Trapezoidal numerical integration without use of function?

267 次查看（过去 30 天）
Weronika Ring
Weronika Ring 2017-11-3
评论： Mpho 2023-4-11
I was trying to do trapezoid integration without actually creating a function, but I seem to have done something wrong.
a=0;
b=1;
f=@(x)x.*sin(x);
n=100;
x=linspace(a,b,n+1);
h=(b-a)/n;
qt=sum((h*f((x(1:n)+x(2:n+1)))/2))
This returns 0.4354, when the actual value of the integral is 0.3012. Where am I going wrong with the code?
  2 个评论
murat onay
murat onay 2019-5-21
编辑：murat onay 2019-5-21
trapezoid matlab code is here. This code return 0.301180
clc;clear;
a=0;
b=1;
n=100;
h=(b-a)/n;
sum=0;
f=@(x) x.*sin(x);
for i=1:1:n-1
sum= sum + f(a+i*h);
end
result = h/2*(f(a)+f(b)+2*sum);
fprintf('%f',result);
Mpho
Mpho 2023-4-11
Hi, how would I achieve this if I want to integrate a constant matrix?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（3 个）

John D'Errico
John D'Errico 2017-11-3
编辑：John D'Errico 2017-11-7
Trapezoidal rule is easy enough. It depends on whether the step is constant or not. The entire point of my response is you need to get the weights correct. If not, then of course your code must fail.
For a constant step size, you need to remember that the weights look like this:
h*[1 2 2 2 2 2 ... 1]/2
So in any interval, we have a trapezoid. The area of a trapezoid is the average of the heights at each end, then multiply by the width.
h*(f(x(i)) + f(x(i+1)))/2
But each trapezoid shares its endpoint with the neighbors. So that gives the first and last points half the weight of the rest. It also gives us a simple way to do trapezoidal rule.
n = 20;
x = linspace(0,pi,n);
dx = x(2) - x(1);
f_x = sin(x);
trapint = (sum(f_x) - (f_x(1) + f_x(end))/2)*dx
trapint =
1.9954
Did we get it correct?
trapz(x,f_x)
ans =
1.9954
Yes. And both agree with the actual result.
syms u
int(sin(u),0,pi)
ans =
2
Could I have done this for unequal spacing? Still easy enough. I'll pick a random spacing here, using a few more points because a random spacing can have some wide places where nothing fell.
n = 50;
x = [0,sort(rand(1,n-2))*pi,pi];
f_x = sin(x);
dx = diff(x);
trapint = dot(dx,(f_x(1:end-1) + f_x(2:end))/2)
trapint =
1.9966
Note that on a function like sin(x) over that interval, trapezoidal rule will tend to underestimate the integral. As you can see, this is exactly what happened, and will always happen for that function, on that interval.
Hiren Rana
Hiren Rana 2021-11-11
a=0; b=1; n=100; h=(b-a)/n; sum=0; f=@(x) x.*sin(x); for i=1:1:n-1 sum= sum + f(a+i*h);
end result = h/2*(f(a)+f(b)+2*sum); fprintf('%f',result);
VBBV
VBBV 2021-12-29
编辑：VBBV 2022-2-5
Ran in:
a=0;
b=1;
f=@(x)x.*sin(x);
n=100;
x=linspace(a,b,n+1);
h=(b-a)/n;
qt=sum(h*(f(x(1:n))+f(x(2:n+1)))/2)
qt = 0.3012
You did not use the function correctly. Try above

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by