clc
clear
format long
d = [0.05 0.025 0.0125 0.00625 0.003125];
syms x y
f= 10000.*exp(y)/(1+abs(x)./2) ;
I_t_x = int(f,x,-5,0);
I_t = double(int(I_t_x,y,-2,2))
for i = 1:length(d)
x = -5 : d(i) : 0 ;
y = -2 : d(i) : 2 ;
[X,Y] = meshgrid(x,y);
F = 10000*exp(y)/(1+abs(x)/2) ;
I = trapz(y,trapz(x,F,2))
end

2 Kommentare

Jan
Jan am 25 Mai 2021
Please post code as text, because then it can be reused easily to write an answer.
Whenever you mention an error in the forum, post a complete copy of the message. Then the readers can concentrate on solving the problem, instead on investing time in guessing, what the error is. You do have this important information on the screen already, so sharing it is the way to go.
jeaho jang
jeaho jang am 25 Mai 2021
I fixed it.

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 Akzeptierte Antwort

Jan
Jan am 25 Mai 2021

0 Stimmen

Which error message do you get?
This should fail, if x and y are symbolic variables:
[X,Y] = meshgrid(x,y);
And you do not use the output X and Y at all:
F = 10000*exp(y)/(1+abs(x)/2) ;
I = trapz(y,trapz(x,F,2))

5 Kommentare

jeaho jang
jeaho jang am 25 Mai 2021
Bearbeitet: jeaho jang am 25 Mai 2021
I want to solve this function with trapezoid rule.
-5 <= x <= 0
-2 <=y <= 2
how can i do?
del x and del y = 0.05 0.025 0.0125 0.00625 0.003125
Torsten
Torsten am 25 Mai 2021
Bearbeitet: Torsten am 25 Mai 2021
In your code, use
F = 10000*exp(Y)./(1+abs(X)/2)
And save the integration results in a vector:
I(i) = trapz(y,trapz(x,F,2))
This modification will also result in an error.
clc
clear
format long
d = [0.05 0.025 0.0125 0.00625 0.003125];
syms x y
f= 10000.*exp(y)/(1+abs(x)./2) ;
I_t_x = int(f,x,-5,0);
I_t = double(int(I_t_x,y,-2,2))
for i = 1:length(d)
x = -5 : d(i) : 0 ;
y = -2 : d(i) : 2 ;
[X,Y] = meshgrid(x,y);
F = 10000*exp(Y)/(1+abs(X)/2);
I(i) = trapz(y,trapz(x,F,2))
end
Torsten
Torsten am 25 Mai 2021
In the definition of F, I wrote ./, not only /
jeaho jang
jeaho jang am 25 Mai 2021
Bearbeitet: jeaho jang am 25 Mai 2021
Thank you torsten

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