any clues on how to solve these types of problems within 2-3 minutes for competitive examsHow to show this...

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


How to show this function is not in $L^{p}$ for any $p neq 2$?How do I solve these definite integrals?How to find a bound for these (simple) integralsHow to solve problems of this type?How to solve for an integral equation from already having the value of the integral?Evaluating a double integral of a complicated rational functionCopula: How to solve Integral with minimum for computation of Spearmans rhoHow to solve for a function an equation with integrals?How to solve for $y$ in $int_{0}^{y} frac{A + t}{B-t} dt = N$Integration of a function approximated by a nth order polynomial













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.










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
    19 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
    13 mins ago










  • $begingroup$
    How about using the reverse product rule?
    $endgroup$
    – Paras Khosla
    10 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.










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
    19 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
    13 mins ago










  • $begingroup$
    How about using the reverse product rule?
    $endgroup$
    – Paras Khosla
    10 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.










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 24 mins ago









Parcly Taxel

42.6k1372101




42.6k1372101










asked 27 mins ago









HOME WORK AND EXERCISESHOME WORK AND EXERCISES

397




397












  • $begingroup$
    My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
    $endgroup$
    – Lord Shark the Unknown
    19 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
    13 mins ago










  • $begingroup$
    How about using the reverse product rule?
    $endgroup$
    – Paras Khosla
    10 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
    19 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
    13 mins ago










  • $begingroup$
    How about using the reverse product rule?
    $endgroup$
    – Paras Khosla
    10 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
19 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
19 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
13 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
13 mins ago












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




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










2 Answers
2






active

oldest

votes


















5












$begingroup$

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
    9 mins ago






  • 1




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



















2












$begingroup$

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











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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
      9 mins ago






    • 1




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
















    5












    $begingroup$

    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
      9 mins ago






    • 1




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














    5












    5








    5





    $begingroup$

    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 16 mins ago









    Jonathan LevyJonathan Levy

    964




    964




    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
      9 mins ago






    • 1




      $begingroup$
      I think Paras said it--the product rule gives it to you.
      $endgroup$
      – Jonathan Levy
      4 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
      9 mins ago






    • 1




      $begingroup$
      I think Paras said it--the product rule gives it to you.
      $endgroup$
      – Jonathan Levy
      4 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
    9 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
    9 mins ago




    1




    1




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




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











    2












    $begingroup$

    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











    $endgroup$


















      2












      $begingroup$

      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











      $endgroup$
















        2












        2








        2





        $begingroup$

        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











        $endgroup$



        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














        share|cite



        share|cite








        edited 2 mins ago

























        answered 8 mins ago









        Paras KhoslaParas Khosla

        1,202216




        1,202216






























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