Question 8.T.14: (Second Substitution Rule) Let the function φ: [a, b] → R ha...
(Second Substitution Rule)
Let the function φ : [a, b] → \mathbb{R} have a continuous derivative that does not vanish anywhere in (a, b) . If f is continuous on the range of φ, and ψ is the inverse of φ, then
\int_{a}^{b}{f (φ (t)) dt} = \int_{φ(a)}^{φ(b)}{f(x) ψ^{\prime} (x) dx}. (8.19)
Once again, we note that the range of φ is an interval I that contains the interval J with end points φ (a) and φ (b) . Since φ^{\prime} (t) ≠ 0 for all t ∈ (a, b), the inverse function theorem (Theorem 7.11) ensures the existence of ψ = φ^{−1} and the continuity of ψ^{\prime} on I. To prove the equality (8.19), we use (8.18) with f \circ φ in place of f, ψ in place of φ, φ (a) in place of a and φ (b) in place of b. We then obtain
\int_{a}^{b}{f (\varphi (t)) \varphi^{\prime}(t) dt} = \int_{\varphi(a)}^{\varphi(b)} {f (x) dx} (8.18)
\int_{\varphi(a)}^{\varphi(b)} {(f \circ φ) (ψ (t)) ψ^{\prime} (t) dt} = \int_{ψ(\varphi(a))}^{ψ(\varphi(b))}{(f \circ φ) (x) dx}.
This gives
\int_{\varphi(a)}^{\varphi(b)}{f (t) ψ^{\prime} (t) dt} = \int_{a}^{b}{f (φ (x)) dx},
which coincides with (8.19) after interchanging the dummy indices t and x.