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trying to proove this limit equality to a specific integral [closed]

0

$begingroup$

can you help me prove this equality?

I tried to use Riemann sums but I haven't succeeded to find something useful.

$$lim_ntoinfty sum_k=1^n fleft(fracknright)frac1n
= int_0^1f(x)dx $$

Thank you very much.

analysis riemann-integration riemann-sum

share|cite|improve this question

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

$endgroup$

closed as off-topic by RRL, NCh, Eevee Trainer, Adrian Keister, Leucippus Apr 6 at 4:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, NCh, Eevee Trainer, Adrian Keister, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What did you get with Riemann sums? Also, summing from $n=1$ to $n$ doesn't make sense
  $endgroup$
  – J. W. Tanner
  Apr 2 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an immediate consequence of the definition of Riemann integral. Here the partition is uniform with points $x_k=k/n$ and tags $t_k$ are choosen the same as $x_k$ so the Riemann sum becomes $sum_k=1^nf(t_k)(x_k-x_k-1)$.
  $endgroup$
  – Paramanand Singh
  Apr 3 at 17:24
  

  
  
  
  

add a comment | 

0

$begingroup$

can you help me prove this equality?

I tried to use Riemann sums but I haven't succeeded to find something useful.

$$lim_ntoinfty sum_k=1^n fleft(fracknright)frac1n
= int_0^1f(x)dx $$

Thank you very much.

analysis riemann-integration riemann-sum

share|cite|improve this question

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

$endgroup$

closed as off-topic by RRL, NCh, Eevee Trainer, Adrian Keister, Leucippus Apr 6 at 4:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, NCh, Eevee Trainer, Adrian Keister, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What did you get with Riemann sums? Also, summing from $n=1$ to $n$ doesn't make sense
  $endgroup$
  – J. W. Tanner
  Apr 2 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an immediate consequence of the definition of Riemann integral. Here the partition is uniform with points $x_k=k/n$ and tags $t_k$ are choosen the same as $x_k$ so the Riemann sum becomes $sum_k=1^nf(t_k)(x_k-x_k-1)$.
  $endgroup$
  – Paramanand Singh
  Apr 3 at 17:24
  

  
  
  
  

add a comment | 

$begingroup$

can you help me prove this equality?

I tried to use Riemann sums but I haven't succeeded to find something useful.

$$lim_ntoinfty sum_k=1^n fleft(fracknright)frac1n
= int_0^1f(x)dx $$

Thank you very much.

analysis riemann-integration riemann-sum

share|cite|improve this question

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

share|cite|improve this question

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

edited Apr 2 at 17:11

gt6989b

36k22557

edited Apr 2 at 17:11

gt6989b

36k22557

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

asked Apr 2 at 16:55

Omer GaflaOmer Gafla

91

1 Answer
1

active

oldest

votes

1

$begingroup$

Your approach using Riemann sums is probably correct.
Try to think it this way, if you were given an integral of f(x) from 0 to 1, using Riemann sums you would break the interval in n equal delta intervals (where each delta interval is of length 1/n). And the value of the function f(x) at each subsequent interval would be f(1/n), f(2/n), ...., f(n/n). Summing the f(i/n).1/n for i=1 to n will give you the area under the curve f(x) from 0 to 1. Hence taking the limit n tending to infinity will give this summation an integral approximation.

share|cite|improve this answer

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

$endgroup$

add a comment | 

1

$begingroup$

Your approach using Riemann sums is probably correct.
Try to think it this way, if you were given an integral of f(x) from 0 to 1, using Riemann sums you would break the interval in n equal delta intervals (where each delta interval is of length 1/n). And the value of the function f(x) at each subsequent interval would be f(1/n), f(2/n), ...., f(n/n). Summing the f(i/n).1/n for i=1 to n will give you the area under the curve f(x) from 0 to 1. Hence taking the limit n tending to infinity will give this summation an integral approximation.

share|cite|improve this answer

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

$endgroup$

add a comment | 

1

$begingroup$

Your approach using Riemann sums is probably correct.
Try to think it this way, if you were given an integral of f(x) from 0 to 1, using Riemann sums you would break the interval in n equal delta intervals (where each delta interval is of length 1/n). And the value of the function f(x) at each subsequent interval would be f(1/n), f(2/n), ...., f(n/n). Summing the f(i/n).1/n for i=1 to n will give you the area under the curve f(x) from 0 to 1. Hence taking the limit n tending to infinity will give this summation an integral approximation.

share|cite|improve this answer

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

$endgroup$

add a comment | 

$begingroup$

Your approach using Riemann sums is probably correct.
Try to think it this way, if you were given an integral of f(x) from 0 to 1, using Riemann sums you would break the interval in n equal delta intervals (where each delta interval is of length 1/n). And the value of the function f(x) at each subsequent interval would be f(1/n), f(2/n), ...., f(n/n). Summing the f(i/n).1/n for i=1 to n will give you the area under the curve f(x) from 0 to 1. Hence taking the limit n tending to infinity will give this summation an integral approximation.

share|cite|improve this answer

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

$endgroup$

share|cite|improve this answer

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113

answered Apr 2 at 17:06

Nisarg JainNisarg Jain

113