Definition D1
G is an antiderivative of h on an interval if G′ = h at every point of that interval.
Proof.
(1)This definition does not assert that such a G exists. If one exists, it is so named.convention
∎
Definition D2
The indefinite integral ∫h(x)dx of h denotes the collection of all antiderivatives of h. If G is one of them, write ∫h(x)dx = G(x) + C, where C stands for an arbitrary constant.
Proof.
(1)For any constant c, (G + c)′ = G′ = h, so G + c is also an antiderivative.A3, D1
(2)Conversely, if H is an antiderivative, then H′ = G′, so H − G is a constant.D1, A2
(3)Hence the antiderivatives are exactly those of the form G + C. Both sides of the equation are collections of functions, not functions.(1), (2)
∎
Lemma L1
If F′ = f and g is differentiable, then [F(g(x))]′ = f(g(x)) · g′(x).
Proof.
(1)By the chain rule, [F(g(x))]′ = F′(g(x)) · g′(x).assumption, A1
(2)F′ = f is an equation at every point, so F′(g(x)) = f(g(x)).assumption, E
(3)Substituting (2) into (1) gives [F(g(x))]′ = f(g(x)) · g′(x).(1), (2), E
∎
Theorem T1
If F is an antiderivative of f and g is differentiable, then ∫ f(g(x)) g′(x) dx = F(g(x)) + C.
Proof.
(1)[F(g(x))]′ = f(g(x))g′(x).assumption, L1
(2)Hence F∘g is an antiderivative of f(g(x))g′(x).(1), D1
(3)The indefinite integral is the collection of all antiderivatives, so ∫f(g(x))g′(x)dx = F(g(x)) + C.(2), D2
(4)Meanwhile, F is an antiderivative of f, so ∫f(u)du = F(u) + C.assumption, D1, D2
(5)Putting g(x) in the place of u gives the right side of (3). That is, ∫f(g(x))g′(x)dx is ∫f(u)du with u = g(x) substituted.(3), (4), E
(6)Carrying g′(x)dx along is not a notational convention. Without it, (1) fails.(1), (5)
∎
Corollary C
If f is continuous, g is differentiable, and g′ is continuous on [a,b], then ∫ₐᵇ f(g(x)) g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du.
Proof.
(1)f∘g is continuous and g′ is continuous, so f(g(x))g′(x) is continuous on [a,b] and integrable.assumption, A6, A7
(2)Since f is continuous, F(u) = ∫_{g(a)}^{u} f(t)dt is an antiderivative of f.assumption, A4, D1
(3)By T1, F∘g is an antiderivative of f(g(x))g′(x).(2), assumption, T1, D1
(4)Applying the fundamental theorem on [a,b] gives ∫ₐᵇ f(g(x))g′(x)dx = F(g(b)) − F(g(a)).(1), (3), A5
(5)Applying the fundamental theorem to F gives ∫_{g(a)}^{g(b)} f(u)du = F(g(b)) − F(g(a)).(2), assumption, A5
(6)Even when g(a) > g(b), the convention ∫ᵤᵛ f = −∫ᵥᵘ f keeps (5) valid as it stands.convention, (5), A5
(7)The two values are equal, so ∫ₐᵇ f(g(x))g′(x)dx = ∫_{g(a)}^{g(b)} f(u)du.(4), (5), (6), E
(8)The reason the limits of integration change lies in (5). What entered F was not a and b but g(a) and g(b).(5), (7)
∎