We begin with a simple example illustrated in Figure 7.4.

The simplest form of Gaussian Integration is based on the use of an
optimally chosen polynomial to approximate the integrand *f*(*t*) over
the interval [-1,+1]. The details of the determination of this
polynomial, meaning determination of the coefficients of *t* in this
polynomial, are beyond the scope of this presentation. The simplest
form uses a uniform weighting over the interval, and the particular
points at which to evaluate *f*(*t*) are the roots of a particular
class of polynomials, the Legendre polynomials, over the interval. It
can be shown that the best estimate of the integral is then:

where

Gaussian quadrature formulae are evaluating using abscissae and weights
from a table like that included here. The choice of value of *n* is
not always clear, and experimentation is useful to see the influence
of choosing a different number of points. When choosing to use *n*
points, we call the method an ``*n*-point Gaussian'' method.

Gauss-Legendre Abscissae and Weights |
|||

n | Values of t | Weights | Degree |

2 | 1.0 | 3 | |

3 | 0.0 | 0.88888889 | |

0.77459667 | 0.55555555 | 5 | |

4 | 0.33998104 | 0.65214515 | 7 |

0.86113631 | 0.34785485 | ||

5 | 0.0 | 0.56888889 | 9 |

0.53846931 | 0.47862867 | ||

0.90617985 | 0.23692689 | ||

6 | 0.23861918 | 0.46791393 | 11 |

0.66120939 | 0.36076157 | ||

0.93246951 | 0.17132449 | ||

7 | 0.0 | 0.41795918 | 13 |

0.40584515 | 0.38183005 | ||

0.74153119 | 0.27970539 | ||

0.94910791 | 0.12948497 | ||

8 | 0.18343464 | 0.36268378 | 15 |

0.52553241 | 0.31370665 | ||

0.79666648 | 0.22238103 | ||

0.96028986 | 0.10122854 | ||

10 | 0.14887434 | 0.29552422 | 19 |

0.43339539 | 0.26926672 | ||

0.67940957 | 0.21908636 | ||

0.86506337 | 0.14945135 | ||

0.97390653 | 0.06667134 |

The Gauss-Legendre integration formula given here evaluates an estimate
of the required integral on the interval for *t* of [-1,+1]. In most
cases we will want to evaluate the integral on a more general interval,
say
.
We will use the variable *x* on this more general
interval, and linearly map the
interval for *x* onto
the [-1,+1] interval for *t* using the linear tranformation:

It is easily verified that substituting

The factor of

Consider the evaluation of the integral:

whose value is 1, as we can obtain by explicit integration. Applying the 2-point Gaussian method, and noting that both

While this example is quite simple, the following table of values obtained for

N | Gauss-Legendre | Simpson's 1/3 |

2 | 0.9984726135 | 1.0022798775 |

4 | 0.9999999770 | 1.0001345845 |

6 | 0.9999999904 | 1.0000263122 |

8 | 1.0000000001 | 1.0000082955 |

10 | 0.9999999902 | 1.0000033922 |