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Simpson rule integrates a quadratic exactly on one two-panel block

Example

On [0,2], the function f(x)=x2 has exact integral 8/3. With n=2, h=1, and the composite Simpson rule,

QS(f;0,2,2)=13(f(0)+4f(1)+f(2))=83.

This is the quadratic case of exactness through degree three.

Facts & Assumptions

Given: f(x)=x2 on [0,2] and the composite Simpson rule with n=2.

[F1]

Composite Simpson rule on an even uniform grid: for n=2, the returned value is h3(f(x0)+4f(x1)+f(x2)).

[F3]

Composite Simpson rule is exact through cubic degree: Simpson is exact for every displayed polynomial of degree at most three.

[F5]

Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable: x↦x2 is Riemann integrable on [0,2] because it is continuous.

Verification

1.1givenF2F4F5

The function x2 is continuous and integrable by [F5]. By [F4], x3/3 is an antiderivative, so [F2] gives ∫02x2 dx=8/3.

2.1F1step 1.1algebra

Here h=1 and the nodes are 0,1,2. By [F1], QS=13(0+4⋅1+4)=8/3. Thus the computed rule value equals the integral.

3.1F3step 2.1∎

The equality also follows from the degree-at-most-three exactness theorem [F3], since x2 has degree two.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources