Definition D1
G is an antiderivative of f if G′(x) = f(x) for every x in [a,b].
Proof.
(1)This definition does not assert that such a G exists. It only names one if it does.convention
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Definition D2
For a ≤ x ≤ b, set F(x) = ∫ₐˣ f(t) dt.
Proof.
(1)f is continuous on [a,x], hence integrable, and F(x) has a value at each x.assumption, A5
(2)When the upper limit is smaller than the lower, we adopt the convention ∫ᵤᵛ f = −∫ᵥᵘ f for u > v.convention
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Definition D3
F′(x) = L means: for every ε > 0 there is δ > 0 such that 0 < |h| < δ implies |(F(x+h) − F(x))/h − L| < ε.
Proof.
(1)The difference quotient (F(x+h) − F(x))/h has meaning only for h ≠ 0. We abbreviate the numerator as ΔF.convention
(2)At the endpoints, h is taken only toward the interior of the interval.convention
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Lemma L1
If x and x+h both lie in [a,b], then F(x+h) − F(x) = ∫ₓˣ⁺ʰ f.
Proof.
(1)Suppose h > 0. ∫ₐˣ⁺ʰ f = ∫ₐˣ f + ∫ₓˣ⁺ʰ fassumption, A3
(2)F(x+h) = F(x) + ∫ₓˣ⁺ʰ f, which gives the conclusion.(1), D2, A1
(3)Suppose h < 0. ∫ₐˣ f = ∫ₐˣ⁺ʰ f + ∫ₓ₊ₕˣ fassumption, A3
(4)Inserting the convention ∫ₓ₊ₕˣ f = −∫ₓˣ⁺ʰ f gives F(x) = F(x+h) − ∫ₓˣ⁺ʰ f, and the conclusion is the same.(3), D2, A1
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Lemma L2
If x and x+h lie in [a,b], 0 < |h|, and |f(t) − f(x)| ≤ η for every t between them, then |(F(x+h) − F(x))/h − f(x)| ≤ η.
Proof.
(1)By assumption, f(x) − η ≤ f(t) ≤ f(x) + η for every t in that interval.assumption, A8
(2)Suppose h > 0. The interval [x, x+h] has length h.assumption
(3)(f(x) − η)h ≤ ∫ₓˣ⁺ʰ f ≤ (f(x) + η)h(1), (2), A4
(4)Since ∫ₓˣ⁺ʰ f = F(x+h) − F(x), −ηh ≤ F(x+h) − F(x) − f(x)h ≤ ηh(3), L1, A1, A2
(5)Since h > 0, dividing by h gives −η ≤ ΔF/h − f(x) ≤ η(4), (2), A1, A2
(6)Suppose h < 0. The interval [x+h, x] has length −h.assumption
(7)(f(x) − η)(−h) ≤ ∫ₓ₊ₕˣ f ≤ (f(x) + η)(−h)(1), (6), A4
(8)Since ∫ₓ₊ₕˣ f = −(F(x+h) − F(x)), ηh ≤ F(x+h) − F(x) − f(x)h ≤ −ηh(7), L1, D2, A1, A2
(9)Since h < 0, dividing by h reverses the inequalities: −η ≤ ΔF/h − f(x) ≤ η(8), (6), A1, A2
(10)In either case, |ΔF/h − f(x)| ≤ η.(5), (9), A8
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Theorem T1
If f is continuous on [a,b] and F(x) = ∫ₐˣ f, then F′(x) = f(x) for every x in [a,b].
Proof.
(1)Let ε > 0 be given.assumption
(2)There is δ > 0 such that |t − x| < δ implies |f(t) − f(x)| < ε/2.A6, A1, A2
(3)Shrinking δ if necessary so that x + h ∈ [a,b], for 0 < |h| < δ every t between x and x+h satisfies |t − x| ≤ |h| < δ.(2), A2, A8
(4)Applying L2 with η = ε/2 gives |ΔF/h − f(x)| ≤ ε/2.(2), (3), L2
(5)Since ε/2 < ε, |ΔF/h − f(x)| < ε.(4), A1, A2
(6)Such a δ exists for every ε > 0, so F′(x) = f(x).(5), D3
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Theorem T2
If G is any antiderivative of f on [a,b], then ∫ₐᵇ f = G(b) − G(a).
Proof.
(1)Setting F(x) = ∫ₐˣ f, we have F′(x) = f(x) for every x.D2, T1
(2)Since G is an antiderivative, G′(x) = f(x).assumption, D1
(3)Since F′ = G′ on [a,b], F − G is constant. Call its value C.(1), (2), A7
(4)By A3, ∫ₐᵃ f = ∫ₐᵃ f + ∫ₐᵃ f, so ∫ₐᵃ f = 0, i.e. F(a) = 0.A3, A1, D2
(5)Since F(a) − G(a) = C, we have C = −G(a).(3), (4), A1
(6)Since F(b) − G(b) = C, we have F(b) = G(b) − G(a).(3), (5), A1
(7)F(b) = ∫ₐᵇ f.D2
(8)∫ₐᵇ f = G(b) − G(a)(6), (7)
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