First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 144
and the integral simplifies as − h5
90 f (4) (ξ) for some ξ ∈ (a, b). In summary, we obtain
where ξ ∈ (a, b).
∫ b
a
f(x)dx = h 3 [f(x 0)+4f(x 1 )+f(x 2 )] − h5
90 f (4) (ξ)
Exercise 4.1-1:
any n.
Prove that the sum of the weights in Newton-Cotes rules is b − a, for
Definition 66. The degree of accuracy, or precision, of a quadrature formula is the largest
positive integer n such that the formula is exact for f(x) =x k , when k =0, 1, ..., n, or
equivalently, for any polynomial of degree less than or equal to n.
Observe that the trapezoidal and Simpson’s rules have degrees of accuracy of one and
three. These two rules are examples of closed Newton-Cotes formulas; closed refers to the
fact that the end points a, b of the interval are used as nodes in the quadrature rule. Here is
the general definition.
Definition 67 (Closed Newton-Cotes). The (n+1)-point closed Newton-Cotes formula uses
nodes x i = x 0 + ih, for i =0, 1, ..., n, where x 0 = a, x n = b, h = b−a,
and n
w i =
∫ xn
x 0
l i (x)dx =
∫ xn
x 0
n
∏
j=0,j≠i
x − x j
x i − x j
dx.
The following theorem provides an error formula for the closed Newton-Cotes formula.
A proof can be found in Isaacson and Keller [12].
Theorem 68. For the (n +1)-point closed Newton-Cotes formula, we have:
•ifn is even and f ∈ C n+2 [a, b]
∫ b
a
f(x)dx =
n∑
i=0
w i f(x i )+ hn+3 f (n+2) (ξ)
(n +2)!
∫ n
0
t 2 (t − 1) ···(t − n)dt,
•ifn is odd and f ∈ C n+1 [a, b]
∫ b
a
f(x)dx =
n∑
i=0
w i f(x i )+ hn+2 f (n+1) (ξ)
(n +1)!
∫ n
0
t(t − 1) ···(t − n)dt