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Topic: integrating the symbolic normcdf
Replies: 5   Last Post: Jun 27, 2010 10:39 PM

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Roger Stafford

Posts: 5,929
Registered: 12/7/04
Re: integrating the symbolic normcdf
Posted: Jun 25, 2010 8:19 PM
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"Angie" <> wrote in message <i03d7t$3tu$>...
> First: In the above post, I should have normcdf(x,0,1) not normcdf(x,1,0)
> Second: I tried this approach in another post, yet I am not sure if this is correct. Can anyone tell me if this is correct?
> fun1 = @(x) x;
> fun2 = @(x) normcdf(x,0,1);
> myintegrand = @(x) fun1(x).*fun2(x);
> quad(myintegrand,0,1)
> ans =
> 0.3710

- - - - - - - - -
Yes, that agrees with the answer I got using integration by parts with old-fashioned pen and paper, namely:

1/4 + 1/2/sqrt(2*pi*exp(1)) = 0.370985362

I would have thought the symbolic toolbox could also solve this.

Roger Stafford

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