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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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If u,vu,v are differentiable on [a,b][a,b] with u,vu',v' integrable, then abuv=u(b)v(b)u(a)v(a)abuv\int_a^b u v' = u(b)v(b)-u(a)v(a) - \int_a^b u'v

Statement

Let a<ba < b be reals and let u,v:[a,b]Ru, v : [a,b] \to \mathbb{R} be differentiable at every point of [a,b][a,b] as functions on [a,b][a,b] (The derivative f(c)=limxcf(x)f(c)xcf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} of f:ARf : A \to \mathbb{R} at a point cAc \in A that is a limit point of AA, and differentiability on a set). Suppose uu' and vv' are integrable on [a,b][a,b] (The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f). Then uvuv' and uvu'v are integrable and

abuv  =  u(b)v(b)u(a)v(a)    abuv.\int_a^b u\,v' \;=\; u(b)v(b) - u(a)v(a) \;-\; \int_a^b u'\,v .

The integrability of uu' and vv' is a hypothesis, not a formality. Without it the two integrals in the display need not exist at all, and the identity is then not false but ill-formed; that is the false statement that deletes it on the companion page. The hypothesis is automatic when uu and vv are continuously differentiable, since a continuous function on [a,b][a,b] is integrable (A continuous function on [a,b][a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Facts & Assumptions

Given: Reals a<ba < b and functions u,v:[a,b]Ru, v : [a,b] \to \mathbb{R}, differentiable at every point of [a,b][a,b], with uu' and vv' integrable on [a,b][a,b].

[L5]

Sums of integrable functions are integrable, and ab(w1+w2)=abw1+abw2\int_a^b(w_1+w_2) = \int_a^b w_1 + \int_a^b w_2 (Integrable functions on [a,b][a,b] form a set closed under sums and scalar multiples, and ab(λf+μg)=λabf+μabg\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g).

[L6]

If HH is differentiable at every point of [a,b][a,b] with HH' integrable there, then abH=H(b)H(a)\int_a^b H' = H(b)-H(a) (The second fundamental theorem: if GG is differentiable on [a,b][a,b] with G=fG' = f and ff is integrable, then abf=G(b)G(a)\int_a^b f = G(b)-G(a)).

Proof

technique · direct
1.1

uu and vv are continuous on [a,b][a,b] by [L2], hence integrable there by [L3].

givenL2L3
1.2

uvuv is differentiable at every point of [a,b][a,b] with (uv)=uv+uv(uv)' = u'v + uv' by [L1].

givenL1
2.1

uvu'v and uvuv' are integrable on [a,b][a,b] by [L4], being products of the integrable uu' with vv and of uu with the integrable vv'.

step 1.1givenL4
3.1

Hence (uv)=uv+uv(uv)' = u'v + uv' is integrable by [L5], and ab(uv)=abuv+abuv\int_a^b (uv)' = \int_a^b u'v + \int_a^b uv'.

step 1.2step 2.1L5
4.1

By [L6] applied to H:=uvH := uv, ab(uv)=u(b)v(b)u(a)v(a)\int_a^b (uv)' = u(b)v(b) - u(a)v(a).

step 1.2step 3.1L6
5.1

Comparing steps 3.1 and 4.1 and subtracting abuv\int_a^b u'v gives abuv=u(b)v(b)u(a)v(a)abuv\int_a^b uv' = u(b)v(b)-u(a)v(a) - \int_a^b u'v.

step 3.1step 4.1algebra

Remarks

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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