Getting 0 when using quad

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Aldo
Aldo el 11 de Nov. de 2016
Comentada: Walter Roberson el 11 de Nov. de 2016
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

Respuestas (2)

Walter Roberson
Walter Roberson el 11 de Nov. de 2016
Editada: Walter Roberson el 11 de Nov. de 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 comentarios
Aldo
Aldo el 11 de Nov. de 2016
And how should you go about to solve it with Quad?
Walter Roberson
Walter Roberson el 11 de Nov. de 2016
I wouldn't solve it with quad(). I might solve it with quadgk
quadgk(fun,0,6, 'waypoints', pi)

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KSSV
KSSV el 11 de Nov. de 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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