MATLAB: Evaluating the integral of a function

mathematicssignal processingtime series

I want to evaluate integral of (Hr(w)*cos(n*w*t)-Hi(w)sin(n*w*t))dw with limits 1 to 2 where t=0.2 and T=2000. where Hr(w) is the real part of a function H(w)=10*(((i*w)/0.006)-1)*((-(w^2)/0.64)+(((2*(i*w))/0.8)+1))*(((i*w)/10)-1)*((-(w^2)/2250000)+(((2*(i*w))/1500)+1))/((((w^2)/0.000004)+(((2*(i*w))/0.002)+1))*((-(w^2)/49)+(((2*(i*w))/7)+1))*((-(w^2)/400000000)+(((2*(i*w))/20000)+1))*(((i*w)/100)-1))? And w is a vecor from [0.0005 to 5].
I am trying the following codes but the integral is not happening and instead it shows the error Undefined function 'int' for input arguments of type 'double
w=[0.0005:0.1:5]; n=1:length(w); t=0.1; T=2000;
for j=1:length(w) y(j)=10*(((i*w(j))./0.006)-1)*((-(w(j).^2)./0.64)+(((2*(j.*w(j)))./0.8)+1))*(((j.*w(j))./10)-1)*((-(w(j).^2)./2250000)+(((2*(j.*w(j)))./1500)+1))/((((w(j).^2)./0.000004)+(((2*(j.*w(j)))./0.002)+1))*((-(w(j).^2)./49)+(((2*(j.*w(j)))./7)+1))*((-(w(j).^2)./400000000)+(((2*(j.*w(j)))./20000)+1))*(((j.*w(j))./100)-1)); rr(j)=real(y(j)) ii(j)=imag(y(j)) xc(j)=cos(n(j)*w(j)*t)*rr(j) xs(j)=sin(n(j)*w(j)*t)*ii(j) ad(j)=xc(j)-xs(j) ri(j)=int(ad(j),w,1,2) end
Please help me and even let me know is my method of finding the integral of the function correct?

Best Answer

Use INT to integrate an expression written using symbolic variables created using SYM or SYMS from Symbolic Math Toolbox.
Use INTEGRAL to integrate a function whose handle (which could be an anonymous function) you specify as the first input.
Use TRAPZ to integrate a function represented by a set of (X, Y) data.
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