Integration with negative support
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Hi, I would like to integrate something like this
z=@(x) x<0;
integral(exppdf(z,10),-0.5, 0.5);
Any help will be useful. Thanks
回答（1 个）
Rik
2017-2-21
You are aware that you are not creating a bound, but a function that says that z is either 1 or 0?
And that there is a vital difference between integrate() and integral()?
Anyway, I copied out the function that exppdf calculates from its help, which resulted in the code below:
mu=10;
z=@(x) (1/mu)*exp(-x/mu);
integral(z,-0.5, 0.5);
4 个评论
Nikolas
2017-2-21
Rik
2017-2-21
Sure, but not if you are integrating for x from -0.5 to 0.5
integral(z,-0.5, 0);
Nikolas
2017-2-21
Rik
2017-2-21
Why do you want to use this function? And if you use exppdf as an input, the integral for 0 to 0.5 should not be 0. There is another thing that may be what you mean:
z2=@(x) exppdf(-x,mu);%turn positive x into negative and vice versa
This will return 0 for integral(z2,0,0.5)
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