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Showing $int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx=a-b$

2

1

$begingroup$

I want to show

$$int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx=a-b$$

I know that
begin{equation}
leftlfloor -x rightrfloor = begin{cases} -leftlfloor x rightrfloor & text{if } x in mathbb{Z} \ -leftlfloor x rightrfloor-1 & text{if } x notin mathbb{Z}. end{cases}
end{equation}

In this case do I use $-leftlfloor x rightrfloor$ or $-leftlfloor x rightrfloor-1$? I think I am confused about some definitions, one of the solutions said $leftlfloor x rightrfloor$ is constant on the open subintervals of the partition
$$P=left(a, leftlfloor a rightrfloor+1 cdots leftlfloor a rightrfloor + leftlfloor b-a rightrfloor, bright)$$
and since there are no integers in the open subintervals of P, then we would use $-leftlfloor x rightrfloor - 1$.. I don't think I quite understand this point here. I know I can solve it and say
begin{align*}
int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx= int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} -leftlfloor x rightrfloor -1 ; dx= a-b
end{align*}

but I don't understand why.

calculus floor-function

share|cite|improve this question

asked Mar 22 at 13:15

D. QaD. Qa

1656

$endgroup$

add a comment | 

2

1

$begingroup$

I want to show

$$int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx=a-b$$

I know that
begin{equation}
leftlfloor -x rightrfloor = begin{cases} -leftlfloor x rightrfloor & text{if } x in mathbb{Z} \ -leftlfloor x rightrfloor-1 & text{if } x notin mathbb{Z}. end{cases}
end{equation}

In this case do I use $-leftlfloor x rightrfloor$ or $-leftlfloor x rightrfloor-1$? I think I am confused about some definitions, one of the solutions said $leftlfloor x rightrfloor$ is constant on the open subintervals of the partition
$$P=left(a, leftlfloor a rightrfloor+1 cdots leftlfloor a rightrfloor + leftlfloor b-a rightrfloor, bright)$$
and since there are no integers in the open subintervals of P, then we would use $-leftlfloor x rightrfloor - 1$.. I don't think I quite understand this point here. I know I can solve it and say
begin{align*}
int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx= int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} -leftlfloor x rightrfloor -1 ; dx= a-b
end{align*}

but I don't understand why.

calculus floor-function

share|cite|improve this question

asked Mar 22 at 13:15

D. QaD. Qa

1656

$endgroup$

add a comment | 

$begingroup$

I want to show

$$int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx=a-b$$

I know that
begin{equation}
leftlfloor -x rightrfloor = begin{cases} -leftlfloor x rightrfloor & text{if } x in mathbb{Z} \ -leftlfloor x rightrfloor-1 & text{if } x notin mathbb{Z}. end{cases}
end{equation}

In this case do I use $-leftlfloor x rightrfloor$ or $-leftlfloor x rightrfloor-1$? I think I am confused about some definitions, one of the solutions said $leftlfloor x rightrfloor$ is constant on the open subintervals of the partition
$$P=left(a, leftlfloor a rightrfloor+1 cdots leftlfloor a rightrfloor + leftlfloor b-a rightrfloor, bright)$$
and since there are no integers in the open subintervals of P, then we would use $-leftlfloor x rightrfloor - 1$.. I don't think I quite understand this point here. I know I can solve it and say
begin{align*}
int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} leftlfloor -x rightrfloor dx= int_{a}^{b} leftlfloor x rightrfloor dx + int_{a}^{b} -leftlfloor x rightrfloor -1 ; dx= a-b
end{align*}

but I don't understand why.

calculus floor-function

share|cite|improve this question

asked Mar 22 at 13:15

D. QaD. Qa

1656

$endgroup$

share|cite|improve this question

asked Mar 22 at 13:15

D. QaD. Qa

1656

share|cite|improve this question

asked Mar 22 at 13:15

D. QaD. Qa

1656

asked Mar 22 at 13:15

D. QaD. Qa

1656

asked Mar 22 at 13:15

D. QaD. Qa

1656

2 Answers
2

active

oldest

votes

5

$begingroup$

What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $lfloor x rfloor = n$, and $-n > x > -(n+1)$, so $lfloor -x rfloor = -(n+1)$. Adding gives $lfloor xrfloor + lfloor -x rfloor = -1$.

share|cite|improve this answer

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This cleared my doubts, thanks!
  $endgroup$
  – D. Qa
  Mar 22 at 13:59
  

  
  
  
  

add a comment | 

1

$begingroup$

Because you can simply ignore the intergers, sice they do not contribute to the integral. Think of a rectangle with height 1 and width 0, it has zero area, and finitely (in the interval $(a,b)$) many of them still sum to zero.

share|cite|improve this answer

answered Mar 22 at 13:42

TreborTrebor

1,01315

$endgroup$

add a comment | 

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5

$begingroup$

What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $lfloor x rfloor = n$, and $-n > x > -(n+1)$, so $lfloor -x rfloor = -(n+1)$. Adding gives $lfloor xrfloor + lfloor -x rfloor = -1$.

share|cite|improve this answer

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This cleared my doubts, thanks!
  $endgroup$
  – D. Qa
  Mar 22 at 13:59
  

  
  
  
  

add a comment | 

5

$begingroup$

What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $lfloor x rfloor = n$, and $-n > x > -(n+1)$, so $lfloor -x rfloor = -(n+1)$. Adding gives $lfloor xrfloor + lfloor -x rfloor = -1$.

share|cite|improve this answer

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This cleared my doubts, thanks!
  $endgroup$
  – D. Qa
  Mar 22 at 13:59
  

  
  
  
  

add a comment | 

$begingroup$

What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $lfloor x rfloor = n$, and $-n > x > -(n+1)$, so $lfloor -x rfloor = -(n+1)$. Adding gives $lfloor xrfloor + lfloor -x rfloor = -1$.

share|cite|improve this answer

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

$endgroup$

share|cite|improve this answer

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

answered Mar 22 at 13:29

ncmathsadistncmathsadist

43.2k261103

1

$begingroup$

Because you can simply ignore the intergers, sice they do not contribute to the integral. Think of a rectangle with height 1 and width 0, it has zero area, and finitely (in the interval $(a,b)$) many of them still sum to zero.

share|cite|improve this answer

answered Mar 22 at 13:42

TreborTrebor

1,01315

$endgroup$

add a comment | 

1

$begingroup$

Because you can simply ignore the intergers, sice they do not contribute to the integral. Think of a rectangle with height 1 and width 0, it has zero area, and finitely (in the interval $(a,b)$) many of them still sum to zero.

share|cite|improve this answer

answered Mar 22 at 13:42

TreborTrebor

1,01315

$endgroup$

add a comment | 

$begingroup$

Because you can simply ignore the intergers, sice they do not contribute to the integral. Think of a rectangle with height 1 and width 0, it has zero area, and finitely (in the interval $(a,b)$) many of them still sum to zero.

share|cite|improve this answer

answered Mar 22 at 13:42

TreborTrebor

1,01315

$endgroup$

share|cite|improve this answer

answered Mar 22 at 13:42

TreborTrebor

1,01315

answered Mar 22 at 13:42

TreborTrebor

1,01315

answered Mar 22 at 13:42

TreborTrebor

1,01315