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.
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
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.
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.