How to calculate the integral of a function with a spline in it

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Renan Kops 2017-4-21

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https://ww2.mathworks.cn/matlabcentral/answers/336542-how-to-calculate-the-integral-of-a-function-with-a-spline-in-it

评论： Andrew Newell 2017-4-25

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Hello Everybody,

So i have a function called FB, and it is the product of a couple of functions

nB = spline(Wavelength,B3);
emi = @(x) 1.989e-6*(10^9*x).^2-0.002;
FB = @(x,T) nB(x).*emi(x)./(x.^5.*(exp((h.*c)./(x.*k.*T))-1))
%integration:
RaBG(Counts) = integral(@(x)FB(x,T),400e-9,720e-9)

nB is a function i must adquire from data. When fitting it with a polinomial function i get a small error which i believe is making me get wrong results, though the code works. I'm trying to fit it with a spline but haven't been able to get the code to work. I also tryied calling FB as this, which didn't work:

FB = @(x,T) ppval(nB,x).*emi(x)./(x.^5.*(exp((h.*c)./(x.*k.*T))-1));

Can anyone please help me?

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Renan Kops 2017-4-24

Some of my data is evenly spaced so I can use this, good idea!

Any alternatives for the data that isn't evenly spaced?

John D'Errico 2017-4-25

I'd like to chime in here,but without knowing what wavelength, B3, and G3 are, it is impossible to give a useful answer.

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回答（1 个）

Andrew Newell 2017-4-24

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编辑：Andrew Newell 2017-4-24

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For evenly or unevenly spaced data, you could use the trapezoidal rule (MATLAB function trapz).

Interpolation is also reasonable. How exactly isn't the code working?

I don't see any values assigned to c, k or T. Are you doing that earlier? If so, it would make sense to define

FB = @(x) ppval(nB,x).*emi(x)./(x.^5.* (exp((h.*c)./(x.*k.*T))-1));

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Renan Kops 2017-4-25

Well it is possible, i will try to rescale it.

But, if that was the case, shouldn't i get wrong answers when integrating a 10th degree polinomial or with the Newton-Cotes method? Because both these methods worked and gave me a nice approximation of the real answer (For now i am trying to replicate the results of a paper to check if my code is performing as expected).

Thank you for all the help Andrew.

Andrew Newell 2017-4-25