MATLAB Problems using integral function in Matlab

AI Thread Summary
The discussion revolves around a function defined for integration, which is expected to yield a positive integral over the interval [0, 1.1225]. The function, defined as "integrando," is dependent on parameters lambda and t, and appears to be constant before decreasing exponentially. However, when the integral is computed using MATLAB's integral function, it returns a negative value, which is unexpected given the function's positive nature. The confusion arises particularly when using different values for x0, with one user noting that using x0 equal to the upper limit of integration (1.1225) leads to negative results. The conversation highlights the need to clarify the choice of x0 and its impact on the function's behavior and the resulting integral.
cacosomoza
Messages
11
Reaction score
0
Hi there!

i have this function handle:

integrando= @(x,landa,t) 1-exp(-((1/(landa*t))*((exp(-landa*x)/x)+((x-x0).*(1+landa*x0)-x0)*(exp(-landa*x0)/(x0^2)))));

which as you can see using

fplot(@(x) integrando(x,3,0.00063),[0 1.1225])

its value is constant (actually 1, and then starts decreasing exponentially until drops to zero at 1.1225). Landa and t are parameters that change the curvature of this drop. Since this is a positive and finite function, I'm waiting a positive integral. However, integrating this same function on this same interval, I get negative figures!

>> q1=integral(@(x) integrando(x,3,0.00063),0,1.1225)

q1 =

-20.3981
This is getting me nuts! I would very much appreciate any help. Thanks in advance!
 
Physics news on Phys.org
What value are you using for x0?

I used 2 just to see if the code works, and I got:

integrando= @(x,landa,t) 1-exp(-((1/(landa*t))*((exp(-landa*x)/x)+((x-2).*(1+landa*2)-2)*(exp(-landa*2)/(2^2)))));

fplot(@(x) integrando(x,3,0.00063),[0 1.1225])

q1=integral(@(x) integrando(x,3,0.00063),0,1.1225)

q1 =

1.1225
 
hi kreil,

I'm using x0=1.1225 ,which is the same value as the limit of integration. If you try again using that value on 'integrando', you'll see the result of the integral is a negative number which doesn't make any sense to me
 

Similar threads

Back
Top