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

조회 수: 6 (최근 30일)
Anke Kügler
Anke Kügler 2015년 9월 26일
댓글: Star Strider 2015년 9월 26일
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!

답변 (1개)

Star Strider
Star Strider 2015년 9월 26일
편집: Star Strider 2015년 9월 26일
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.)
  댓글 수: 2
Star Strider
Star Strider 2015년 9월 26일
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 2015년 9월 26일
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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