Definite integral of greatest integer function

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Why is π/23π/2[cos-1t].dt = π/23π/2-1.dt?

[.] denotes the greatest integer function. Is it because [cos-1t] = -1 in t ∈ [π/2,3π/2)? But [cos-1t] = 0 for x = 3π/2. So we have one 0 along with all the -1's in (π/2,3π/2]. So how can we substitute [cos-1t] = -1 for every x in (π/2,3π/2] even though [cos-1t] = 0 at x = 3π/2?

Similarly, how can we substitute 3π/2[cos-1t].dt = 3π/20.dt even though [cos-1t] = 1 at x = 2π?
 
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  • #2
Svein
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So how can we substitute [cos-1t] = -1 for every x in (π/2,3π/2] even though [cos-1t] = 0 at x = 3π/2?
Removing one point (end point) does not change the value of the integral.
 
  • #3
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Removing one point (end point) does not change the value of the integral.
Why? The integral is made of infinitely small rectangular strips. If we remove one of those strips won't the area change? Or would it not?
 
  • #4
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If we remove one of those strips won't the area change?
The easiest way for me is to say "the measure of a single point is 0, so the measure of [0, 1] = the measure of [0, 1).
Another way: Let f(x)=1 for 0≤x<1 and f(1)=2. Then, obviously [itex] \int_{0}^{1-\frac{1}{n}}f(x)dx=1-\frac{1}{n}[/itex] for all n. Take it to the limit...
 
  • #5
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The easiest way for me is to say "the measure of a single point is 0, so the measure of [0, 1] = the measure of [0, 1).
The measure of all single points is 0 except the last one.
 
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Anybody there?
 
  • #7
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The measure of all single points is 0 except the last one.
What's different about the "last one"?
 
  • #8
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What's different about the "last one"?
That everything before the last one had area 0.dx. But the last one had area 1.dx
 
  • #9
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But the last one had area 1.dx
Sorry. Let us go back to my function and extend it, so that f(x)=1 for 0≤x<1 and f(x)=2 for 1≤x<2. This is by definition a step function, used to define Riemann integrals.For each step, the integral of the step function is (the value of the step function in that interval) times (the length of the interval). Observe that no "dx" is involved. Thus the integral of the step function f(x) over the interval [0, 1) is 1⋅1=1.

Now, if you really want to involve the point "1", observe that [itex] \int_{0}^{1+\frac{1}{n}}f(x)dx=1+\frac{2}{n}[/itex]. This is valid for all n. Since [itex][0, 1]=\bigcap_{n}[0, \frac{1}{n}] [/itex], the integral over [0, 1] is 1.
 
  • #10
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[0,1]=⋂n[0,1n]
What is this notation? I didn't understand what this means.
 
  • #11
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[0,1]=⋂n[0,1n]

What is this notation? I didn't understand it.
 
  • #12
Svein
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What is this notation? I didn't understand it.
Sorry, typo. It should be [itex][0, 1] = \bigcap_{n}[0, 1+\frac{1}{n}] [/itex] which means "the intersection of all intervals of type [0, 1 + 1/n]". Given ε>0, it is possible to find an N such that the integral value is less than 1 + ε (and ≥1) for all n>N.
That everything before the last one had area 0.dx. But the last one had area 1.dx
Sorry. By "measure" we mean "the length of the interval". The interval [a, b] has measure b - a. Thus for a single point, say p, the interval [p, p] has length p - p = 0.
 
  • #13
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Sorry, typo. It should be [itex][0, 1] = \bigcap_{n}[0, 1+\frac{1}{n}] [/itex] which means "the intersection of all intervals of type [0, 1 + 1/n]". Given ε>0, it is possible to find an N such that the integral value is less than 1 + ε (and ≥1) for all n>N.
Sorry. By "measure" we mean "the length of the interval". The interval [a, b] has measure b - a. Thus for a single point, say p, the interval [p, p] has length p - p = 0.
I think you are not getting what I am trying to say. The last point has the length of the interval dx and not 0. That's because we have divided the area under the curve in infinite strips of width dx each. So of those infinite strips, infinite -1 have area 0 and 1 strip has area 1.dx So why is the area of the whole interval comprising of infinite dx lengths still coming out to be 0? Is it because dx → 0, therefore 1.dx → 0? But then if this was not the case and f(x) would actually have some definite value which was not zero at all points in the interval (a,b], then shouldn't the area even then be 0? That's because the area would be f(a+dx).dx + f(a+2dx).dx + f(a+3dx).dx .... f(b).dx. Now since dx → 0, therefore all the terms like: f(a+dx).dx, f(a+2dx).dx, f(a+3dx).dx tend to zero too. So the area in the end should be 0. But clearly that is not the case. So why is that? What am I overlooking/missing?
 
  • #14
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I think you are not getting what I am trying to say. The last point has the length of the interval dx and not 0. That's because we have divided the area under the curve in infinite strips of width dx each. So of those infinite strips, infinite -1 have area 0 and 1 strip has area 1.dx So why is the area of the whole interval comprising of infinite dx lengths still coming out to be 0? Is it because dx → 0, therefore 1.dx → 0? But then if this was not the case and f(x) would actually have some definite value which was not zero at all points in the interval (a,b], then shouldn't the area even then be 0? That's because the area would be f(a+dx).dx + f(a+2dx).dx + f(a+3dx).dx .... f(b).dx. Now since dx → 0, therefore all the terms like: f(a+dx).dx, f(a+2dx).dx, f(a+3dx).dx tend to zero too. So the area in the end should be 0. But clearly that is not the case. So why is that? What am I overlooking/missing?
You are trying to reason using infinitesimals, which is very tricky, and therefore not used any more in integration theory. For a short introduction, see http://en.wikipedia.org/wiki/Lebesgue_integration.
 
  • #15
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You are trying to reason using infinitesimals, which is very tricky, and therefore not used any more in integration theory. For a short introduction, see http://en.wikipedia.org/wiki/Lebesgue_integration.
So how would you explain why [cos-1t] was substituted as 0 all throughout? In fact, the question reduces down to: while computing definite integral with lower limit a and upper limit b, do we mean (a,b] or [a,b] or [a,b) or (a,b)?
 
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  • #16
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So how would you explain why [cos-1t] was substituted as 0 all throughout?
This was your last integral. As I have explained, the measure of a single point is 0. Therefore, [cos-1t] is 0 almost everywhere (which means except for a set of measure 0) in the interval [3π/2, 2π].
 
  • #17
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almost everywhere (which means except for a set of measure 0)
What does that mean? And please read post #15 again. I edited it a just a second ago.
 
  • #18
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What does that mean?
Exactly what it says.
In fact, the question reduces down to: while computing definite integral with lower limit a and upper limit b, do we mean (a,b] or [a,b] or [a,b) or (a,b)?
It does not matter, since the difference between these are a set of measure 0. And please, read up on integration theory instead of asking the same questions again and again.
 
  • #19
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Exactly what it says.
It does not matter, since the difference between these are a set of measure 0. And please, read up on integration theory instead of asking the same questions again and again.
I don't think I need to read the integration theory because what I am asking is just high school stuff. And the link you posted is way beyond the scope of knowledge required for high school. What does set of measure 0 mean?
 
  • #21
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I explained it above (post #12).
The terminologies you are involving are way beyond the requirements of high school integration. For now I think I will go with the fact that integration happens over [a,b)
 

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