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Irregular shape area calculation using "integral"

조회 수: 9 (최근 30일)
friet
friet 2016년 12월 12일
댓글: Star Strider 2016년 12월 12일
I have a a graph that is irregular shape with many data points. However, for the sake of simplicity I am gona ask here simplified version. I have x and Y data points and I want to find the area inclosed by the graph between x=2 and x=3. The code is below. It didnt work. Can anyone help me.
clc
clear all
x=[0 1 2 2.5 3 4 5 6] ;
yf=@(x)[0 -1 0 0.10 1 0 -1 0];
y=yf(x);
plot(x,y)
grid on
area_1= integral(yf, x(3), x(4));

답변 (1개)

Star Strider
Star Strider 2016년 12월 12일
You cannot use the integral function on data such as you presented. You must use trapz.
Another example:
x=[0 1 2 2.5 3 4 5 6] ;
yf=[0 -1 0 0.10 1 0 -1 0];
y=yf;
plot(x,y)
grid
min_y = min(y);
xidx = find((x >= 2) & (x <= 3));
int_1 = trapz(x(xidx), y(xidx)) % Area From ‘y = 0’
int_2 = trapz(x(xidx), y(xidx)-min_y) % Area From ‘y = min(y)’
int_1 =
0.3
int_2 =
1.3
  댓글 수: 2
friet
friet 2016년 12월 12일
The area above y=0 should be a little bit less than 1. I dont think int_1=0.3 is write? When you use
int_1 = trapz(x(xidx), y(xidx))
you specify the x intervals as,. How can I specify my y interval?
Star Strider
Star Strider 2016년 12월 12일
First, I do not agree that between 2 and 3 the area ‘should be a little bit less than 1. The height of the triangle from 2 to 3 with a height of 1 (from a baseline of 0) would be ½*b*h or 0.5, so it should be a bit less than 0.5, and (with a value of 0.3), it is.
I specify the intervals for both ‘x’ and ‘y’ with respect to the index values in ‘xidx’. Explore the code and the results to see that it is correct.

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