Question regarding integration by substitution. The theorem on integration by substitution says that 
$$\int_{\phi(a)}^{\phi(b)}f(x)dx=\int_{a}^{b}f(\phi(t))\phi'(t)dt$$
provided that $\phi$ has an integrable derivative. My question is, shouldn't $\phi$ be monotonic on $[a,b]$? I have this doubt as I am unable to prove this using Riemann Sums. Can someone tell how this works?
The proof on Wikipedia assumes that $$F(\phi(b))-F(\phi(a))=\int_{\phi(a)}^{\phi(b)}f(x)dx$$
but for this to happen, $\phi(a)<\phi(b)$, and $\phi$ should be increasing on $[a,b]$, isn't it?

You do not need that $\phi $ is bijective.


*

*$\phi $ continuously differentiable at $[a,b] $

*$f $ continuous at $\phi ([a,b]) $.


For the proof, consider
$$g (x)=\int_{\phi (a)}^{\phi (x)} f(t)dt-\int_a^x f (\phi (t))\phi'(t)dt $$
and you show by FTC , that $g'(x)=0$.
thus
$$g (b)=g (a)=0$$
