I have the integral $\int_{a}^{b} f(x) dx$ and I wish to make a substitution $x=g(u)$. On making this substitution, I evaluate the integral $\int_{g^{-1}(a)}^{g^{-1}(b)} f(g(u) ) g'(u) du$ not in the Cartesian $x$-$y$ plane but in the Cartesian $u$-$ f(g(u) ) g'(u) $ plane. We assume that the mapping $g: u \rightarrow x$ be one-to-one. I am finding it quite hard to find a reason behind this assumption. If the mapping were not one-to-one, what could happen while evaluating the integral?
An example will also help me understand.
Forget about $g^{-1}$ The theorem says if a=g(c) and b=g(d) ,g with a continuous derivative , then
$\int_a^bf(x)dx =\int_c^df(g(u))g'(u)du$ . No use of one- one for g is needed .Of course given a and b ,you have to guess c and d . the proof is just by considering d as a variable and checking using the chain rule that both sides have the same derivative with respect to d .and agree at d=c .