Clues on how to solve these types of problems within 2-3 minutes for competitive exams
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$begingroup$
$$int_0^102left(prod_k=1^100(x-k)right)left(sum_k=1^100frac1x-kright),dx$$
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 on how to solve this problem.
integration derivatives definite-integrals
edited Feb 27 at 9:09
Paras Khosla
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
$endgroup$
add a comment |
$begingroup$
$$int_0^102left(prod_k=1^100(x-k)right)left(sum_k=1^100frac1x-kright),dx$$
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 on how to solve this problem.
integration derivatives definite-integrals
edited Feb 27 at 9:09
Paras Khosla
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
$endgroup$
2
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
10
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
7
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20
add a comment |
$begingroup$
$$int_0^102left(prod_k=1^100(x-k)right)left(sum_k=1^100frac1x-kright),dx$$
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 on how to solve this problem.
integration derivatives definite-integrals
edited Feb 27 at 9:09
Paras Khosla
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
$endgroup$
$$int_0^102left(prod_k=1^100(x-k)right)left(sum_k=1^100frac1x-kright),dx$$
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 on how to solve this problem.
integration derivatives definite-integrals
integration derivatives definite-integrals
edited Feb 27 at 9:09
Paras Khosla
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
edited Feb 27 at 9:09
Paras Khosla
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
edited Feb 27 at 9:09
Paras Khosla
2,663323
edited Feb 27 at 9:09
Paras Khosla
2,663323
edited Feb 27 at 9:09
Paras Khosla
2,663323
2,663323
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
asked Feb 27 at 6:50
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
999
999
2
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
10
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
7
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20
add a comment |
2
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
10
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
7
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20
2
2
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
10
10
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
7
7
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
By the product rule you have the following result:
$$dfracmathrm dmathrm dxprod_k=1^100(x-k)=left(prod_k=1^100(x-k)right)left(sum_k=1^100dfrac1(x-k)right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited Feb 27 at 10:14
Community♦
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
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.
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
$endgroup$
$begingroup$
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
Feb 27 at 7:09
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
By the product rule you have the following result:
$$dfracmathrm dmathrm dxprod_k=1^100(x-k)=left(prod_k=1^100(x-k)right)left(sum_k=1^100dfrac1(x-k)right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited Feb 27 at 10:14
Community♦
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
add a comment |
$begingroup$
Hint:
By the product rule you have the following result:
$$dfracmathrm dmathrm dxprod_k=1^100(x-k)=left(prod_k=1^100(x-k)right)left(sum_k=1^100dfrac1(x-k)right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited Feb 27 at 10:14
Community♦
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
add a comment |
$begingroup$
Hint:
By the product rule you have the following result:
$$dfracmathrm dmathrm dxprod_k=1^100(x-k)=left(prod_k=1^100(x-k)right)left(sum_k=1^100dfrac1(x-k)right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited Feb 27 at 10:14
Community♦
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
$endgroup$
Hint:
By the product rule you have the following result:
$$dfracmathrm dmathrm dxprod_k=1^100(x-k)=left(prod_k=1^100(x-k)right)left(sum_k=1^100dfrac1(x-k)right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited Feb 27 at 10:14
Community♦
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
edited Feb 27 at 10:14
Community♦
1
edited Feb 27 at 10:14
Community♦
1
edited Feb 27 at 10:14
Community♦
1
1
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
answered Feb 27 at 7:10
Paras KhoslaParas Khosla
2,663323
2,663323
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– quid♦
Feb 28 at 8:49
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.
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
$endgroup$
$begingroup$
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
Feb 27 at 7:09
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
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.
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
$endgroup$
$begingroup$
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
Feb 27 at 7:09
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
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.
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
$endgroup$
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.
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
answered Feb 27 at 7:01
Jonathan LevyJonathan Levy
1796
1796
$begingroup$
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
Feb 27 at 7:09
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
add a comment |
$begingroup$
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
Feb 27 at 7:09
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
$begingroup$
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
Feb 27 at 7:09
$begingroup$
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
Feb 27 at 7:09
2
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
Feb 27 at 7:13
add a comment |
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$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
Feb 27 at 6:59
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
Feb 27 at 7:05
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
Feb 27 at 7:08
10
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
Feb 27 at 8:58
7
$begingroup$
Anybody gives me such an exercise, I start by replacing 100 by 3 (or even 2), this gives something which is far more easy to handle, and I have something I can learn from.
$endgroup$
– Dominique
Feb 27 at 14:20