Substitution: if $\varphi$ is differentiable on $[c,d]$ with $\varphi'$ integrable and $f$ is continuous on an interval containing $\varphi([c,d])$, then $\int_{\varphi(c)}^{\varphi(d)} f = \int_c^d (f\circ\varphi)\,\varphi'$

Let $c < d$ be reals and let $\varphi : [c,d] \to \mathbb{R}$ be differentiable at every point of $[c,d]$ as a function on $[c,d]$ (The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ 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 $\varphi'$ 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 $\varphi[\,[c,d]\,] \subseteq J$, 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 $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

Then $(f\circ\varphi)\,\varphi'$ is integrable on $[c,d]$ and

$\int_{\varphi(c)}^{\varphi(d)} f \;=\; \int_c^d (f\circ\varphi)\,\varphi' ,$

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 $\varphi$ is assumed, and that is exactly why the left-hand side is written with oriented limits: $\varphi(d)$ may lie below $\varphi(c)$, and $\varphi$ 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 $f \circ \varphi$ 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 $\int_{\varphi(c)}^{\varphi(d)} f$ would be undefined whenever $\varphi(d) \le \varphi(c)$, and the theorem would have to carry a monotonicity hypothesis it does not need.

• Two integrability facts are checked, not assumed. That $f\circ\varphi$ is integrable is If $f$ is integrable on $[a,b]$ with values in $[m,M]$ and $\varphi$ is continuous on $[m,M]$, then $\varphi \circ f$ is integrable with the hypotheses in the order that theorem requires — the continuous function is the outer one — and that the product with $\varphi'$ is integrable is the product clause of If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$, $f^{2}$, $fg$, $\max(f,g)$ and $\min(f,g)$, and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$. 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 $\varphi$ is continuously differentiable then $\varphi'$ is integrable automatically, and if $\varphi$ 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, $f\circ\varphi$ still being integrable ↗ on the companion page; nothing above depends on it.