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












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











$endgroup$












  • $begingroup$
    My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
    $endgroup$
    – Lord Shark the Unknown
    22 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
    17 mins ago










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
















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











$endgroup$












  • $begingroup$
    My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
    $endgroup$
    – Lord Shark the Unknown
    22 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
    17 mins ago










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














1












1








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











$endgroup$




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



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




definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 28 mins ago









Parcly Taxel

42.6k1372101




42.6k1372101










asked 31 mins ago









HOME WORK AND EXERCISESHOME WORK AND EXERCISES

417




417












  • $begingroup$
    My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
    $endgroup$
    – Lord Shark the Unknown
    22 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
    17 mins ago










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


















  • $begingroup$
    My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
    $endgroup$
    – Lord Shark the Unknown
    22 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
    17 mins ago










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
















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




$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
22 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
17 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
17 mins ago












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




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










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











$endgroup$













  • $begingroup$
    Thank you that was rather uncomplicated. :D
    $endgroup$
    – HOME WORK AND EXERCISES
    15 secs ago



















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








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
    13 mins ago






  • 1




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











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






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











$endgroup$













  • $begingroup$
    Thank you that was rather uncomplicated. :D
    $endgroup$
    – HOME WORK AND EXERCISES
    15 secs ago
















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











$endgroup$













  • $begingroup$
    Thank you that was rather uncomplicated. :D
    $endgroup$
    – HOME WORK AND EXERCISES
    15 secs ago














3












3








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











$endgroup$



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














share|cite|improve this answer



share|cite|improve this answer








edited 6 mins ago

























answered 12 mins ago









Paras KhoslaParas Khosla

1,227216




1,227216












  • $begingroup$
    Thank you that was rather uncomplicated. :D
    $endgroup$
    – HOME WORK AND EXERCISES
    15 secs ago


















  • $begingroup$
    Thank you that was rather uncomplicated. :D
    $endgroup$
    – HOME WORK AND EXERCISES
    15 secs ago
















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




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











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








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
    13 mins ago






  • 1




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
















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








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
    13 mins ago






  • 1




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














6












6








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








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.






$endgroup$



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








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.









share|cite|improve this answer



share|cite|improve this answer






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 20 mins ago









Jonathan LevyJonathan Levy

1064




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.





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.






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












  • $begingroup$
    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
    13 mins ago






  • 1




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


















  • $begingroup$
    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
    13 mins ago






  • 1




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
















$begingroup$
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
13 mins ago




$begingroup$
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
13 mins ago




1




1




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




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


















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