I want to define a function in the integral form
f(x)=integral(g(z), z=0, z=x)
as an example:
for which I wrote:
fun = @(z) z^2+3*z-1;
x=linspace(0,1,100);
y = integral(fun,0,@(x) x)
plot(x,y)
But Matlab returned the following error:
Error using integral (line 85)
A and B must be floating-point scalars.
Error in IntegralFunction (line 8)
y = integral(fun,0,@(x) x)
Is it possible at all to do this in Matlab?

 Akzeptierte Antwort

Torsten
Torsten am 6 Sep. 2018

2 Stimmen

fun=@(z)z.^2+3*z-1;
F=@(x)integral(fun,0,x);
F(10)

5 Kommentare

Saeid
Saeid am 6 Sep. 2018
Danke, Torsten!
Saeid
Saeid am 7 Sep. 2018
Bearbeitet: Saeid am 7 Sep. 2018
Hi again Torsten! Actually, my aim to define a function like this was for the final result (in this case F(x)=x^3/3+3*x^2/2-x) to be a coefficient in a differential equation. Something like: y"+F(x).y'=y"+(x^3/3+3*x^2/2-x)*y'=0 Do you know if that is possible in Matlab?
Torsten
Torsten am 7 Sep. 2018
Yes, it's possible just the way I showed you above. What exactly is the problem when you want to use it in your context ?
I tried the following format but it didn't work:
tspan = [0 5];
y0 = 0;
[t,y] = ode45(@(t,y) DYDT(t,y) , tspan, y0);
plot(t,y,'-o')
function dydt=DYDT(t,y)
fun=@(z)z.^2+3*z-1;
dydt=@(y)integral(fun,0,y)
end
Aishwarya Kasarla
Aishwarya Kasarla am 1 Apr. 2023
Hey, I am also stuck with the same proble, did you find the solution? Can you please help me out with this?

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

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

Start Hunting!

Translated by