1. Elements of Numerical Integration
Sec:4.3
Elements of Numerical Integration

2. Sec:4.3 Elements of Numerical Integration
How to integrate a definite integral of a function that has no explicit antiderivative or whose antiderivative is not easy to obtain.
Example
Approximate the integral using the Trapezoidal rule
𝒂 𝒃 𝒇 𝒙 𝒅𝒙
0 2 𝑥 2 𝑑𝑥
𝑓 𝑥 = 𝑥 2 ,ℎ=2, 𝑎=0,𝑏=2
The Trapezoidal Rule
0 2 𝑥 2 𝑑𝑥 ≈ 𝟏 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 ≈ 1 2 ℎ 𝒇 𝒂 +𝒇 𝒃
≈ 𝟏 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
ℎ=b−a
≈ 𝟏 𝟐 (𝟐) 𝟎+𝟒 ≈𝟒

3. Sec:4.3 Elements of Numerical Integration
Derivation for The Trapezoidal Rule
using first Lagrange polynomials
𝑥 0 =𝑎
𝑥 1 =𝑏
ℎ=𝑏−𝑎
𝑓 𝑥 = 𝒑 𝟏 𝒙 + 𝑹 𝟏 (𝒙)
𝑓 𝑥 =𝒇 𝒙 𝟎 𝑳 𝟎 (𝒙)+𝒇 𝒙 𝟏 𝑳 𝟏 (𝒙)+ 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 )
𝑓 𝑥 =𝒇 𝒙 𝟎 𝒙− 𝒙 𝟏 𝒙 𝟎 − 𝒙 𝟏 +𝒇 𝒙 𝟏 𝒙− 𝒙 𝟎 𝒙 𝟏 − 𝒙 𝟎 + 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 )
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝒇 𝒙 𝟎 𝒙 𝟎 𝒙 𝟏 𝑳 𝟎 𝒙 𝒅𝒙 +𝒇 𝒙 𝟏 𝒙 𝟎 𝒙 𝟏 𝑳 𝟏 𝒙 𝒅𝒙+ 𝒙 𝟎 𝒙 𝟏 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 1 2 ℎ 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 + 𝒙 𝟎 𝒙 𝟏 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx
𝒙 𝟎 𝒙 𝟏 𝑳 𝟎 𝒙 𝒅𝒙= 𝒙 𝟎 𝒙 𝟏 𝒙− 𝒙 𝟏 𝒙 𝟎 − 𝒙 𝟏 𝒅𝒙= 𝟏 −𝒉 𝒙 𝟎 𝒙 𝟏 𝒙− 𝒙 𝟏 𝒅𝒙= − 𝟏 𝒉 𝒙− 𝒙 𝟏 𝟐 𝟐 𝒙 𝟎 𝒙 𝟏 = − 𝟏 𝒉 𝟎− 𝟏 𝟐 𝒉 𝟐 = 𝟏 𝟐 𝒉
𝒙 𝟎 𝒙 𝟏 𝑳 𝟏 𝒙 𝒅𝒙= 𝒙 𝟎 𝒙 𝟏 𝒙− 𝒙 𝟎 𝒙 𝟏 − 𝒙 𝟎 𝒅𝒙= 𝟏 𝒉 𝒙 𝟎 𝒙 𝟏 𝒙− 𝒙 𝟎 𝒅𝒙= 𝟏 𝒉 𝒙− 𝒙 𝟎 𝟐 𝟐 𝒙 𝟎 𝒙 𝟏 = 𝟏 𝒉 𝟏 𝟐 𝒉 𝟐 −𝟎 = 𝟏 𝟐 𝒉

4. Sec:4.3 Elements of Numerical Integration
Error for The Trapezoidal Rule
The Trapezoidal Rule
Weighted Mean Value Theorem for Integrals
Suppose f ∈ C[a, b], the Riemann integral of g exists on [a, b], and g(x) does not change
sign on [a, b]. Then there exists a number c in (a, b) with
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 1 2 ℎ 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 + 𝒙 𝟎 𝒙 𝟏 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx
𝒙 𝟎 𝒙 𝟏 𝒇 ′′ 𝝃 𝟐! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx = 𝟏 𝟐 𝒙 𝟎 𝒙 𝟏 𝒇 ′′ 𝝃(𝒙) (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx
𝑎 𝑏 𝑓 𝑥 𝑔 𝑥 𝑑𝑥=𝑓(𝑐) 𝑎 𝑏 𝑔 𝑥 𝑑𝑥
The Weighted Mean Value Theorem for Integrals
= 𝒇 ′′ 𝝃 𝟐 𝒙 𝟎 𝒙 𝟏 (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) dx
= 𝒇 ′′ 𝝃 𝟐 − 𝒉 𝟑 𝟔 ==− 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 1 2 ℎ 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃

5. Sec:4.3 Elements of Numerical Integration
Simpson’s Rule
Example
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 ≈ 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐
Approximate the integral using the Simpson’s rule
ℎ= b−a 2
0 2 𝑥 2 𝑑𝑥
𝑓 𝑥 = 𝑥 2 ,ℎ=1, 𝑥 0 =0, 𝑥 1 =1, 𝑥 2 =2
𝑥 0 =𝑎
𝑥 1 = 𝑥 0 +ℎ
𝑥 2 =𝑏
0 2 𝑥 2 𝑑𝑥 ≈
𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐
≈ 𝟏 𝟑 𝒇 𝟎 +𝟒𝒇 𝟏 +𝒇 𝟐
≈ 𝟏 𝟑 𝟎+𝟒+𝟒
≈ 𝟖 𝟑
It is exact

6. Sec:4.3 Elements of Numerical Integration
Simpson’s Rule Derivation
using second Lagrange polynomials
𝑥 0 =𝑎
𝑥 1 = 𝑥 0 +ℎ
𝑥 2 =𝑏
𝑓 𝑥 = 𝒑 𝟏 𝒙 + 𝑹 𝟏 (𝒙)
𝑓 𝑥 =𝒇 𝒙 𝟎 𝑳 𝟎 (𝒙)+𝒇 𝒙 𝟏 𝑳 𝟏 (𝒙)+𝒇 𝒙 𝟐 𝑳 𝟐 (𝒙)+ 𝒇 ′′′ 𝝃 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 ) (𝒙− 𝒙 𝟐 )
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝒇 𝒙 𝟎 𝒙 𝟎 𝒙 𝟐 𝑳 𝟎 𝒙 𝒅𝒙 +𝒇 𝒙 𝟏 𝒙 𝟎 𝒙 𝟐 𝑳 𝟏 𝒙 𝒅𝑥+𝒇 𝒙 𝟐 𝒙 𝟎 𝒙 𝟐 𝑳 𝟐 𝒙 𝒅𝒙+ 𝒙 𝟎 𝒙 𝟐 𝒇 ′′′ 𝝃 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 )(𝒙− 𝒙 𝟐 )dx
= 𝒂 𝟎 𝒇 𝒙 𝟎 + 𝒂 𝟏 𝒇 𝒙 𝟏 + 𝒂 𝟐 𝒇 𝒙 𝟐 + 𝒙 𝟎 𝒙 𝟐 𝒇 ′′′ 𝝃 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 )(𝒙− 𝒙 𝟐 )dx
𝒂 𝟎 = 𝒙 𝟎 𝒙 𝟐 𝑳 𝟎 𝒙 𝒅𝒙= 𝒉 𝟑
𝒂 𝟏 = 𝒙 𝟎 𝒙 𝟐 𝑳 𝟏 𝒙 𝒅𝒙= 𝟒𝒉 𝟑
𝒂 𝟐 = 𝒙 𝟎 𝒙 𝟐 𝑳 𝟐 𝒙 𝒅𝒙= 𝒉 𝟑
Simpson’s Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐 + 𝒙 𝟎 𝒙 𝟐 𝒇 ′′′ 𝝃 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟏 )(𝒙− 𝒙 𝟐 )dx
Deriving Simpson’s rule in this manner, however, provides only an 𝑂( ℎ 4 ) error term involving 𝒇′′′ (𝝃).

7. Sec:4.3 Elements of Numerical Integration
approaching the problem in another way, a higher-order term
𝑥 0 =𝑎
𝑥 1 = 𝑥 0 +ℎ
𝑥 2 =𝑏
Third Taylor polynomial
𝑓 𝑥 =𝑓 𝑥 1 + 𝑓 ′ ( 𝑥 1 ) 𝑥− 𝑥 𝑓 ′ ′( 𝑥 1 ) 2 𝑥− 𝑥 𝑓 ′ ′′( 𝑥 1 ) 6 𝑥− 𝑥 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒
𝑥 0 𝑥 2 𝑓 𝑥 𝑑𝑥 = 𝑥 0 𝑥 2 𝑓 𝑥 1 + 𝑓 ′ ( 𝑥 1 ) 𝑥− 𝑥 𝑓 ′ ′( 𝑥 1 ) 2 𝑥− 𝑥 𝑓 ′ ′′( 𝑥 1 ) 6 𝑥− 𝑥 𝑑𝑥
+ 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
𝒙 𝟎 𝒙 𝟐 𝒇 ′ ( 𝒙 𝟏 ) 𝒙− 𝒙 𝟏 𝒅𝒙 =𝟎
𝑥 0 𝑥 𝑓 ′ ′′( 𝑥 1 ) 6 𝑥− 𝑥 𝑑𝑥=0
𝑥 0 𝑥 2 𝑓 𝑥 𝑑𝑥 = 𝑥 0 𝑥 2 𝑓 𝑥 𝑓 ′ ′( 𝑥 1 ) 2 𝑥− 𝑥 𝑑𝑥 + 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙

8. Sec:4.3 Elements of Numerical Integration
𝑥 0 𝑥 2 𝑓 𝑥 𝑑𝑥 = 𝑥 0 𝑥 2 𝑓 𝑥 𝑓 ′ ′( 𝑥 1 ) 2 𝑥− 𝑥 𝑑𝑥 + 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
=2ℎ𝑓 𝑥 ℎ 3 3 𝑓′′ 𝑥 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
𝒇 ′′ 𝒙 𝟏 = 𝒇 𝒙 𝟎 −𝟐𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) 𝒉 𝟐 − 𝒉 𝟐 𝟏𝟐 𝒇 𝟒 ( 𝝃 𝟐 )
Using centered difference formaula for
=2ℎ𝑓 𝑥 ℎ 𝒇 𝒙 𝟎 −𝟐𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) 𝒉 𝟐 − 𝒉 𝟐 𝟏𝟐 𝒇 𝟒 ( 𝝃 𝟐 ) + 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
=2ℎ𝑓 𝑥 1 + ℎ 3 𝒇 𝒙 𝟎 −𝟐𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − ℎ 3 𝒉 𝟒 𝟏𝟐 𝒇 𝟒 ( 𝝃 𝟐 ) + 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
= ℎ 3 𝟔𝒇 𝒙 𝟏 +𝒇 𝒙 𝟎 −𝟐𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟑𝟔 𝒇 𝟒 ( 𝝃 𝟐 ) + 𝒙 𝟎 𝒙 𝟐 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
𝑥 0 𝑥 2 𝑥− 𝑥 𝑑𝑥 =
= 𝑥− 𝑥 𝑥 0 𝑥 2 = 2 ℎ 5 5
The Weighted Mean Value Theorem for Integrals
= ℎ 3 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟑𝟔 𝒇 𝟒 ( 𝝃 𝟐 ) + 𝒇 (𝟒) (𝝃) 𝟐𝟒 𝒙 𝟎 𝒙 𝟐 𝒙− 𝒙 𝟏 𝟒 𝒅𝒙
= ℎ 3 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟑𝟔 𝒇 𝟒 𝝃 𝟐 + 𝒉 𝟓 𝟔𝟎 𝒇 𝟒 𝝃
= ℎ 3 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗

9. Sec:4.3 Elements of Numerical Integration
Simpson’s Rule
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 )
− 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
− 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
f x^2;
a=0; b=2;
h=(b-a)/2;
x=a:h:b;
integ = (h/3)*(f(x(1))+4*f(x(2))+f(x(3)))
f x^2;
a=0; b=2;
h=b-a;
integ = (h/2)*(f(a)+f(b))

10. Sec:4.3 Elements of Numerical Integration
Notice that in each instance Simpson’s Rule is significantly superior.
Simpson’s Rule
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 )
− 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
− 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃

11. Sec:4.3 Elements of Numerical Integration
Definition
The degree of precision of a quadrature formula is n if and only if the error is zero for all polynomials
𝑞 0 𝑥 =1, 𝑞 1 𝑥 =𝑥, 𝑞 2 𝑥 = 𝑥 2 ,⋯, 𝑞 𝑛 𝑥 = 𝑥 𝑛
but is not zero for the polynomial
𝑞 𝑛+1 𝑥 = 𝑥 𝑛+1
0 1 1 𝑑𝑥 =1
The Trapezoidal Rule
the Trapezoidal has degree of precision one
exact
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
− 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
0 1 𝑥 𝑑𝑥 = 1 2
Also, it is clear from the error term (i.e 𝑓′′(𝜉)
exact
0 1 𝑥 2 𝑑𝑥 = 1 2
not exact

12. Sec:4.3 Elements of Numerical Integration
Definition
The degree of precision of a quadrature formula is n if and only if the error is zero for all polynomials
𝑞 0 𝑥 =1, 𝑞 1 𝑥 =𝑥, 𝑞 2 𝑥 = 𝑥 2 ,⋯, 𝑞 𝑛 𝑥 = 𝑥 𝑛
but is not zero for the polynomial
𝑞 𝑛+1 𝑥 = 𝑥 𝑛+1
Simpson’s Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 )
− 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
Simpson’s Rule has degree of precision three 𝑓′′′′(𝜉)

13. Sec:4.3 Elements of Numerical Integration
Definition
The degree of precision of a quadrature formula is n if and only if the error is zero for all polynomials
𝑞 0 𝑥 =1, 𝑞 1 𝑥 =𝑥, 𝑞 2 𝑥 = 𝑥 2 ,⋯, 𝑞 𝑛 𝑥 = 𝑥 𝑛
but is not zero for the polynomial
𝑞 𝑛+1 𝑥 = 𝑥 𝑛+1

14. Sec:4.3 Elements of Numerical Integration
The Trapezoidal Rule (2pts)
Simpson’s Rule (3pts)
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
𝑓 𝑥 ≈𝒇 𝒙 𝟎 𝑳 𝟎 (𝒙)+⋯+𝒇 𝒙 𝒏 𝑳 𝒏 (𝒙)
𝒉= 𝒃−𝒂 𝒏
𝑓 𝑥 ≈ 𝑖=0 𝑛 𝒇 𝒙 𝒊 𝑳 𝒊 (𝒙)
𝑥 0 𝑥 𝑛 𝑓 𝑥 𝑑𝑥 ≈ 𝑖=0 𝑛 𝒇 𝒙 𝒊 𝒙 𝟎 𝒙 𝒏 𝑳 𝒊 (𝒙)𝒅𝒙
𝑥 0 𝑥 𝑛 𝑓 𝑥 𝑑𝑥 ≈ 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊
Closed Newton-Cotes Formulas
The (n+1)-point closed Newton-Cotes formula

15. 𝑹 𝑹 𝑹 Sec:4.3 Elements of Numerical Integration
Closed Newton-Cotes Formulas
The (n+1)-point closed Newton-Cotes formula
𝑥 0 𝑥 𝑛 𝑓 𝑥 𝑑𝑥 = 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊 + 𝑹
𝒂 𝒊 = 𝒙 𝟎 𝒙 𝒏 𝑳 𝒊 (𝒙)𝒅𝒙 = 𝒙 𝟎 𝒙 𝒏 𝒋=𝟎 𝒋≠𝒊 𝒏 (𝒙− 𝒙 𝒋 ) ( 𝒙 𝒊 − 𝒙 𝒋 ) 𝒅𝒙
where
Theorem 4.2
There exists 𝜉∈ 𝑎,𝑏 such that
𝒏 is odd
𝒏 is even 𝑹
= ℎ 𝑛+2 𝑓 𝑛+1 (𝜉) 𝑛+1 ! 0 𝑛 𝑡 (𝑡−1)⋯ 𝑡−𝑛 𝑑𝑡 𝑹
= ℎ 𝑛+3 𝑓 𝑛+2 (𝜉) 𝑛+2 ! 0 𝑛 𝑡 2 (𝑡−1)⋯ 𝑡−𝑛 𝑑𝑡
the degree of precision is 𝑛
the degree of precision is 𝑛+1
The Trapezoidal Rule (2pts, n=1)
Simpson’s Rule (3pts, n=2)
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗

16. 𝑹 Sec:4.3 Elements of Numerical Integration
n=1 : The Trapezoidal Rule
The (n+1)-point closed Newton-Cotes formula
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
𝑥 0 𝑥 𝑛 𝑓 𝑥 𝑑𝑥 = 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊 + 𝑹
n=2 : Simpson’s Rule
It is called closed because the endpoints are included as nodes.
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃
n = 3: Simpson’s Three-Eighths rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟑𝒉 𝟖 𝒇 𝒙 𝟎 +𝟑𝒇 𝒙 𝟏 +𝟑𝒇 𝒙 𝟐 +𝒇( 𝒙 𝟑 ) − 𝟑𝒉 𝟓 𝟖𝟎 𝒇 𝟒 𝝃
n = 4:
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟐𝒉 𝟒𝟓 𝟕𝒇 𝒙 𝟎 +𝟑𝟐𝒇 𝒙 𝟏 +𝟏𝟐𝒇 𝒙 𝟐 +𝟑𝟐𝒇 𝒙 𝟑 +𝟕𝒇( 𝒙 𝟒 ) − 𝟖𝒉 𝟕 𝟗𝟒𝟓 𝒇 𝟔 𝝃

17. 𝑹 Sec:4.3 Elements of Numerical Integration Open Newton-Cotes Formulas
𝒉= 𝒃−𝒂 𝒏+𝟐
The open Newton-Cotes formulas do not include the endpoints as nodes.
Open Newton-Cotes Formulas
𝑥 −1 𝑥 𝑛+1 𝑓 𝑥 𝑑𝑥 = 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊 +
𝒂 𝒊 = 𝒙 −𝟏 𝒙 𝒏+𝟏 𝑳 𝒊 (𝒙)𝒅𝒙 = 𝒙 −𝟏 𝒙 𝒏+𝟏 𝒋=𝟎 𝒋≠𝒊 𝒏 (𝒙− 𝒙 𝒋 ) ( 𝒙 𝒊 − 𝒙 𝒋 ) 𝒅𝒙 𝑹
where

18. 𝑹 𝑹 𝑹 Sec:4.3 Elements of Numerical Integration
Open Newton-Cotes Formulas
𝑥 −1 𝑥 𝑛+1 𝑓 𝑥 𝑑𝑥 = 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊 +
𝒂 𝒊 = 𝒙 −𝟏 𝒙 𝒏+𝟏 𝑳 𝒊 (𝒙)𝒅𝒙 = 𝒙 −𝟏 𝒙 𝒏+𝟏 𝒋=𝟎 𝒋≠𝒊 𝒏 (𝒙− 𝒙 𝒋 ) ( 𝒙 𝒊 − 𝒙 𝒋 ) 𝒅𝒙 𝑹
where
Theorem 4.2
There exists 𝜉∈ 𝑎,𝑏 such that
𝒏 is odd
𝒏 is even 𝑹
= ℎ 𝑛+2 𝑓 𝑛+1 (𝜉) 𝑛+1 ! −1 𝑛+1 𝑡 (𝑡−1)⋯ 𝑡−𝑛 𝑑𝑡 𝑹
= ℎ 𝑛+3 𝑓 𝑛+2 (𝜉) 𝑛+2 ! −1 𝑛+1 𝑡 2 (𝑡−1)⋯ 𝑡−𝑛 𝑑𝑡
n=1 :
n=0 : The midpoint Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟑𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝟑𝒉 𝟑 𝟒 𝒇 ′′ 𝝃
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃

19. 𝑹 Sec:4.3 Elements of Numerical Integration Open Newton-Cotes Formulas
n=0 : The midpoint Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
𝑥 −1 𝑥 𝑛+1 𝑓 𝑥 𝑑𝑥 = 𝑖=0 𝑛 𝒂 𝒊 𝒇 𝒙 𝒊 + 𝑹
n=1 :
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟑𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝟑𝒉 𝟑 𝟒 𝒇 ′′ 𝝃
n = 2:
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟒𝒉 𝟑 𝟐𝒇 𝒙 𝟎 −𝒇 𝒙 𝟏 +𝟐𝒇 𝒙 𝟐 − 𝟏𝟒𝒉 𝟓 𝟒𝟓 𝒇 𝟒 𝝃
n = 3:
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝟓𝒉 𝟐𝟒 𝟏𝟏𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐 +𝟏𝟏𝒇 𝒙 𝟑 − 𝟗𝟓𝒉 𝟕 𝟏𝟒𝟒 𝒇 𝟔 𝝃

20. Sec:4.3 Elements of Numerical Integration
Example

21. Sec:4.3 Elements of Numerical Integration
Example

22. Sec:4.3 Elements of Numerical Integration

23. Sec:4.3 Elements of Numerical Integration
summary