TheoremBase

Integration by Parts on a Closed Interval

For differentiable F and G on [a,b] with continuous derivatives, FG' and F'G are integrable and the integral of FG' equals F(b)G(b) - F(a)G(a) minus the integral of F'G.

Statement

In the setting of The Real Line: Standing Notation and Background for Calculus on Intervals, let a,b∈Ra,b\in\mathbb{R} with a<ba<b. Derivatives of maps [a,b]β†’R[a,b]\to\mathbb{R} are taken relative to [a,b][a,b], an interval containing two distinct points by Basic Facts about Intervals of the Real Line (v). Let F,G:[a,b]β†’RF,G:[a,b]\to\mathbb{R} be differentiable, and suppose that their derivatives Fβ€²,Gβ€²:[a,b]β†’RF',G':[a,b]\to\mathbb{R} are continuous on [a,b][a,b]. Let FGβ€²FG' and Fβ€²GF'G be the pointwise products, and let integrability on [a,b][a,b] and the integral be as in The Riemann Integral on a Closed Interval via Lower and Upper Integrals.

FGβ€²FG' and Fβ€²GF'G are bounded and integrable on [a,b][a,b].

∫abF(t) Gβ€²(t) dt=F(b) G(b)βˆ’F(a) G(a)βˆ’βˆ«abFβ€²(t) G(t) dt.\int_{a}^{b}F(t)\,G'(t)\,dt=F(b)\,G(b)-F(a)\,G(a)-\int_{a}^{b}F'(t)\,G(t)\,dt .

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