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:

enter image description here

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?

  • 1
    $\begingroup$ Probably : Numerical integration or numerical quadrature. $\endgroup$ – Mauro ALLEGRANZA Dec 23 '16 at 12:25
  • $\begingroup$ See also Gaussian quadrature. $\endgroup$ – Mauro ALLEGRANZA Dec 23 '16 at 12:27
  • $\begingroup$ Thank you. good clue. I need somebody explain it in a simple manner to me and especially by that figure. $\endgroup$ – Roh Dec 23 '16 at 12:31
  • $\begingroup$ 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. $\endgroup$ – Mauro ALLEGRANZA Dec 23 '16 at 12:35
  • $\begingroup$ 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,... $\endgroup$ – Roh Dec 23 '16 at 12:37

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 $f_1=f(x_1), f_2=f(x_2),\dots, f_n=f(x_n)$:

enter image description here

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.

  • $\begingroup$ Thank you. what is n? why do we write the width of each rectangle in this manner? 1/(n-1) $\endgroup$ – Roh Dec 23 '16 at 12:59
  • $\begingroup$ @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. $\endgroup$ – KittyL Dec 23 '16 at 13:09

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.


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