Substitution: if $\phi$ is differentiable on $[c,d]$ with ${\phi }^{\mathrm{\prime }}$ integrable and $f$ is continuous on an interval containing $\phi ([c,d])$, then ${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }}$

Let $c<d$ be reals and let $\phi :[c,d]\to \mathbb{R}$ be differentiable at every point of $[c,d]$ as a function on $[c,d]$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set), with ${\phi }^{\mathrm{\prime }}$ integrable on $[c,d]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$). Let $J\subseteq \mathbb{R}$ be order-convex with at least two elements (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with $\phi [[c,d]]\subseteq J$, and let $f:J\to \mathbb{R}$ be continuous on $J$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point).

Then $(f\circ \phi ){\phi }^{\mathrm{\prime }}$ is integrable on $[c,d]$ and

${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }},$

the left-hand integral being the oriented one of The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$.

Neither injectivity nor monotonicity of $\phi$ is assumed, and that is exactly why the left-hand side is written with oriented limits: $\phi (d)$ may lie below $\phi (c)$, and $\phi$ may return to the same value many times. The proof runs through a primitive of $f$ and the chain rule, and no inverse function is ever formed.

Continuity of $f$ is a hypothesis and cannot be weakened to integrability. With $f$ merely integrable the composite $f\circ \phi$ need not be integrable at all, so the right-hand side need not exist; that is the false statement that weakens it on the companion page.

• The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$ is what makes step 2.2 legal. Without the orientation convention the symbol ${\int }_{\phi (c)}^{\phi (d)}f$ would be undefined whenever $\phi (d)\le \phi (c)$, and the theorem would have to carry a monotonicity hypothesis it does not need.

• Two integrability facts are checked, not assumed. That $f\circ \phi$ is integrable is If $f$ is integrable on $[a,b]$ with values in $[m,M]$ and $\phi$ is continuous on $[m,M]$, then $\phi \circ f$ is integrable with the hypotheses in the order that theorem requires — the continuous function is the outer one — and that the product with ${\phi }^{\mathrm{\prime }}$ is integrable is the product clause of If $f,g$ are integrable on $[a,b]$ then so are $\mid f\mid$, $f^2$, $fg$, $\max (f,g)$ and $\min (f,g)$, and $\mid {\int }_a^{b}f\mid \le {\int }_a^b\mid f\mid$. Neither is automatic, and the companion page's false statement is exactly the claim that the first of them survives weakening $f$ to an integrable function.

• Where the more familiar hypotheses sit. If $\phi$ is continuously differentiable then ${\phi }^{\mathrm{\prime }}$ is integrable automatically, and if $\phi$ is in addition strictly monotone then the substitution can be read in either direction; neither refinement is needed above, and neither is claimed.

• Forward reference, orientation only. The false statement that weakens the continuity of $f$ to integrability is FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f\circ \phi$ still being integrable ↗ on the companion page; nothing above depends on it.