(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)

Question

(Second Substitution Rule)

Let the function [latex]φ : [a, b] → \mathbb{R}[/latex] 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

[latex]\int_{a}^{b}{f (φ (t)) dt} = \int_{φ(a)}^{φ(b)}{f(x) ψ^{\prime} (x) dx}.[/latex] (8.19)

Accepted Answer

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 [latex]φ^{\prime} (t) ≠ 0[/latex] for all t ∈ (a, b), the inverse function theorem (Theorem 7.11) ensures the existence of [latex]ψ = φ^{−1}[/latex] and the continuity of [latex]ψ^{\prime}[/latex] on I. To prove the equality (8.19), we use (8.18) with [latex]f \circ φ[/latex] 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

[latex]\int_{a}^{b}{f (\varphi (t)) \varphi^{\prime}(t) dt} = \int_{\varphi(a)}^{\varphi(b)} {f (x) dx}[/latex] (8.18)

[latex]\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}.[/latex]

This gives

[latex]\int_{\varphi(a)}^{\varphi(b)}{f (t) ψ^{\prime} (t) dt} = \int_{a}^{b}{f (φ (x)) dx},[/latex]

which coincides with (8.19) after interchanging the dummy indices t and x.

Question 8.T.14: (Second Substitution Rule) Let the function φ: [a, b] → R ha...

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