1. Trapezoidal method

In this method, we divide the area under the curve in trapezes.

A = Σ A_i
h = (b - a)/n
A_i = h [f(x_i) + f(x_i+1)]/2 I = A₀ + A₁ + A₂ + ... + A_n-1
= h [f(a) + f(x1)]/2 + h [f(x1) + f(x2)]/2 + h [f(x2) + f(x3)]/2 + ... + h [f(xn-1) + f(b)]/2
= (h/2) [f(a) + f(x1) + f(x1) + f(x2) + f(x2) + f(x3) + ... + f(xn) + f(b)]
= (h/2) [f(a) + 2f(x1) + 2f(x2) + 2f(x3) + ... + 2f(xn-1) + f(b)]
= h [(f(a) + f(b))/2 + f(x1) + f(x2) + f(x3) + ... + f(xn-1)]
= h [(f(a) + f(b))/2 + Σf(xi) + ... + f(xn-1)]
i runs from 1 to n-1

A = h [(f(a) + f(b))/2 + Σ f(xi)]
i runs from 1 to n-1
x_i = a + i h
x₀ = a
x_n = b

2. Rectangle method

A = Σ A_i
h = (b - a)/n
A_i = h f(x_i)
I = A₀ + A₁ + A₂ + ... + A_n-1
= h f(x₀) + h f(x₁) + ... + h f(x_n-1)
= h Σ f(x_i) i runs from 0 to n-1

A = h Σ f(x_i)
i runs from 0 to n-1
xi = a + i h
x₀ = a
x_n = b

3. Midpoint method

A = Σ A_i
h = (b - a)/n
A_i = h f(x_i) I = A₀ + A₁ + A₂ + ... + A_n
= h Σ f(x_i)
i runs from 0 to n
x_i = a + (i + 1/2) x h

A = = h{[f(a) + f(b)]/2 + Σ f(x_i)}
i runs from 0 to n-1
x_i = a + (i + 1/2) h
x₀ = a + h/2
x_n-1 + (n - 1/2) h
x₀ - h/2 = a
x_n-1 + h/2 = b

4. Simpson method

The area under the curve on the interval [x0, x2] is approximated by a parabola y = ax² + bx + c. The total integral will be equal to the sum of all the area under the set of parabolic arches.



A₁ = ∫ (ax² + bx + c)dx
from - h to + h
= (a/3)[+ h³ + h³ ] + (a/2)[h² - h²] + c[h + h] =
(2a/3)h³ + 0 + 2ch = (2a/3)h³ + 2ch

A₁ = (h/3)[2ah² + 6c ]       (1)

We have also:
f(x₀) = f(-h) = ah² - bh + c
f(x₁) = f(0) = c
f(x₂) = f(+h) = ah² + bh + c

Then:
A₁ = (h/3)[2ah² + 6c ] = (h/3)[ f(x₀) + f(x₂) - 2c + 6f(x₁)] =
(h/3)[ f(x₀) + f(x₂) - 2f(x₁) + 6f(x₁)] = (h/3)[ f(x₀) + 4f(x₁) + f(x₂) ] =

A₁ = (h/3)[ f(x₀) + 4f(x₁) + f(x₂) ]       (2)

A = Σ A_i = A₁ + A₃ + ... + A_n-1 =
(h/3)[ f(x₀) + 4f(x₁) + f(x₂) ] + (h/3)[ f(x₂) + 4f(x₃) + f(x₄) ] + (h/3)[ f(x₄) + 4f(x₅) + f(x₆) ] + ... +
(h/3)[ f(x_n-2) + 4f(x_n-1) + f(x_N) ]
Simpson’s rule requires an even number "N" of subintervals.
A = (h/3)[ f(x₀) + 4f(x₁) + 2f(x₂) + + 4f(x₃) + 2f(x₄) + + 4f(x₅) + f(x₆) + ... + 2f(x_n-2) + 4f(x_n-1) + f(x_N)]

A_i = (h/3)[ f(x_i-1) + 4f(x_i) + f(x_i+1) ]       (3)

Therefore:

A = Σ A_i = (h/3)Σ[ f(x_i-1) + 4f(x_i) + f(x_i+1) ]
i runs from: 1 to n

5.Gauss Quadrature rule

In this method, we equate the function to integrate f(x) to a polynomial and integrate this polynomial that will be equal to a linear combination of other simple functions.

Thar is:
∫f(x) dx (from a to b) = ∫ Σ a_ixⁱ) dx = Σ c_if(x_i)
i runs from 0 to n

∫ f(x) dx from a to b = Σ c_i f(x_i) i from 0 to n
x_i is called weight, and x_i: argument.


f(x) = a₀ + a₁ x¹ + a₂ x² + a₃ x³ + ... + a_n xⁿ = Σ a_i xⁱ i from 0 to n
Σ c_i f(x_i) i from 0 to n = Σ a_i xⁱ i from 0 to n
c₀ f(x₀) = a₀ x⁰
c₁ f(x₁) = a₁ x¹
c₂ f(x2) = a₂ x²
...
c_n f(x_n) = a_n xⁿ

∫ f(x) dx from a to b = Σ c_i f(x_i) = ∫ [Σ a_i xⁱ] dx

= a₀(b - a) + (a₁/2)(b² - a²) + (a₂/3)(b³ - a³) + ... + (a_n/(n+1))(bⁿ⁺¹ - aⁿ⁺¹)
Then:
a₀(b - a) = c₀ f(x₀)
(a₁/2)(b² - a²) = c₁ f(x₁)
...
(a_n/(n+1))(bⁿ⁺¹ - aⁿ⁺¹) = c_n f(x_n)

Example:
I = ∫ (2 + 3 x²) dx from -1 to +1
f(x) = 2 + 3 x²
. Litterally, we have:
∫ f(x) dx = 2 [1 + 1] + 0 + 3 (1/3)[1³ - (-1)³] = 4 + 0 + 2 = 6

1. One point quadrature rule:
∫ f(x) dx = c₁f(x₁)
f(x) = a₀x⁰ + a₁x¹
∫ f(x) dx from -1 to + 1 = a₀ x 2 + a₁ (1/2) [(1)² - (-1)²] = 2a₀ + 0 a₁ = 2a₀

c₁f(x₁) =
c₁a₀x⁰ + c₁a₁x¹ = 2a₀
c₁a₀x⁰ = 2 a₀
c₁a₁x¹ = 0
c₁x⁰ = 2
c₁x¹ = 0
c₁ = 2
x¹ = 0
For our example:
∫ f(x) dx = 2 f(0) = 2 x 2 = 4
The order one is not sufficient ...

2. Two points quadrature rule :
∫ f(x) dx = c₁f(x₁) + c₂f(x₂)
f(x) = a₀x⁰ + a₁x¹ + a₂x² + a₃x³

∫ f(x) dx from a to b = a₀ (b - a) + (a₁/2) [b² - a²] + (a₂/3) [b³ - a³] + + (a₃/4) [b⁴ - a⁴]

c₁f(x₁) + c₂f(x₂) = c₁ [a₀ + a₁x₁+ a₂x₁² + a₃x₁³] +
c₂[a₀ + a₁x₂ + a₂x₂² + a₃x₂³]

(c₁ + c₂)a₀ + (c₁x₁ + c₂ x₂)a₁ + (c₁x₁² + c₂x₂²)a₂ + (c₁x₁³+ c₂x₂³)a₃

(c₁ + c₂)a₀ + (c₁x₁ + c₂ x₂)a₁ + (c₁x₁² + c₂x₂²)a₂ + (c₁x₁³+ c₂x₂³)a₃

a₀ (b - a) + (a₁/2) (a² - b²) + (a₂/3) (a³ - b³) + + (a₃/4) (a⁴ - b⁴)
Therefore:
c₁ + c₂ = (b - a)
c₁x₁ + c₂ x₂ = (1/2) (b² - a²)
c₁x₁² + c₂x₂² = (1/3) (b³ - a³)
c₁x₁³+ c₂x₂³ (1/4) (b⁴ - a⁴)

Example:
a = -1 and b = +1, then:
c₁ + c₂ = 2     (1)
c₁x₁ + c₂ x₂ = 0   (2)
c₁x₁² + c₂x₂² = 2/3   (3)
c₁x₁³+ c₂x₂³ = 0   (4)

c₁x₁ = - c₂ x₂    (2)
c₁x₁³ = - c₂x₂³    (4)

Dividing (4)by (2) yields:
x₁² = x₂²
Then:
x₁ = +/- x₂
If x₁ = x₂ then c₁ - c₂ = 0 No according to (2) and we want these two arguments not null. Therefore:
x₁ = - x₂ and c₁ = c₂ = 1
Using (3)
c₁x₁² + c₂x₁² = 2/3   (3)
[c₁ + c₂]x₁² = 2/3
Using (1)
x₁² = 1/3 and x₁= 1/3^1/2

Then:
c₁ = c₂ = 1
x₁ = - x₂ = 1/3^1/2


For our example:
f(x) = 2 + 3 x²
∫ f(x) dx = c₁f(x₁) + c₂f(x₂) = f(1/3^1/2) + f( - 1/3^1/2)
= 2 + 1 + 2 + 1 = 6
That is the exact result.

First order : 4
second order: 6


3. General case:
∫ f(x) dx = Σ c_if(x_i) i from 1 to n
f(x) = Σ a_ixⁱxⁱ i from 1 to n
Then:
Σ c_i = (b - a)
Σ c_ix_i = (1/2) (b² - a²)
Σ c_ix_i² = (1/3) (b³ - a³)
Σ c_ix_i³ = (1/4) (b⁴ - a⁴)
....
Σ c_ix_i^k = (1/(k+1)) (b^k+1 - a^k)

ΣΣ c_ix_i^k = (1/(k+1)) (b^k+1 - a^k)
i from: 1 to n and k from: 0 to n+1

4. General case for integrations limits:
∫ f(x) dx from a to b, can be written as:

∫ f(x) dx from -1 to +1 by changing the variable x to X
x = (b - a)X/2 + (a + b)/2
Then:
dx = (b - a)dX/2, then:
∫ f(x) dx from a to b = [(b - a)/2]∫ f((b - a)X/2 + (a + b)/2) dX from -1 to +1

The Gauss n quardature rule is:
ΣΣ c_ix_i^k = (1/(k+1)) (b^k+1 - a^k)
i from: 1 to n and k from: 0 to n+1
= ∫ f(x) dx from a to b = [(b - a)/2]∫ f((b - a)X/2 + (a + b)/2) dX from -1 to +1
With x = (b - a)X/2 + (a + b)/2

∫ f(x) dx from a to b = [(b - a)/2]∫ f((b - a)x/2 + (a + b)/2)dx
x runs from -1 to +1
= [(b - a)/2] Σ c_ix_i
Where c_i and x_i are solved with a = -1 and b = +1 in the equation:
ΣΣ c_ix_i^k = (1/(k+1)) (b^k+1 - a^k)
i runs from 1 to n and k from: 0 to n+1

Example

I = ∫ (2 + 3 x²) , dx from -2 to +5
= [(5 + 2)/2]∫ f((5 + 2)X/2 + (5 - 2)/2), dX from -1 to +1
= (7/2)∫ f(7X/2 + (3/2)) dX from -1 to +1
= (7/2)∫ (2 + 3 [7X/2 + (3/2)²]) dX from -1 to +1 = (7/2)∫ (2 + 3 [(49/4)x² + (21/2)X +(9/4)]) dX from -1 to +1 =
(7/2)∫ (2 + (147/4)x² + (63/2)X +(27/4)) dX from -1 to +1 = (7/2)∫ ((147/4)x² + (63/2)X +(35/4)) dX from -1 to +1 =
(7/2)((147/12)x³ + (63/4)X² +(35/4)X) = (7/2)((147/12)[1 + 1] + (63/4)[1 - 1] +(35/4)[1 +1]) = (7/2)(252/6) = (7/2)(42) = 147

Litterally,
I = ∫ (2 + 3 x²) dx from -2 to +5 =
2[7] + (3/3)[5³ - (-2)x³] = 14 + [125 + 8] = 147
We obtain the same result.
Now we will integrate by using Gauss' two quadrature order method: I = ∫ (2 + 3 x²) , dx from -2 to +5
= (7/2)∫ (2 + 3X²), dX from -1 to +1
with X = (5 + 2)/2 x + (-2 + 5)/2 = 3.5 x + 1.5
Then:
c₁ = c₂ = 1
X₁ = - X₂ = 1/3^1/2


∫ f(x) dx = c₁f(x₁) + c₂f(x₂)
= (7/2) [f(3.5/3^1/2 + 1.5 ) + f( - 3.5/3^1/2 + 1.5)]
= (7/2) [2 + 3 (3.5/3^1/2 + 1.5)² + 2 + 3(- 3.5/3^1/2 + 1.5)²]

= (7/2) [4 + 2 (3.5² + 6 (1.5)²]
= (7/2) [4 + 24.50 + 13.50>] = 147
We obtain the same result.