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
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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
$begingroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
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
edited 32 mins ago
Parcly Taxel
42.6k1372101
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
$endgroup$
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
edited 32 mins ago
Parcly Taxel
42.6k1372101
asked 36 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
27 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
21 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 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
edited 32 mins ago
Parcly Taxel
42.6k1372101
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
$endgroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
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
edited 32 mins ago
Parcly Taxel
42.6k1372101
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
edited 32 mins ago
Parcly Taxel
42.6k1372101
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
edited 32 mins ago
Parcly Taxel
42.6k1372101
edited 32 mins ago
Parcly Taxel
42.6k1372101
edited 32 mins ago
Parcly Taxel
42.6k1372101
42.6k1372101
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
asked 36 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
asked 36 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
27 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
21 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 mins ago
add a comment |
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
27 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
21 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 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
27 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
27 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
21 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
21 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 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.
answered 25 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
17 mins ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 mins ago
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. Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
edited 11 mins ago
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
answered 16 mins ago
Paras KhoslaParas Khosla
1,227216
1,227216
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 mins ago
add a comment |
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 mins ago
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 mins ago
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
4 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.
answered 25 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
17 mins ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 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.
answered 25 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
17 mins ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 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.
answered 25 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.
$endgroup$
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 25 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 25 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 25 mins ago
Jonathan LevyJonathan Levy
1064
answered 25 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
17 mins ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 mins ago
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
17 mins ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
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
17 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
17 mins ago
2
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 mins ago
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
13 mins ago
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
27 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
21 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
18 mins ago