma \circ g) \cdot (-\gamma ' \circ g)$$
$$=\int_{-b}^{-a} \big((f \circ \gamma) \circ g \big) \cdot (-\gamma' \circ g)$$
$$=\int_{g(-b)}^{g(-a)}(f \circ \gamma ) \cdot (-\gamma')\cdot (-1)=\int_{b}^{a}(f \circ \gamma) \cdot \gamma'=-\int_a^b (f \circ \gamma) \cdot \gamma'=-\int_{\gamma} f.$$
