What is Integral Quadrature?

Question

I'm reading this book:

Differential Quadrature and Its Application in Engineering

trying to find out what is "Integral Quadrature". here is a snapshot of the page which is about Integral Quadrature:

My questions is what are w(weighting coefficiont) and f(functional value) and could you please show me them on the figure 1.1?

It's so close to the concept of "Integration" which we learnt at school but I don't know why it make me crazy. this is what we learnt about integrating at school:

Introduction to Integration

P.S. I don't have enough reputation to add these tages: "Integral Quadrature" and "Differential Quadrature" would you please do it for me?

Thank you. good clue. I need somebody explain it in a simple manner to me and especially by that figure. — Roh, Dec 23 '16 at 12:31
What figure ? The integral is the area... The numerical techniques are used to approximate the curve (by way of polynomials or "simple" curves in general) in order to compute the integral. — Mauro ALLEGRANZA, Dec 23 '16 at 12:35
The figure I posted in the question. it's a part of the book. e.g. could you show me w1, w2,w3.... and f1, f2, f3,... — Roh, Dec 23 '16 at 12:37
See following page Trapezoidal rule : choose on the $x$-axis uniformly spaced point $x_i$ with step $h$ ($= x_{i+1}-x_i$) and use the values of the function $f(x_i)$ in those points. — Mauro ALLEGRANZA, Dec 23 '16 at 13:01
From here: "...there was generally geometric interest in tangents and in "squaring" regions, also known as finding the "quadrature" of a region, that is, finding a way to construct a square or rectangle that had the same area as the region you were considering..." — J. M. ain't a mathematician, Dec 23 '16 at 13:32

KittyL · Accepted Answer · 2016-12-23T13:10:41.227

I changed a little of one of the pictures in your link so you see those $x_1, x_2, \dots, x_n$ and their corresponding $y$-values $f_1=f(x_1), f_2=f(x_2),\dots, f_n=f(x_n)$:

Suppose the original interval is $[a,b]$, that is , you want to find the area under the curve $f(x)$ in the interval $[a,b].$ Now if $w_1=(b-a)/(n-1), w_2=(b-a)/(n-1), \dots, w_{n-1}=(b-a)/(n-1), w_n=0$, we get $$\frac{b-a}{n-1}(f_1+\cdots+f_{n-1}),$$ which is the sum of the area of the rectangles using the height of the left boundaries. Notice that $(b-a)/(n-1)$ is the width of each rectangle.

Otherwise, if $w_1=0, w_2=(b-a)/(n-1), \dots, w_{n-1}=(b-a)/(n-1), w_n=(b-a)/(n-1)$, we get $$\frac{b-a}{n-1}(f_2+\cdots+f_{n}),$$ which is the sum of the area of the rectangles using the height of the right boundaries.

The integration quadrature is a generalization of this idea, with different weights on different $y$-values.

Thank you. what is n? why do we write the width of each rectangle in this manner? 1/(n-1) — Roh, Dec 23 '16 at 12:59
@Roh: $n$ is the number of points when you divide the interval $[a,b]$ into $n-1$ subintervals, so that you can sum up the area of the rectangles to get an approximation of the area under the curve. Sorry $1/(n-1)$ should be $(b-a)/(n-1)$. It is the length of the subinterval. — KittyL, Dec 23 '16 at 13:09

score 0 · Answer 2 · answered Dec 23 '16 at 12:55

Maybe a good thing to think about is why you would want to weight the areas rather than just use a simpler method like the trapezium rule where you simply average all the trapeziums. For instance, if you have a very 'wavy' function you might want to use non-equally spaced points and weight them accordingly.

What is Integral Quadrature?

2 Answers2