Getting 0 when using quad
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fun = @(x) 80*exp(-((x-pi)/0.002).^2);
I = quad(fun,0,6)
I2= integral(fun,0,6)
I =
0
I2 =
1.464129900321425e-69
Why do I get 0 when using quad?
Best regards Aldo
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Antworten (2)
Walter Roberson
am 11 Nov. 2016
Bearbeitet: Walter Roberson
am 11 Nov. 2016
The two use different adaptive techniques. The function has a sharp peak at Pi and one technique misses it completely and the other barely catches it. The actual integral over that range is (2/25*(erf(500*Pi)+erf(3000-500*Pi)))*sqrt(Pi) which is about 0.284
To get a better answer:
integral(fun,0,6,'Waypoints', pi)
2 Kommentare
Walter Roberson
am 11 Nov. 2016
I wouldn't solve it with quad(). I might solve it with quadgk
quadgk(fun,0,6, 'waypoints', pi)
KSSV
am 11 Nov. 2016
You are trying to calculate the area under the curve as shown in the attached image. See the values on the axes, so I think getting zero is legitimate.

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