Part A

(5.1)    In this problem we construct a non-linear function for which the trapezium rule gives a more accurate result than Simpson’s rule.

Calculate the errors in the trapezium rule and Simpson’s rule approximations of

1                                       1

:4 d:?    and         :5 d:!

尸                                       尸

Find the value of a constant 扌 such that the trapezium rule gives the exact value for

1

Ⅰ =       ╱:5 _ 扌:4、d:!

尸

Determine also an interval for 扌, such that the trapezium rule gives a more accurate result for Ⅰ than Simpson’s rule.

(5.2)    The Newton-Cotes formula with n = 3 on the interval [_1? 1] is

1

f (:) d: ≈ Ⅰ3 (f) = 2尸 f (_1) + 21 f (_1}3) + 22 f (1}3) + 23 f (1)?      (1)

- 1

where the weights 2k  are the integrals of the Lagrange basis functions Lk (:), and the quadrature nodes are :尸  = _1, :1  = _1}3, :2  = 1}3, and :3  = 1. Using the fact that formula (1) is exact for polynomials of degree 3, show that the weights satisfy the relations

2尸 + 21 = 1?

2尸 + 1 21 = 1

and hence determine all four of the weights.

This approach provides an alternative to evaluation of the integrals  1-1 Lk (:) d:.