First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 143
role of f(x). Applying the theorem, we get
∫ b
(x − a)(x − b)f ′′ (ξ(x))dx = f ′′ (ξ)
∫ b
a
a
(x − a)(x − b)dx
where we kept the same notation ξ, somewhat inappropriately, as we moved f ′′ (ξ(x))
from inside to the outside of the integral. Finally, observe that
∫ b
a
(x − a)(x − b)dx =
(a − b)3
6
= −h3
6 ,
where h = b − a. Putting all the pieces together, we have obtained
∫ b
a
f(x)dx = h 2 [f(x 0)+f(x 1 )] − h3
12 f ′′ (ξ).
• Simpson’s rule
Let f ∈ C 4 [a, b]. Take three equally-spaced nodes, x 0 = a, x 1 = a + h, x 2 = b, where
h = b−a , and use the second degree Lagrange polynomial
2
P 2 (x) = (x − x 1)(x − x 2 )
(x 0 − x 1 )(x 0 − x 2 ) f(x 0)+ (x − x 0)(x − x 2 )
(x 1 − x 0 )(x 1 − x 2 ) f(x 1)+ (x − x 0)(x − x 1 )
(x 2 − x 0 )(x 2 − x 1 ) f(x 2)
to estimate f(x). Substitute n =2in Equation (4.1) toget
∫ b
a
f(x)dx =
∫ b
and then substitute for P 2 to obtain
∫ b
a
f(x)dx =
+
∫ b
a
∫ b
a
a
P 2 (x)dx + 1 3!
∫ b
∫
(x − x 1 )(x − x 2 )
b
(x 0 − x 1 )(x 0 − x 2 ) f(x 0)dx +
(x − x 0 )(x − x 1 )
(x 2 − x 0 )(x 2 − x 1 ) f(x 2)dx + 1 6
a
2∏
(x − x i )f (3) (ξ(x))dx,
i=0
a
∫ b
(x − x 0 )(x − x 2 )
(x 1 − x 0 )(x 1 − x 2 ) f(x 1)dx+
a
(x − x 0 )(x − x 1 )(x − x 2 )f (3) (ξ(x))dx.
The sum of the first three integrals on the right-hand side simplify as:
h
[f(x 3 0)+4f(x 1 )+f(x 2 )]. The last integral cannot be evaluated using Theorem 65
directly, like in the trapezoidal rule, since the function (x − x 0 )(x − x 1 )(x − x 2 ) changes
sign on [a, b]. However, a clever application of integration by parts transforms the
integral to an integral where Theorem 65 is applicable (see Atkinson [3] for details),