1. ## An interesting integration

This question is one I came up with after considering the rules of integration.

We know that the integration of a function from a to b such that a = b is 0: INT{n:n} f(x) = 0

BUT, is this always true? By this I mean, would this be true in a function where f(n) is undefined?

For example:
Does INT{1:1} 1/(1-x)^2 = 0? I will work this out. I just want someone to tell me if my work is correct.

Since f(1) in this case is undefined, I'll use limits:
lim{n->1} INT{n:n} 1/(1-x)^2
lim{n->1} [1/(1-x)] from {n:n}
lim{n->1} 1/(1-n) - 1/(1-n) ... which is 0/0, so using L'Hopital's Rule:
lim{n->1} 0/(-n) = 0

As we see, this does equal 0, so the rule seems to work. However, if I try a slightly different equation, such that f(n) is undefined and lim{x->n} f(x) is undefined, does the rule still work?

Does INT{1:1} 1/(1-x) = 0?

lim{n->1} INT{n:n} 1/(1-x)
lim{n->1} -ln(1-x) from {n:n}
lim{n->1} -ln(1-n) + ln(1-n) ... which is 'negative' undefined plus undefined.
lim{n->1} ln[(1-n)/(1-n)]
lim{n->1} ln(1) = 0 ... but can I do this???

Since
lim{n->1} ln(1-n) is not defined

Because
lim{n->1-} ln(1-n) does not exist

Is it legal to say
lim{n->1} ln[(1-n)/(1-n)] = lim{n->1} ln(1) ???

2. Originally Posted by ecMathGeek
This question is one I came up with after considering the rules of integration.

We know that the integration of a function from a to b such that a = b is 0: INT{n:n} f(x) = 0

BUT, is this always true? By this I mean, would this be true in a function where f(n) is undefined?
We define,
INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
For function f defined on closed interval [a,b].

Note, under this strict definition, we require that f to be defined on [a,b] (and to be integratable). Hence a<b.

We define

INT(a,b)f(x) dx=-INT(b,a)f(x)dx.
Again, this IS NOT a theorem.
This is a definition.

And if a=b we define INT(a,b)f(x)dx=0.
For f defined at "a".

3. Originally Posted by ThePerfectHacker
We define,
INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
For function f defined on closed interval [a,b].

Note, under this strict definition, we require that f to be defined on [a,b] (and to be integratable). Hence a<b.

We define

INT(a,b)f(x) dx=-INT(b,a)f(x)dx.
Again, this IS NOT a theorem.
This is a definition.

And if a=b we define INT(a,b)f(x)dx=0.
For f defined at "a".
Thank you for responding, but I already know ALL of those definitions. All you needed to point out was this:

Originally Posted by ThePerfectHacker
We define,
INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
For function f defined on closed interval [a,b].
I merely forgot that f(x) MUST be defined on the interval [a,b] for the integration to be defined. Everything else I did was irrelevant.