Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses

Refine at the breakpoints of $h$ and at their preimages of the breakpoints of $\gamma$. On each resulting interval, [L5] gives $(\gamma \circ h{)}^{\mathrm{\prime }}=({\gamma }^{\mathrm{\prime }}\circ h)h^{\mathrm{\prime }}$. The refinements do not alter either line integral by [L3].

For the scalar integrand, step 1.1 gives $f(\gamma (h(t)))\parallel (\gamma \circ h{)}^{\mathrm{\prime }}(t){\parallel }_2=f(\gamma (h(t)))\parallel {\gamma }^{\mathrm{\prime }}(h(t)){\parallel }_2\mathrm{\mid }{h}^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{.}$

For the vector integrand, step 1.1 and bilinearity give $⟨F(\gamma (h(t))),(\gamma \circ h{)}^{\mathrm{\prime }}(t)⟩=⟨F(\gamma (h(t))),{\gamma }^{\mathrm{\prime }}(h(t))⟩h^{\mathrm{\prime }}(t)\mathrm{.}$

When $h$ is increasing, $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }=h^{\mathrm{\prime }}$ and [L4] identifies the sum of these integrals with ${\int }_{\gamma }fds$. When $h$ is decreasing, $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }=-h^{\mathrm{\prime }}$ and the reversal of the oriented substitution limits supplies the second minus sign. Thus the scalar equality holds in both cases.

Applying [L4] piece by piece to step 2.2 gives the same oriented integral when $h(c)=a,h(d)=b$, and its negative when $h(c)=b,h(d)=a$. These are respectively the orientation-preserving and orientation-reversing cases in [L1].

Steps 3.1 and 3.2 prove all three formulas. The nondegenerate-interval, nonzero-derivative, and bijectivity hypotheses in [L1] rule out singleton reparametrizations, pauses, and multiple traversals.