Integral -> "First input argument must be a function handle"

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Anke Kügler
Anke Kügler 2015-9-26
评论： Star Strider 2015-9-26
Dear all,
I'm completly new to MATLAB and am learning as I'm going. I need it for class and homework. No I have the following issue. I have a function and need to calculate the integral, but it returns me an error and I don't know why. Here is the code:
b=(cn-c0)/(zn-z0);
%w=zn+c0/b;
xmin=c0/b;
xmax=zn+c0/b;
r=(a*b*(zn+cn/b))/sqrt(1-power(a,2)*power(b,2)*power(zn+cn/b,2));
rx=integral(r,xmin,xmax)
c0, cn, z0, zn and a will all bit entered (I saved the above part as a script).
Thank you very much in advance!

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回答（1 个）

Star Strider
Star Strider 2015-9-26
编辑：Star Strider 2015-9-26
From what you wrote, you’re integrating over ‘b’, so convert your ‘r’ to an anonymous function:
r = @(b) (a.*b.*(zn+cn./b))./sqrt(1-power(a,2).*power(b,2).*power(zn+cn./b,2));
rx = integral(r,xmin,xmax)
Assuming I guessed correctly, that should work. I vectorised your equation as well. (I did not test this code.)
  2 个评论
Star Strider
Star Strider 2015-9-26
Anke Kügler’s Answer moved here...
Thank you, that helped a lot! It also made me relize I made a mistake in my function ;)
Thanks again. As I said, I'm still learning MATLAB.
Star Strider
Star Strider 2015-9-26
My pleasure.
The sincerest expression of appreciation here on MATLAB Answers is to Accept the Answer that most closely solves your problem.

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