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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.6 - Problem 57ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.6 - Problem 57ae

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# Exact Simpson’s Rule Prove that Simpson’s Rule is exact

ISBN: 9780321570567 2

## Solution for problem 57AE Chapter 7.6

Calculus: Early Transcendentals | 1st Edition

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Problem 57AE

Problem 57AE

Exact Simpson’s Rule

Prove that Simpson’s Rule is exact (no error) when approximating the definite integral of a linear function and a quadratic function.

Step-by-Step Solution:
Step 1 of 3

Solution:-

Step1

Simpson’s rule

Three points integration rule derived using the method of undetermined coefficients.

Suppose that we add a quadrature point at the middle of the interval [a, b],

To avoid algebra,

substitute x =  and define h =

so that the integral becomes

Since we have three unknowns,

we can make this formula exact for all quadratic functions;

so, let us use,

e.g., F ≡ 1, F ≡ u and F ≡ u 2 .

F ≡ 1 ⇒  = 2h = w1 + w2 + w3;

F ≡ u ⇒   = 0 = −hw1 + hw3 ⇒ w1 = w3;

F ≡  ⇒   =  = w1 + w3 ⇒ w1 = w3 =  ; w2 = 2h − w1 − w3 = 2h − h =

Hence we obtain the approximation

which translates into

This is called Simpson’s rule.

Step2

As this formula is exact for all quadratic functions,

we expect the error to be of the form KF ′′′(ξ).

However, if we examine F ≡ u 3 ,

the integral

= 0 and

Simpson’s rule gives

= 0 ⇒ K ≡ 0.

Step3

Therefore

Simpson’s rule is exact for cubic functions as well.

Consequently, we try an error term of the form OF (iv)(ξ), ξ ∈ (−h, h).

To find the value of K use,

e.g., F(u) ≡  (with F (iv)(ξ) = 4!, independent of ξ).

!K

4!K=

Simpson’s rule is fifth-order accurate: the error varies as the fifth power of the width of the interval. Simpson’s rule is given by

f (ξ), ξ ∈ (a, b)

Step 2 of 3

Step 3 of 3

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