I could not integrate using MatLab, Can you please help me?
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Tika Ram Pokhrel
le 2 Mai 2021
Commenté : Walter Roberson
le 6 Mai 2021
In solving a problem I need to integrate the following function with respect to 't' from the limit 0 to t.
3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
I used the following commands but got the same result as given herewith.
>> syms a c t real
mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
>> int(mag_dr,t,0,t)
ans =
int(3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2), t, 0, t)
Let me know the best way(s) to tackle this type of problem.
1 commentaire
Réponse acceptée
Dyuman Joshi
le 5 Mai 2021
Modifié(e) : Dyuman Joshi
le 6 Mai 2021
syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
%result of integration, z = -(3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2));
t=0;
res = z - subs(z);
%obtain final result by evaluating the integral, z(t)-z(0), by assigning t & using subs()
1 commentaire
Walter Roberson
le 5 Mai 2021
Modifié(e) : Walter Roberson
le 5 Mai 2021
Not quite.
syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
char(z)
z0 = limit(z, t, 0, 'right');
char(z0)
res = simplify(z - z0);
char(res)
fplot(subs(z, [a,c], [1 2]), [-5 5])
fplot((subs(fun,[a,c], [1 2])), [-5 5])
That is, the problem is that the integral is discontinuous at t = 0 and that is why int() cannot resolve it.
Plus de réponses (2)
Walter Roberson
le 5 Mai 2021
syms a c t real
mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
z = int(mag_dr, t)
z - limit(z, t, 0, 'right')
The integral is discontinuous at 0, which is why it cannot be resolved by MATLAB.
4 commentaires
Dyuman Joshi
le 6 Mai 2021
The wrong substitution was a mistake on my part, mostly cause I did it in a hurry. I have edited my nswer accordingly as well. Other than that, is subs() a good approach or would you recommend otherwise?
Walter Roberson
le 6 Mai 2021
limit() is more robust than subs() for cases like this. But limit() is sometimes quite expensive to calculate, or is beyond MATLAB's ability to calculate, even in some finite cases.
Sindhu Karri
le 5 Mai 2021
Hii
The "int" function cannot solve all integrals since symbolic integration is such a complicated task. It is also possible that no analytic or elementary closed-form solution exists.
For definite integrals, a numeric approximation can be performed by using the "integral" function.
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