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any clues on how to solve these types of problems within 2-3 minutes for competitive exams

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any clues on how to solve these types of problems within 2-3 minutes for competitive exams

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1

$begingroup$

$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$

I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but i'd be glad if i could get some clues to how do i solve this problem.

definite-integrals

share|cite|improve this question

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
  $endgroup$
  – Lord Shark the Unknown
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer given is 101!-100! but no solutions also i can't find such problem online to learn
  $endgroup$
  – HOME WORK AND EXERCISES
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the reverse product rule?
  $endgroup$
  – Paras Khosla
  25 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$

I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but i'd be glad if i could get some clues to how do i solve this problem.

definite-integrals

share|cite|improve this question

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
  $endgroup$
  – Lord Shark the Unknown
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer given is 101!-100! but no solutions also i can't find such problem online to learn
  $endgroup$
  – HOME WORK AND EXERCISES
  28 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the reverse product rule?
  $endgroup$
  – Paras Khosla
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$

I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but i'd be glad if i could get some clues to how do i solve this problem.

definite-integrals

share|cite|improve this question

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

$endgroup$

share|cite|improve this question

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

share|cite|improve this question

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

edited 39 mins ago

Parcly Taxel

42.6k1372101

edited 39 mins ago

Parcly Taxel

42.6k1372101

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

asked 42 mins ago

HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417

2 Answers
2

active

oldest

votes

3

$begingroup$

Hint:

By the product rule you have the following result. Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.

$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$

share|cite|improve this answer

edited 17 mins ago

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you that was rather uncomplicated. :D
  $endgroup$
  – HOME WORK AND EXERCISES
  11 mins ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.

share|cite|improve this answer

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
  $endgroup$
  – HOME WORK AND EXERCISES
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think Paras said it--the product rule gives it to you.
  $endgroup$
  – Jonathan Levy
  19 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

Hint:

By the product rule you have the following result. Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.

$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$

share|cite|improve this answer

edited 17 mins ago

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you that was rather uncomplicated. :D
  $endgroup$
  – HOME WORK AND EXERCISES
  11 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint:

By the product rule you have the following result. Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.

$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$

share|cite|improve this answer

edited 17 mins ago

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you that was rather uncomplicated. :D
  $endgroup$
  – HOME WORK AND EXERCISES
  11 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

By the product rule you have the following result. Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.

$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$

share|cite|improve this answer

edited 17 mins ago

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

$endgroup$

share|cite|improve this answer

edited 17 mins ago

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

answered 23 mins ago

Paras KhoslaParas Khosla

1,227216

6

$begingroup$

Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.

share|cite|improve this answer

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
  $endgroup$
  – HOME WORK AND EXERCISES
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think Paras said it--the product rule gives it to you.
  $endgroup$
  – Jonathan Levy
  19 mins ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.

share|cite|improve this answer

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
  $endgroup$
  – HOME WORK AND EXERCISES
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think Paras said it--the product rule gives it to you.
  $endgroup$
  – Jonathan Levy
  19 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.

share|cite|improve this answer

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 31 mins ago

Jonathan LevyJonathan Levy

1064

New contributor

Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 31 mins ago

Jonathan LevyJonathan Levy

1064