# How is integration by substitution related to the chain rule?

May 28, 2018

Let $f \left(x\right)$ be defined and continuous in $\left[a , b\right]$ and $g \left(x\right)$ defined and differantiable in $\left[c , d\right]$ with values in $\left[a , b\right]$, such that $g \left(c\right) = a$ and $g \left(d\right) = b$.
Suppose also for simplicity that $g ' \left(x\right) > 0$.

The chain rule states that:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

Consider now the definite integral as limit of the Riemann sum:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f \left(g \left({\xi}_{k}\right)\right) g ' \left({\xi}_{k}\right) \left({x}_{k + 1} - {x}_{k}\right)$

where ${x}_{0} < {x}_{1} < \ldots < {x}_{N} \in \left[c , d\right]$ and ${\xi}_{k} \in \left({x}_{k} , {x}_{k + 1}\right)$.

As the choice of ${\xi}_{k}$ in the interval is arbitrary we can choose them as the points for which, based on Lagrange's theorem:

$g ' \left({\xi}_{k}\right) = \frac{g \left({x}_{k + 1}\right) - g \left({x}_{k}\right)}{{x}_{k + 1} - {x}_{k}}$

so:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f ' \left(g \left({\xi}_{k}\right)\right) \left(g \left({x}_{k + 1}\right) - g \left({x}_{k}\right)\right)$

Let now: ${y}_{k} = g \left({x}_{k}\right)$. As $g ' \left(x\right) > 0$ the function is strictly increasing so: ${y}_{0} < {y}_{1} < \ldots < {y}_{N}$

Then:

${\int}_{a}^{b} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\lim}_{N \to \infty} {\sum}_{n = 0}^{N} f ' \left({y}_{k}\right) \left({y}_{k + 1} - {y}_{k}\right)$

but this is a Riemann sum of $f \left(y\right)$ for $y \in \left[a , b\right]$, so:

${\int}_{c}^{d} f ' \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx} = {\int}_{a}^{b} f ' \left(y\right) \mathrm{dy} = f \left(b\right) - f \left(a\right)$