Integral -> "First input argument must be a function handle"
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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!
0 Kommentare
Antworten (1)
Star Strider
am 26 Sep. 2015
Bearbeitet: Star Strider
am 26 Sep. 2015
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 Kommentare
Star Strider
am 26 Sep. 2015
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
am 26 Sep. 2015
My pleasure.
The sincerest expression of appreciation here on MATLAB Answers is to Accept the Answer that most closely solves your problem.
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!