First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 142
The first integral on the right-hand side gives the quadrature rule:
⎛ ⎞
∫ b
∫ (
b n∑
)
n∑
∫ b
P n (x)dx = f(x i )l i (x) dx = ⎜
⎝ l i (x)dx⎟
⎠ f(x i)=
a
a i=0
i=0 a
} {{ }
w i
n∑
w i f(x i ),
i=0
and the second integral gives the error term.
We obtain different quadrature rules by taking different nodes, or number of nodes. The
following result is useful in the theoretical analysis of Newton-Cotes formulas.
Theorem 65 (Weighted mean value theorem for integrals). Suppose f ∈ C 0 [a, b], the Riemann
integral of g(x) exists on [a, b], and g(x) does not change sign on [a, b]. Then there
exists ξ ∈ (a, b) with ∫ b
f(x)g(x)dx = f(ξ) ∫ b
g(x)dx.
a a
Two well-known numerical quadrature rules, trapezoidal rule and Simpson’s rule, are
examples of Newton-Cotes formulas:
• Trapezoidal rule
Let f ∈ C 2 [a, b]. Take two nodes, x 0 = a, x 1 = b, and use the linear Lagrange
polynomial
P 1 (x) = x − x 1
f(x 0 )+ x − x 0
f(x 1 )
x 0 − x 1 x 1 − x 0
to estimate f(x). Substitute n =1in Equation (4.1) toget
∫ b
a
f(x)dx =
∫ b
a
P 1 (x)dx + 1 2
∫ b
and then substitute for P 1 to obtain
∫ b ∫ b
∫
x − x b
1
x − x 0
f(x)dx =
f(x 0 )dx +
f(x 1 )dx + 1
a
a x 0 − x 1 a x 1 − x 0 2
a
1∏
(x − x i )f ′′ (ξ(x))dx,
i=0
∫ b
a
(x − x 0 )(x − x 1 )f ′′ (ξ(x))dx.
The first two integrals on the right-hand side can be evaluated easily: the first integral
is hf(x 2 0) and the second one is hf(x 2 1), where h = x 1 − x 0 = b − a. Let’s evaluate
∫ b
a
(x − x 0 )(x − x 1 )f ′′ (ξ(x))dx.
We will use Theorem 65 for this computation. Note that the function (x−x 0 )(x−x 1 )=
(x − a)(x − b) does not change sign on the interval [a, b] and it is integrable: so this
function serves the role of g(x) in Theorem 65. The other term, f ′′ (ξ(x)), serves the