how do I store the value of integrals of function k for all values of T for the below code??

T=[0:20:1000];
k=@(x)((x.^2).*exp((4*x)/3))./((exp(x)-1).^2)
q=integral(k,0,-td./T)

I'm getting the following error:

Error using integral (line 85)
A and B must be floating-point scalars.

Error in hw4 (line 12)
q=integral(k,0,-td./T)

T is an array ,k is a function, td is constant

[Torsten]

Is td positive or negative ?

First remove the singularity of your function ; you divide by 0 at x=0.

Best wishes

Torsten.

Answer 1

[Walter Roberson]

You could do

q = arrayfun(@(endpoint) integral(k,0,endpoint), -td./T)

However, this would require that the integration over each segment be done multiple times. It would be more efficient to do something like

endpoints = [0, -td./T(:).'];
qt = arrayfun(@(IDX) integral(k,endpoints(IDX),endpoints(IDX+1)), 1:length(endpoints)-1);
q = cumsum(qt);

[Aqeel Mohammed Abdul Mageeth]

I'm again getting the following error please help

A and B must be floating-point scalars

Error in hw4>@(IDX)integral(k,endpoints(IDX),endpoints(IDX+1))

Error in hw4 (line 23) qt = arrayfun(@(IDX) integral(k,endpoints(IDX),endpoints(IDX+1)), 1:length(endpoints)-1);

[Walter Roberson]

What is class(td) and class(T) ?

[Aqeel Mohammed Abdul Mageeth]

td is a constant with given value and T is an array [1:20:1001]

[Walter Roberson]

What is class(td) ? Is it double, or is possible it is integer data type? Is td positive or negative?

I notice that you start T at 0 and your endpoints are -td./T which implies that your first endpoint is either -inf or +inf. I did not break up the endpoints the right way for that case; the proper break-up will depend upon whether td is positive or negative.