What happens when the integrand approaches zero in an integral?

[webbster]

Hi All,

i got a short question. if i have an integral an let the integrand approach zero, how do i handle that?

e.g.:
[int]$$\int_a^b \! f(c+x) \, dx.$$

what happens if lim(x->0) ?
Maple still computes the integral as if nothing has happened...

thanks all

edit: can't get the integral sign to work...

 

[CompuChip]

There is no variable x to take the limit of.
Note that in
$$\int_a^b \! f(c+x) \, dx,$$
x is a dummy variable. When you evaluate the integral, you will get something which doesn't depend on x anymore, just on - in this case - a, b and c.

It's like asking, what happens when you take the limit of x -> 0 of 1/2, and wondering why Maple still gives 1/2.

 

[webbster]

Hi again,

im still a bit puzzled by this. what happens when
\int f(c+x)h(x)dx is the according integral? Consider f and h to be pdfs.

thanks again

 

[webbster]

more detailed i got a probability:

P=$$\int_{a}^{b}f(c+x)h(x)dx/(\int f(c+x)h(x)dx)$$

I was now told that as x approaches zero the denominator approaches one because we can eliminate the resulting f(c)...

or is it just a misstatement?

 

[wayneckm]

Note that $$x$$ is a dummy variable in the integral.

Equivalently, writing as follows may make it clearer:
$$\lim_{x \rightarrow 0} \int_{a}^{b} f(c + x) dx = \lim_{x \rightarrow 0} \int_{a}^{b} f(c + y) dy$$

That's why "there is no $$x$$ for you to take limit".

Or, loosely speaking, after performing integration, you will "remove the dependence of the integrator variable, leaving the integral depending on the upper and lower bound", i.e. $$\int_{a}^{b} f(x)dx = g(a,b)$$.

Hope this helps.