Solution
= Solution
{c}
First let $b>0$. Substitute $x=-y$ in the negative half of the <integral>. Since $g$ is an <odd function>,
$$
\int_{-b}^0g(x)\,dx=\int_0^b g(-y)\,dy=-\int_0^b g(y)\,dy.
$$
The two halves cancel, so
$$
\boxed{\int_{-b}^b g(x)\,dx=0.}
$$
For $b=0$ both limits coincide. For $b<0$, reverse the limits and apply the result with $-b>0$. Thus the identity holds for every real $b$.