# Substitution is Just the Chain Rule Backwards

The substitution rule is the chain rule backwards: I'd prefer that when we see an integral,

${\displaystyle \left.\int _{a}^{b}f(x)dx\right.}$

we ask ourselves this: can I think of ${\displaystyle \left.f\right.}$ as a derivative? In particular, can I think of it as a result of the application of the chain rule? That is, ${\displaystyle \left.f(x)=h'(g(x))g'(x)\right.}$?

If so, then we're done: we can solve that integral:

${\displaystyle \left.h(g(b))-h(g(a))=\int _{a}^{b}h'(g(x))g'(x)dx\right.}$

and, even better,

${\displaystyle \left.h(g(b))-h(g(a))=\int _{a}^{b}h'(g(x))g'(x)dx=\int _{g(a)}^{g(b)}h'(u)du\right.}$

Forgetting for a moment that we might know how to solve this!;), we can always do the change of variables

${\displaystyle \left.\int _{a}^{b}f(g(x))g'(x)dx=\int _{g(a)}^{g(b)}f(u)du\right.}$

and hope that the integral on the right is easier to solve (certainly less cluttered).

Writing it in this last way may be mysterious, because of the change of variable to u (and the change in the limits); but it's the disappearance of g'(x) that's really curious. It falls right out of the change of variables, however:

${\displaystyle \left.u=g(x)\Longrightarrow {\frac {du}{dx}}=g'(x)\right.}$

Solving for ${\displaystyle \left.du\right.}$, we get ${\displaystyle \left.du=g'(x)dx\right.}$, and that piece disappears.