Numerical Integration

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Numerical Integration

We use three types of approximation for integral
  1. Piecewise constant approximation = Rectangular Rule [upper, lower, middle, left, right]
  2. Piecewise linear approximation = Trapezium Rule
  3. Piecewise quadratic approximation = Simpson’s 1/3 Rule
  4. Piecewise cubic approximation = Simpson’s 3/8 Rule
Numerical Integration is a process to compute an approximate solution to a definite integral \(\displaystyle \int _a^b f(x) dx \) to a given degree of accuracy. It is the process of evaluating a definite integral from a set of tabulated values of the integrand f(x).
There are several reasons for carrying out numerical integration, some of them are
  1. The integrand f(x) does not exist, but it is known at certain points, as sampling.
  2. The integrand f(x) may exist, but we have a discrete data points only
  3. The integrand f(x) exist, but formula for antiderivative is unknown
  4. The integrand f(x) exist, the formula for antiderivative exist, but complicated to execute

Newton’s-Cotes quadrature formula

Suppose\(y=f( x )\)is tabulated for equally spaced valued
\( x=x_0,x_1,x_2,\ldots ,x_n \)
given by
\( y=y_0,y_1,y_2,\ldots ,y_n \)
Now assuming that, \(y=f( x )\) be a polynomial of degree n in x,
We integrate
\(I= \int_a^b f(x) dx\)
Let, interval [a,b] is divided into n equal subintervals of width h so that
\(x_i=x_0+ih\) for \(i=0,1,2,\ldots ,n\) with \(x_0=a,x_n=b\)
Then
\(I= \int_a^bydx\)
or \(I= \int_{x_0}^{x_n}f(x)dx\)
or \(I= \int_{x_0}^{x_n}[ y_0+p\Delta y_0+\frac{p( p-1 )}{2!}\Delta ^2 y_0+\frac{p( p-1 )( p-2 )}{3!}\Delta ^3 y_0+\ldots ]dx\)
As we know
\(x=x_0+ph\)
We have
\(dx=h \times dp\)
Here
when \(x=x_0\), then p=0
when \(x=x_n\), then p=n
Hence, we have
\(I=h \int_0^n[ y_0+p\Delta y_0+\frac{p( p-1 )}{2!}\Delta ^2 y_0+\frac{p( p-1 )( p-2 )}{3!}\Delta ^3 y_0+\ldots ]dp\)
or \(I=h \left [ y_0p+\frac{p^2}{2}\Delta y_0+\frac{( 2p^3-3p^2 )}{12}\Delta ^2 y_0+\frac{p^4-4p^3+4p^2}{24}\Delta ^3y_0+\ldots \right ]_0^n\)
or \(I=h \left [ y_0n+\frac{n^2}{2}\Delta y_0+\frac{( 2n^3-3n^2 )}{12}\Delta ^2 y_0+\frac{n^4-4n^3+4n^2}{24}\Delta ^3y_0+\ldots \right ]\)
or \(I=nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0+\frac{n( n-2 )^2}{24}\Delta ^3y_0+\ldots ]\)
This is known as Newton’s-Cotes quadrature formula.

Numerical Integration: Trapezoidal rule [n=1]

Let y=f(x) is defined for a tabulated values within the interval [a,b], then we divide [a,b] to tagged partitions given by
\( a=x_0 \leq x_1 \leq x_2 \leq x_3 \leq ... \leq x_n=b\)
This partitions the interval [a, b] into n sub-intervals\( [x_{i−1}, x_i] \) of length \(\Delta x=\frac{b-a}{n}\)
Then Trapezoidal rule approximate the area under f(x) by a sum of n trapezia (instead of rectangles).
There are an infinite number of straight lines through any given points, but only one straight line through two given points, so the trapezoidal rule is used for estimating definite integrals with such unique trapezoids to approximate the area under a curve.

Suppose\(y=f( x )\)is tabulated for equally spaced valued
\( x=x_0,x_1,x_2,\ldots ,x_n \)
given by
\( y=y_0,y_1,y_2,\ldots ,y_n \)
Then, Newton’s-Cotes quadrature formula for\(I= \int_a^b f(x) dx\)is
\(I=nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0+\frac{n( n-2 )^2}{24}\Delta ^3y_0+\ldots ]\)
Taking n=1,
The polynomial is straight line
All higher differences greater than one are zero.
Thus,
\( \int_{x_0}^{x_1}f( x )dx\)
or \(nh[ y_0+\frac{n}{2}\Delta y_0]\)
or \(h[ y_0+\frac{1}{2}\Delta {y_0} ]\)
or \( h [ y_0+ \frac{1}{2} (y_1-y_0) ]\)
or \( h [ \frac{2y_0+y_1-y_0}{2}] \)
or \( \frac{h}{2} ( y_0+y_1 )\)
Here

\( \int_{x_0}^{x_1}f( x )dx=\frac{h}{2} ( y_0+y_1 )\)
Consider the trapezoids shown below, here we assume that the length of each subinterval is h.
First, recall that the area of a trapezoid with a height of h and bases of length \(b_1\) and \(b_2\) is given by
Area=\(\frac{1}{2} h(b_1+b_2)\)
In our formula, we see that the first trapezoid has a height h and parallel bases of length \(f(x_0)\) and \(f(x_1)\) . Thus, the area of the first trapezoid is
Area=\(\frac{1}{2} h[f(x_0)+f(x_0)]\)
or Area=\(\frac{1}{2} h(y_0+y_1)\)


Similarly, we get
\( \int_{x_1}^{x_2}f( x )dx=\frac{h}{2}[ {y_1}+{y_2} ]\)
or \( \int_{x_{n-1}}^{x_n}f( x )dx=\frac{h}{2}[ {y_{n-1}}+{y_n} ]\) Adding all, we get
\( \int_{x_0}^{x_n}f( x )dx=\frac{h}{2}[ {y_0}+{y_n}+2( {y_1}+{y_2}+\ldots +{y_{n-1}} ) ]\)
This is known as Trapezoidal Rule
=================================

Numerical Integration: Simpson’s 1/3 rule [n=2]

Suppose \( y=f(x) \) is tabulated for equally spaced valued
\( x=x_0,x_1,x_2,\ldots ,x_n \)
given by
\( y=y_0,y_1,y_2,\ldots ,y_n \)
Then, Newton’s-Cotes quadrature formula for\(I= \int_a^b f(x) dx\)is
\(I=nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0+\frac{n( n-2 )^2}{24}\Delta ^3y_0+\ldots ]\)
Taking n=2,
The polynomial is formed as parabolic curve
All higher differences greater than two are zero Thus,
\( \int_{x_0}^{x_2}f( x )dx\)
or \(nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0]\)
\( 2h[ y_0+\Delta y_0+\frac{1}{6}\Delta ^2y_0 ]\)
or \( 2h[ y_0+(y_1-y_0)+\frac{1}{6}(\Delta y_1-\Delta y_0 )\)
or \( 2h[ y_1+\frac{1}{6}\left \{(y_2-y_1)-(y_1-y_0) \right \} \)
or \( 2h[ y_1+\frac{1}{6}(y_2-2y_1+y_0) \)
or \( \frac{2h}{6}( 6y_1+y_2-2y_1+y_0) \)
or \( \frac{h}{3}( y_0+4y_1+y_2) \)
=====================================
Here
\( \int_{x_0}^{x_2}f( x )dx=\frac{h}{3}[ {y_0}+4{y_1}+{y_2} ]\)
Similarly, we get
\( \int_{x_2}^{x_4}f( x )dx=\frac{h}{3}[ {y_2}+4{y_3}+{y_4} ]\)
………………………………
\( \int_{x_{n-2}}^{x_n}f( x )dx=\frac{h}{3}[ {y_{n-2}}+4{y_{n-1}}+{y_n} ]\)
=====================================
Adding all we get
\( \int_{x_0}^{x_n}f( x )dx=\frac{h}{3}[ y_0+y_n+4( {y_1}+{y_3}+\ldots +{y_{n-1}} )+2( {y_2}+{y_4}+\ldots +{y_{n-2}} ) ]\)
This is known as Simpson’s 1/3 rule
=====================================

Numerical Integration: Simpson’s 3/8 rule [n=3]

Suppose\(y=f( x )\)is tabulated for equally spaced valued
\( x=x_0,x_1,x_2,\ldots ,x_n \)
given by
\( y=y_0,y_1,y_2,\ldots ,y_n \)
Then, Newton’s-Cotes quadrature formula for is
\(I=nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0+\frac{n( n-2 )^2}{24}\Delta ^3y_0+\ldots ]\)
Taking n=3,
The polynomial is formed as cubic curve
All higher differences greater than three are zero

Thus,
\( \int_{x_0}^{x_3}f( x )dx \)
or \( nh[ y_0+\frac{n}{2}\Delta y_0+\frac{n( 2n-3 )}{12}\Delta ^2y_0+\frac{n( n-2 )^2}{24}\Delta ^3y_0 ]\)
or \(3h[ y_0+\frac{3}{2}\Delta y_0+\frac{3}{4}\Delta ^2y_0+\frac{1}{8}\Delta ^3y_0 ]\)
or \(3h[ y_0+\frac{3}{2}(y_1-y_0)+\frac{3}{4}(y_2-2y_1+y_0)+\frac{1}{8}(y_3-3y_2-3y_1+y_0) ]\)
or \(\frac{3h}{8}[ 8y_0+12(y_1-y_0)+6(y_2-2y_1+y_0)+(y_3-3y_2+3y_1-y_0) ]\)
or \(\frac{3h}{8}[ y_0+3y_1+3y_2+y_3 ]\)
=====================================
Here
\( \int_{x_0}^{x_3}f( x )dx=\frac{3h}{8}[ y_0+3y_1+3y_2+y_3 ]\)
Similarly, we get
\( \int_{x_3}^{x_6}f( x )dx=\frac{3h}{8}[ y_3+3y_4+3y_5+y_6 ]\)
………………………………
\( \int_{x_{n-3}}^{x_n}f( x )dx=\frac{3h}{8}[ y_{n-3}+3y_{n-2}+3y_{n-1}+y_n ]\)
=====================================
Adding all we get
\( \int_{x_0}^{x_n}f( x )dx=\frac{3h}{8}[ y_0+y_n+3( y_1+y_2+y_4+y_5+\ldots ) +2( y_3+y_6+\ldots ) ]\)
This is known as Simpson’s 3/8 rule
=====================================

Which is better

The answer depends on function type and step value, however:
  1. Trapezoidal rule (better for linear approximation)
  2. Simpson’s 1/3 rule (better for even number of data points)
  3. Simpson’s 3/8 rule [better for (mod 3)+2 { for example 5, 8, 11,...} number of data points]

Numerical Integration: Example

  1. Evaluate\( \int_{0}^{1}\frac{1}{1+x}dx\) with h= 1/6
    Solution
    Taking \(h=\frac{1}{6}, a=0,b=1\), the functional values for the partition are given below.
    x 0 1/6 2/6 3/6 4/6 5/6 1
    \(f(x)=\frac{1}{1+x}\) 1 6/7 6/8 6/9 6/10 6/11 1/2
    Now, the numerical integral are as below.
    1. Trapezoidal rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or \(\frac{h}{2}[ {y_0}+{y_n}+2( {y_1}+{y_2}+\ldots +{y_{n-1}} ) ]\)
      or \( \frac1{12}[ 1+\frac{1}{2}+2( \frac6{7}+\frac6{8}+\frac6{9}+\frac{6}{10}+\frac{6}{11} ) ]\)
      or \( 0.695\)
    2. Simpson’s 1/3 rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or \(\frac{h}{3}[ {y_0}+{y_n}+4( {y_1}+{y_3}+\ldots +{y_{n-1}} ) +2( {y_2}+{y_4}+\ldots +{y_{n-2}} ) ]\)
      or \( \frac1{18}[ 1+\frac{1}{2}+4( \frac{6}{7}+\frac6{9}+\frac{6}{11} )+2( \frac{6}{8}+\frac6{10} ) ]\)
      or \( 0.693\)
    3. Simpson’s 3/8 rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or \(\frac{3h}{8}[ {y_0}+{y_n}+3( {y_1}+{y_2}+{y_4}+{y_5}+\ldots ) +2( {y_3}+{y_6}+\ldots ) ]\)
      or \( \frac{1}{16}[ 1+\frac{1}{2}+3( \frac6{7}+\frac{6}{8}+\frac{6}{10}+\frac{6}{11} )+2( \frac{6}{9} ) ]\)
      or \(0.693\)
  2. Evaluate \( \int_{0}^{1}\frac1{1+x}dx\) correct to three decimal place with h= 0.5.
    Solution
    Taking \(h=0.5, a=0,b=1\), the functional values for the partition are given below.
    x 0 0.5 1
    \(f(x)=\frac{1}{1+x}\) 1 1/1.5 1/2
    Now, the numerical integral are as below.
    1. Trapezoidal rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or \(\frac{h}{2}[ y_0+y_n+2( y_1+y_2+\ldots +y_{n-1} ) ]\)
      or \( \frac{1}{4}[ 1+0.5+2( 0.6667 ) ]\)
      or \( 0.708\)
    2. Simpson’s 1/3 rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or\(\frac{h}{3}[ y_0+y_n+4( y_1+y_3+\ldots +y_{n-1} ) +2( y_2+y_4+\ldots +y_{n-2} ) ]\)
      or \( \frac{1}{6}[ 1+0.5+4( 0.6667 ) ]\)
      or \(0.694\)
    3. Simpson’s 3/8 rule
      \( \int_{0}^{1}\frac{1}{1+x}dx\)
      or \( \int_{x_0}^{x_n}f( x )dx\)
      or \( \frac{3h}{8}[ y_0+y_n+3( y_1+y_2+y_4+y_5+\ldots ) +2( y_3+y_6+\ldots ) ]\)
      or \( \frac{3}{16}[ 1+0.5+3( 0.666 ) ]\)
      or \( 0.655\)

Exercise

Evaluate following integrals
  1. \( \displaystyle \int_0^3(5x+4)\) with n=6
  2. \( \displaystyle \int_0^1 x^2\) with n=4
  3. \( \displaystyle \int_0^1 (1-x^2)\) with n=8
  4. \( \displaystyle \int_0^4 x^3\) with n=4
  5. \( \displaystyle \int_1^2 \frac{1}{x}\) with n=5

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