Solution
= Solution
{c}
The <complex line integral> along the smooth parametrized curve is
$$
\int_\gamma f(z)dz:=\int_a^b f(\gamma(t))\gamma'(t)dt.
$$
For the reversed curve, the <chain rule> gives $\widetilde\gamma'(t)=-\gamma'(a+b-t)$. Substitution $s=a+b-t$ yields
$$
\int_{\widetilde\gamma}f(z)dz
=-\int_a^b f(\gamma(a+b-t))\gamma'(a+b-t)dt
=-\int_a^b f(\gamma(s))\gamma'(s)ds.
$$
Consequently \b[reversing orientation negates the complex line integral].