Need help in using trapz to integrate a definite function

I have a simple function, A2 to integrate (from x = 3 um to 5 um) using the trapezoidal method so I used the "trapz" but it didn't work. I would be grateful if someone could assist. My code is below. Thank you in advance.

clear all
c=3*10^8; % speed of light in vaccum
h=6.626*10.^-34; %  Planck constant 
k=1.38*10.^-23; %   Boltzmann constant
T=700; % Temperatures in Kelvin
x=(0.0:0.01:50).*1e-6; % in meters                                                                                                   
A2=(2*h*c*c)./((x.^5).*(exp((h.*c)./(k.*T.*x))-1));
q=trapz(A2,x,3,5)

Error using trapz
Too many input arguments.

Answer 1

[the cyclist]

trapz is probably not the best way to do this. You could use integral instead

c=3*10^8; % speed of light in vaccum
h=6.626*10.^-34; %  Planck constant 
k=1.38*10.^-23; %   Boltzmann constant
T=700; % Temperatures in Kelvin

A2 = @(x) (2*h*c*c)./((x.^5).*(exp((h.*c)./(k.*T.*x))-1));

% Unclear what you really want for x values. Here are two versions.
q1 = integral(A2,3.e-6,5.e-6)

q1 = 1.2934e+03

q2 = integral(A2,3,5)

q2 = 5.6099e-14

[Emmanuel Sarpong]

Thanks cyclist.

Answer 2

[the cyclist]

Did you read the documentation for the trapz function? It expects either two or three inputs, not the four you provided.

I am going to guess that your intention is to only integration your A2 function in the range x in (3,5). In that case, you could do

c=3*10^8; % speed of light in vaccum
h=6.626*10.^-34; %  Planck constant 
k=1.38*10.^-23; %   Boltzmann constant
T=700; % Temperatures in Kelvin
x=(0.0:0.01:50).*1e-6; % in meters                                                                                                   
A2=(2*h*c*c)./((x.^5).*(exp((h.*c)./(k.*T.*x))-1));

inRange = (x >= 3) & (x<=5);

q=trapz(x(inRange),A2(inRange))

q = 0

That's probably not the result you are expecting, but I am not going to debug this too much, since it is not actually clear what you are trying to achieve. Please explain more.

[Emmanuel Sarpong]

Hi cyclist, you're right! that's not the answer I am expecting. Thank you anyway.

The function decribes a curve and I want to find the area under a part of the curve from x =3 to 5 using the trapezoidal method of integration. Thanks

[the cyclist]

Do you mean x = 3, or x = 3e-6?

Answer 3

[Torsten]

Maybe you mean

clear all
c=3*10^8; % speed of light in vaccum
h=6.626*10.^-34; %  Planck constant 
k=1.38*10.^-23; %   Boltzmann constant
T=700; % Temperatures in Kelvin
x=(0.0:0.01:50).*1e-6; % in meters                                                                                                   
A2=(2*h*c*c)./((x.^5).*(exp((h.*c)./(k.*T.*x))-1));
%q=trapz(A2,x,3,5)
idx = x>=3e-6 & x<=5e-6;
x = x(idx);
A2 = A2(idx);
plot(x,A2)

trapz(x,A2)

ans = 1.2934e+03

6e8*2e-6    % Approximate value of the integral

ans = 1200

[Emmanuel Sarpong]

Hi Torsten, the function, A2 plots a curve and I want to find the area under a part of the curve from x = 3 to 5 using the trapezoidal method of integration. I hope this helps. Thanks for your help anyway

[Torsten]

If you define x as

x=(0.0:0.01:50).*1e-6; % in meters

and you evaluate A2 with this vector, you cannot determine the value of the integral between 3 and 5, but only between 3e-6 and 5e-6. And that's what the code above does.