any clues on how to solve these types of problems within 2-3 minutes for competitive examsHow do I solve...
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any clues on how to solve these types of problems within 2-3 minutes for competitive exams
How do I solve these definite integrals?How to find a bound for these (simple) integralsDefinite integral of Bessel function product over fourth powerHow to solve problems of this type?Prove that $lim_{n toinfty} int_0^1 frac{x^n}{sqrt{1+x^n}}, mathrm dx=0$.Copula: 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 polynomialDecreasing function with a target integral?
$begingroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
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 25 mins ago
Parcly Taxel
42.6k1372101
asked 29 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 25 mins ago
Parcly Taxel
42.6k1372101
asked 29 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
20 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
14 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 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 25 mins ago
Parcly Taxel
42.6k1372101
asked 29 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 25 mins ago
Parcly Taxel
42.6k1372101
asked 29 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
edited 25 mins ago
Parcly Taxel
42.6k1372101
asked 29 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
edited 25 mins ago
Parcly Taxel
42.6k1372101
edited 25 mins ago
Parcly Taxel
42.6k1372101
edited 25 mins ago
Parcly Taxel
42.6k1372101
42.6k1372101
asked 29 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
asked 29 mins ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
417
asked 29 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
20 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
14 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 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
20 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
14 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 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
20 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
20 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
14 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
14 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 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 9 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
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 18 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
10 mins ago
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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 9 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
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 9 mins ago
Paras KhoslaParas Khosla
1,227216
$endgroup$
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 9 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 9 mins ago
Paras KhoslaParas Khosla
1,227216
edited 4 mins ago
answered 9 mins ago
Paras KhoslaParas Khosla
1,227216
answered 9 mins ago
Paras KhoslaParas Khosla
1,227216
answered 9 mins ago
Paras KhoslaParas Khosla
1,227216
1,227216
add a comment |
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 18 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
10 mins ago
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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 18 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
10 mins ago
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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 18 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 18 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 18 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 18 mins ago
Jonathan LevyJonathan Levy
1064
answered 18 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
10 mins ago
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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
10 mins ago
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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
10 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
10 mins ago
1
1
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 mins ago
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 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
20 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
14 mins ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
11 mins ago