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
回答（2 个）
Walter Roberson
2016-11-11
编辑：Walter Roberson
2016-11-11
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 个评论
Aldo
2016-11-11
Walter Roberson
2016-11-11
I wouldn't solve it with quad(). I might solve it with quadgk
quadgk(fun,0,6, 'waypoints', pi)
KSSV
2016-11-11
0 个投票
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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2016-11-11
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