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.
𝒂 𝒃 𝒇 𝒙 𝒅𝒙

3. Sec:4.3 Elements of Numerical Integration
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
ℎ=𝑏−𝑎
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 ∗
ℎ=(𝑏−𝑎)/2
The midpoint Rule 𝑎 𝑏
𝑥 0
𝒚=𝒇(𝒙)
How to use
MATLAB
Derivation
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
ℎ=(𝑏−𝑎)/2
𝑥 0 =(𝑏+𝑎)/2

4. Sec:4.3 Elements of Numerical Integration
Example
The Trapezoidal Rule
Approximate the integral using the Trapezoidal rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
0 2 𝑥 2 𝑑𝑥
ℎ=𝑏−𝑎
𝑓 𝑥 = 𝑥 2 ,ℎ=2, 𝑎=0,𝑏=2
0 2 𝑥 2 𝑑𝑥 ≈ 𝟏 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
≈ 𝟏 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏
≈ 𝟏 𝟐 (𝟐) 𝟎+𝟒 ≈𝟒

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

6. Sec:4.3 Elements of Numerical Integration
Example
The midpoint Rule
Approximate the integral using the midpoint rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
0 𝜋 4 sin⁡(𝑥) 𝑑𝑥
ℎ=(𝑏−𝑎)/2
𝑥 0 =(𝑏+𝑎)/2 𝑎 𝑏
𝑥 0
𝒚=𝒇(𝒙)
𝑓 𝑥 =sin⁡(𝑥),ℎ= 𝜋 8 , 𝑎=0, 𝑏= 𝜋 4
𝑥 0 = 𝜋 8 ,
0 𝜋 4 sin⁡(𝑥)𝑑𝑥 ≈
𝟐( 𝜋 8 ) 𝒇 𝜋 8
𝑎=0
𝑥 0 = 𝜋 8
𝑏= 𝜋 4
≈𝟎.𝟑𝟎𝟎𝟓𝟓𝟖𝟖𝟕

7. Sec:4.3 Elements of Numerical Integration
The Trapezoidal Rule
f x^2;
a=0; b=2;
h=b-a;
intg = (h/2)*(f(a)+f(b))
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
ℎ=𝑏−𝑎
Simpson’s Rule
f x^2;
a=0; b=2;
h=(b-a)/2;
x=a:h:b;
intg = (h/3)*(f(x(1))+4*f(x(2))+f(x(3)))
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
ℎ=(𝑏−𝑎)/2
The midpoint Rule
f sin(x);
a=0; b=pi/4;
h=(b-a)/2;
x0= =(b+a)/2;
intg = 2*h*f(x0)
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
ℎ=(𝑏−𝑎)/2
𝑥 0 =(𝑏+𝑎)/2

8. Sec:4.3 Elements of Numerical Integration
Example
Simpson’s Rule
ℎ=(𝑏−𝑎)/2
Approximate the integral using the 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 ∗
0 2 𝑥 2 𝑑𝑥
Definition
The degree of precision (or accuracy) of integral 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
𝑓 𝑥 = 𝑥 2 ,ℎ=1, 𝑥 0 =0, 𝑥 1 =1, 𝑥 2 =2
0 2 𝑥 2 𝑑𝑥 ≈
𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐
≈ 𝟏 𝟑 𝒇 𝟎 +𝟒𝒇 𝟏 +𝒇 𝟐
≈ 𝟖 𝟑
It is exact
The approximation from Simpson's rule is exact because its truncation error involves 𝒇 𝟒 𝝃 ∗ , which is identically 0 when 𝑓 𝑥 = 𝑥 2
Remark
the degree of precision of Simpson’s rule is 3.
The integral of any polynomial of degree 3 or less is exact when using Simpson’s rule

9. How to find degree of precision
Sec:4.3 Elements of Numerical Integration
Definition
Definition
The degree of precision (or accuracy) of integral 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
Find the degree of precision of the formula
−𝟏 𝟏 𝒇(𝒙) 𝒅𝒙≈𝒇 − 𝟑 𝟑 +𝒇( 𝟑 𝟑 )
solution
function:
Exact:
Use the formula:
−𝟏 𝟏 𝟏 𝒅𝒙=𝟐
𝒇 − 𝟑 𝟑 +𝒇 𝟑 𝟑 =𝟏+𝟏=𝟐
𝒇 𝒙 =𝟏
How to find degree of precision
function:
Exact:
Use the formula:
−𝟏 𝟏 𝒙 𝒅𝒙=𝟎
𝒇 − 𝟑 𝟑 +𝒇 𝟑 𝟑 =− 𝟑 𝟑 + 𝟑 𝟑 =𝟎
𝒇 𝒙 =𝒙
𝒇 𝒙 =𝟏
𝒂 𝒃 𝟏 𝒅𝒙
𝒂 𝒃 𝟏 𝒅𝒙
−𝟏 𝟏 𝒙 𝟐 𝒅𝒙 = 𝟐 𝟑
𝒇 − 𝟑 𝟑 +𝒇 𝟑 𝟑 = 𝟏 𝟑 + 𝟏 𝟑 = 𝟐 𝟑
𝒇 𝒙 = 𝒙 𝟐
𝒇 𝒙 =𝒙
𝒂 𝒃 𝒙 𝒅𝒙
𝒂 𝒃 𝒙 𝒅𝒙
−𝟏 𝟏 𝒙 𝟑 𝒅𝒙 =𝟎
𝒇 − 𝟑 𝟑 +𝒇 𝟑 𝟑 =− 𝟑 𝟗 + 𝟑 𝟗 =𝟎
𝒇 𝒙 = 𝒙 𝟑
𝒇 𝒙 = 𝒙 𝟐
𝒂 𝒃 𝒙 𝟐 𝒅𝒙
𝒂 𝒃 𝒙 𝟐 𝒅𝒙
−𝟏 𝟏 𝒙 𝟒 𝒅𝒙 = 𝟐 𝟓
𝒇 − 𝟑 𝟑 +𝒇 𝟑 𝟑 = 𝟏 𝟗 + 𝟏 𝟗 = 𝟐 𝟗
𝒇 𝒙 = 𝒙 𝟒 ⋮ ⋮ ⋮
the degree of precision is 3.

10. Derivation Sec:4.3 Elements of Numerical Integration
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
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
ℎ=𝑏−𝑎
𝑎 𝑏 𝑓 𝑥 𝑔 𝑥 𝑑𝑥=𝒇(𝒄) 𝑎 𝑏 𝑔 𝑥 𝑑𝑥

11. 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

12. 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

13. Derivation Sec:4.3 Elements of Numerical Integration 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 ∗
ℎ=(𝑏−𝑎)/2

14. 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).

15. 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

16. 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 ∗

17. 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

18. Sec:4.3 Elements of Numerical Integration
The Trapezoidal Rule
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟐 𝒇 𝒙 𝟎 +𝒇 𝒙 𝟏 − 𝒉 𝟑 𝟏𝟐 𝒇 ′′ 𝝃
The Trapezoidal and Simpson's rules are examples of a class of methods known as closed Newton-cotes methods.
ℎ=𝑏−𝑎
Simpson’s Rule
In the formula information in the endpoints are used
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝒉 𝟑 𝒇 𝒙 𝟎 +𝟒𝒇 𝒙 𝟏 +𝒇( 𝒙 𝟐 ) − 𝒉 𝟓 𝟗𝟎 𝒇 𝟒 𝝃 ∗
ℎ=(𝑏−𝑎)/2
The midpoint Rule
The midpoint rules is an example of a class of methods known as open Newton-cotes methods.
𝑎 𝑏 𝑓 𝑥 𝑑𝑥 =𝟐𝒉𝒇 𝒙 𝟎 + 𝒉 𝟑 𝟑 𝒇 ′′ 𝝃
In the formula information in the endpoints are not used
ℎ=(𝑏−𝑎)/2
𝑥 0 =(𝑏+𝑎)/2

19. 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

20. 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 ∗

21. 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

22. 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

23. 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

24. 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

25. Sec:4.3 Elements of Numerical Integration
Example

26. Sec:4.3 Elements of Numerical Integration
Example

27. Sec:4.3 Elements of Numerical Integration

28. Sec:4.3 Elements of Numerical Integration
summary