Let $\gamma$ be a rectifiable curve and suppose $f$ is a function continuous on $\gamma$. Suppose $\gamma$ is defined on $[a,b]$ and let $\gamma_-(t):=\gamma(-t), -b\leq t\leq -a$. I am trying to understand why $\int_\gamma f=-\int_{-\gamma}f$. My calculation is as follows: $\int_{\gamma_-}f(z)dz=\int_{-a}^{-b}f(\gamma(-t))\gamma'(-t)(-1)dt=\int_a^bf(\gamma(-t))\gamma'(-t)dt$.
What am I missing?
On
First of all, you should be integrating from $-b$ to $-a$. Second of all, I don't know how you arrived at the final equality. You should have $$-\int_{-b}^{-a} f(\gamma(-t))\gamma'(-t)\,dt = \int_{-a}^{-b} f(\gamma(-t))\gamma'(-t)\,dt.$$ Now make the obvious substitution $-t = u$ in the integral.
By the way, for fun, you should note that the usual way of parametrizing $\gamma_-$ is still by the interval $[a,b]$, with $\gamma_-(t) = \gamma(b+a-t)$.
Note that\begin{align}\int_{\gamma_-}f(z)\,\mathrm dz&=\int_{-b}^{-a}f\bigl(\gamma(-t)\bigr)\bigl(-\gamma'(-t)\bigr)\,\mathrm dt\\&=-\int_{-b}^{-a}f\bigl(\gamma(-t)\bigr)\gamma'(-t)\,\mathrm dt.\end{align}Now, if you do $-t=u$ and $\mathrm dt=-\mathrm du$, this integral becomes$$\int_b^af\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt=-\int_a^bf\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt.$$