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
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$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
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
$endgroup$
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
$endgroup$
$begingroup$
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
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
$endgroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
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 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
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
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
$endgroup$
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
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
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.
New contributor
$endgroup$
$begingroup$
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
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered 16 mins ago
Jonathan LevyJonathan Levy
964
964
New contributor
New contributor
$begingroup$
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
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
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
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
$endgroup$
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
$endgroup$
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
$endgroup$
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.
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
edited 2 mins ago
answered 8 mins ago
Paras KhoslaParas Khosla
1,202216
1,202216
add a comment |
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
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