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 Incoming links: Integral remainder in Taylor theorem

Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 12C / Solution Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 19D / Solution Created 2026-09-24 Updated 2026-10-03

Repeated integration by parts in

\frac{1}{k !} \int_{a}^{x} (x - θ)^{k} f^{(k + 1)} (θ) d θ

produces the first

k + 1

terms of the Taylor polynomial and leaves

f (x)

, proving the integral remainder in Taylor theorem

R (x) = \frac{1}{k !} \int_{a}^{x} (x - θ)^{k} f^{(k + 1)} (θ) d θ .

Apply the linear functional

λ_{k}

to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

λ_{k} [f] = \frac{1}{k !} \int_{a}^{b} λ_{k} [(x - θ)_{k}^{+}] f^{(k + 1)} (θ) d θ .

For the midpoint quadrature rule on

[0, 1]

,

e [1] = e [x] = 0

, so

k = 1

. Its Peano kernel is

K (θ) = \int_{0}^{1} (x - θ)_{+} d x - (1/2 - θ)_{+} = {θ^{2} /2, (1 - θ)^{2} /2, 0 \leq θ \leq 1/2, 1/2 \leq θ \leq 1.

For

f (x) = x^{2}

, the formula gives

e [f] = \int_{0}^{1} K (θ) f^{''} (θ) d θ = 2 \int_{0}^{1} K (θ) d θ = \frac{1}{12},

agreeing with

\int_{0}^{1} x^{2} d x - (1/2)^{2} = 1/3 - 1/4

.

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