14.6. INDEPENDENCE OF PARAMETRIZATION∗ 273

This implies that for s ∈ I,

h′i (gi (s)) =1

g′i (s). (14.10)

Now letting s = φ (t) for s ∈ I, it follows t ∈ J1, an open interval. Also, for s and t relatedthis way, f (t) = g (s) and so in particular, for s ∈ I, gi (s) = fi (t) . Consequently,

s = hi (gi (s)) = hi ( fi (t)) = φ (t)

and so, for t ∈ J1,

φ′ (t) = h′i ( fi (t)) f ′i (t) = h′i (gi (s)) f ′i (t) =

f ′i (t)g′i (φ (t))

(14.11)

which shows that φ′ exists and is continuous on J1, an open interval containing φ

−1 (s0).Since s0 is arbitrary, this shows φ

′ exists on [a+δ ,b−δ ] and is continuous there.Now f (t) = g◦

(g−1 ◦f

)(t) = g (φ (t)), and it was just shown that φ

′ is a continuousfunction on [a−δ ,b+δ ]. It follows from the chain rule, f ′ (t) = g′ (φ (t))φ

′ (t) and so, byTheorem 14.6.5,∫

φ(b−δ )

φ(a+δ )

∣∣g′ (s)∣∣ds =∫ b−δ

a+δ

∣∣g′ (φ (t))∣∣ ∣∣φ ′ (t)∣∣dt =

∫ b−δ

a+δ

∣∣f ′ (t)∣∣dt.

Now using the continuity of φ ,g′, and f ′ on [a,b] and letting δ → 0+ in the above, yields∫ d

c

∣∣g′ (s)∣∣ds =∫ b

a

∣∣f ′ (t)∣∣dt. ■