Let

f(x)=

{ 1 if -1 < or equal to x<0;

{-1 if 0 < or equal to x < or equal to 1.

Prove that f(x) is integrable on -1 < or equal to x < or equal to 1.

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- Dec 18th 2006, 11:01 PMSwamifezHelp with Integrable Proof
Let

f(x)=

{ 1 if -1 < or equal to x<0;

{-1 if 0 < or equal to x < or equal to 1.

Prove that f(x) is integrable on -1 < or equal to x < or equal to 1. - Dec 19th 2006, 01:07 AMCaptainBlack
I fail to see how this can be a problem, what do you know about integration?

(what type of integral are we talking about here).

If this is a Riemann integral it is obvious that the integral of $\displaystyle f(x)$ is

defined over $\displaystyle [-1,1]$ from a elementary consideration of the Riemann

sums of $\displaystyle f(x)$ for partitions of this interval, and that the integral is:

$\displaystyle \int_{-1}^1 f(x) dx=\int_{-1}^0 f(x) dx + \int_{0}^1 f(x) dx=1-1=0$

RonL - Dec 19th 2006, 05:50 AMThePerfectHacker
I can actually do the limit of the sum to show it exists.

But the rule is: "*A real-function defined on a closed interval is Riemann integrable if and only if it is continous almost everywhere*".

This function is just discontinous on $\displaystyle x=0$. Thus, it is integrable. - Dec 21st 2006, 03:04 AMSwamifez
Sorry I am posting so much, but these problems are going to be similar to by final. Can someone help me clarify this problem in to one big proof. I also need to pick 5 paritions of this interval and prove it within. Thank you

- Dec 21st 2006, 08:33 AMThePerfectHacker
Okay, for suffiently large $\displaystyle n$.

You want to find,

$\displaystyle \int_{-1}^1 f(x)dx$

I will use right endpoints,

$\displaystyle \lim_{n\to \infty}\sum_{k=1}^n f(-1+k\Delta x)\Delta x$

But,

$\displaystyle \Delta x=\frac{2}{n}$

Thus,

$\displaystyle \lim_{n\to \infty}\sum_{k=1}^n f\left(-1+\frac{2k}{n} \right) \frac{2}{n}$

Note,

$\displaystyle -1\leq -1+\frac{2k}{n} <0$

$\displaystyle -n\leq -n+2k<0$

$\displaystyle 0\leq 2k<n$

$\displaystyle 0\leq k<n/2$

$\displaystyle k=0,1,2,...,[n/2]$ (greatest integer).

Thus, the function is 1 for these values.

And it is -1 for after these values.

$\displaystyle \sum_{k=1}^{[n/2]} \frac{2}{n}-\sum_{k=[n/2]+1}^n \frac{2}{n}$

If $\displaystyle n$ is even we have,

$\displaystyle \sum_{k=1}^{n/2} \frac{2}{n}-\sum_{k=n/2+1}^n \frac{2}{n}=0$

Thus, as $\displaystyle n\to \infty$ the limit of the Riemann sum is zero.

If $\displaystyle n$ is odd we have,

$\displaystyle \sum_{k=1}^{(n-1)/2}\frac{2}{n}-\sum_{k=(n-1)/2+1}^n \frac{2}{n}=-\frac{2}{n}$

Thus, as $\displaystyle n\to \infty$ the limit of the Riemann sum is zero.

Thus, the limit is always zero no matter what.