Integration/Substitution/Fact/Proof

Proof

Since ${}f$ is continuous and ${}g$ is continuously differentiable, both integrals exist. Let ${}F$ denote a primitive function for ${}f$, which exists, due to fact. Because of the chain rule, the composite function

${}t\mapsto F(g(t))=(F\circ g)(t)\,$

has the derivative ${}F'(g(t))g'(t)=f(g(t))g'(t)$. Therefore,

${}\int _{a}^{b}f(g(t))g'(t)\,dt=(F\circ g)|_{a}^{b}=F(g(b))-F(g(a))=F|_{g(a)}^{g(b)}=\int _{g(a)}^{g(b)}f(s)\,ds\,.$

To fact