Theorem. Let the real function $x \mapsto f(x)$ be continuous on the interval $[a,\,b]$ . We introduce via the the equation $$x \,=\, \varphi(t)$$ a new variable $t$ satisfying

• $\varphi(\alpha) = a, \quad \varphi(\beta) = b$ ,
• $\varphi$ and $\varphi'$ are continuous on the interval with endpoints $\alpha$ and $\beta$ .

Proof. As a continuous function, $f$ has an antiderivative $F$ . Then the compound function $F\circ\varphi$ is an antiderivative of $(f\circ\varphi)\cdot\varphi'$ , since by the chain rule we have $$\frac{d}{dt}F(\varphi(t)) \,=\, F'(\varphi(t))\,\varphi'(t) \,=\, f(\varphi(t))\,\varphi'(t).$$ Using the Newton-Leibniz formula we obtain $$\int_a^b\!f(x)\,dx \,=\, F(b)-F(a) \,=\, F(\varphi(\beta))-F(\varphi(\alpha)) \,=\, \int_\alpha^\beta\!f(\varphi(t))\,\varphi'(t)\,dt,$$ Q.E.D.