# Thread: Riemann integral

1. ## Riemann integral

$f(x) = 0$ if 0 <= x < 1 or 1 < x <= 2
$f(x) = 1$ for every other value of x

The small and large f is the same function, I have no idea why the font size changes, sorry about that.

Show that f is integrable on [0, 2] and find $\int_0^2 \! f(x) \, dx$

This is an example from a calculus textbook, the solution is given below. I'm having a problem with understanding how the partition is made as in the solution provided it is simply stated that this is the partition without any reason given for how the partition was decided to be like that.

My question is simply how was it determined that the partition should be as stated? I have no doubt that it is correct, I simply don't understand it.

Solution:

Let $\epsilon > 0$ be given.
Let P = {0, 1 - $\epsilon$/3, 1 + $\epsilon$/3, 2}

Then L(f, P) = 0, since f(x) = 0 at points of each of these subintervals into which P subdivides [0, 2].

Since f(1) = 1, we have:

U(f, P) = 0(1 - $\epsilon$/3) + 1(2 $\epsilon$/3) + 0(2 - (1 + $\epsilon$/3))
U(f, P) = (2 $\epsilon$)/3

Hence, U(f, P) - L(f, P) < $\epsilon$ and f is integrable on [0, 2]. Since L(f, P) = 0 for every partition,
$\int_0^2 \! f(x) \, dx$ = $I_*$ = 0.

2. Originally Posted by posix_memalign
$f(x) = 0$ if 0 <= x < 1 or 1 < x <= 2
$f(x) = 1$ for every other value of x
Show that f is integrable on [0, 2] and find $\int_0^2 \! f(x) \, dx$
My question is simply how was it determined that the partition should be as stated? I have no doubt that it is correct, I simply don't understand it.

Solution:

Let $\epsilon > 0$ be given.
Let P = {0, 1 - $\epsilon$/3, 1 + $\epsilon$/3, 2}

Then L(f, P) = 0, since f(x) = 0 at points of each of these subintervals into which P subdivides [0, 2].

Since f(1) = 1, we have:

U(f, P) = 0(1 - $\epsilon$/3) + 1(2 $\epsilon$/3) + 0(2 - (1 + $\epsilon$/3))
U(f, P) = (2 $\epsilon$)/3

Hence, U(f, P) - L(f, P) < $\epsilon$ and f is integrable on [0, 2]. Since L(f, P) = 0 for every partition,
$\int_0^2 \! f(x) \, dx$ = $I_*$ = 0.
The answer to that question is simple.
Pick a partition that works. i.e makes the difference less than $\epsilon$

3. Originally Posted by Plato
The answer to that question is simple.
Pick a partition that works. i.e makes the difference less than $\epsilon$
So there are a lot of partitions that would work, then?

4. Specifically, why is $\epsilon$ divided by 3 chosen? Wouldn't also any number greater than 2 also yield that U(f, P) - L(f, P) < $\epsilon$?

5. Originally Posted by posix_memalign
Specifically, why is $\epsilon$ divided by 3 chosen? Wouldn't also any number greater than 2 also yield that U(f, P) - L(f, P) < $\varepsilon$?
It could have been any $\dfrac{\varepsilon }{N}$ where $N>2$.