Getting 0 when using quad

7 Ansichten (letzte 30 Tage)
Aldo
Aldo am 11 Nov. 2016
Kommentiert: Walter Roberson am 11 Nov. 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

Antworten (2)

Walter Roberson
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
Aldo
Aldo am 11 Nov. 2016
And how should you go about to solve it with Quad?
Walter Roberson
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)

Melden Sie sich an, um zu kommentieren.


KSSV
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.

Kategorien

Mehr zu Marine and Underwater Vehicles finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by