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
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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.A4, D1
(2)Conversely, if H is an antiderivative, then H′ = G′, so H − G is a constant.D1, A3
(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)
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Lemma L1
If f and g are differentiable and H is an antiderivative of f′g, then (fg − H)′ = fg′.
Proof.
(1)By the product rule, (fg)′ = f′g + fg′.assumption, A1
(2)(fg − H)′ = (fg)′ − H′.A2
(3)H is an antiderivative of f′g, so H′ = f′g.assumption, D1
(4)(fg − H)′ = (f′g + fg′) − f′g = fg′(1), (2), (3), A10, E
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Theorem T1
If f and g are differentiable and f′g has an antiderivative, then ∫ f(x)g′(x) dx = f(x)g(x) − ∫ f′(x)g(x) dx.
Proof.
(1)Let H be an antiderivative of f′g.assumption, D1
(2)(fg − H)′ = fg′, so fg − H is an antiderivative of fg′.(1), L1, D1
(3)Hence the left side ∫fg′dx is the collection of all functions of the form f(x)g(x) − H(x) + C.(2), D2
(4)∫f′g dx is the collection of all functions of the form H(x) + C.(1), D2
(5)Therefore the right side f(x)g(x) − ∫f′g dx is the collection of all functions of the form f(x)g(x) − H(x) − C.(4), A10
(6)As C runs over all constants, so does −C, so (3) and (5) are the same collection.(3), (5), A10
(7)Hence the constant of integration is written only once. Writing one on each side uses the same arbitrariness twice.(6), D2
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Theorem T2
If f and g are differentiable and f′ and g′ are continuous on [a,b], then ∫ₐᵇ f g′ = f(b)g(b) − f(a)g(a) − ∫ₐᵇ f′ g.
Proof.
(1)f and g are differentiable, hence continuous, so f′g and fg′ are continuous on [a,b] and integrable.assumption, A9, A7, A8
(2)H(x) = ∫ₐˣ f′g is an antiderivative of f′g.(1), A5, D1
(3)By L1, fg − H is an antiderivative of fg′.(2), L1, D1
(4)By the fundamental theorem, ∫ₐᵇ fg′ = (fg − H)(b) − (fg − H)(a).(1), (3), A6
(5)H′ = f′g, so by the fundamental theorem, ∫ₐᵇ f′g = H(b) − H(a).(1), (2), A6
(6)Expanding (4) gives ∫ₐᵇ fg′ = f(b)g(b) − f(a)g(a) − (H(b) − H(a)).(4), A10
(7)Substituting (5) gives ∫ₐᵇ fg′ = f(b)g(b) − f(a)g(a) − ∫ₐᵇ f′g.(5), (6), E
(8)The arbitrary constant of T1 is absent here, because taking the difference of two antiderivative values makes the constant vanish.T1, (4), (6), A10
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