Integral inequality with exponential and power [closed]
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Prove that the integral
$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1} le sqrt{2016/2015}$
calculus inequality
edited Nov 22 at 20:43
Mason
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
closed as off-topic by T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797 Nov 23 at 3:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
Prove that the integral
$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1} le sqrt{2016/2015}$
calculus inequality
edited Nov 22 at 20:43
Mason
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
closed as off-topic by T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797 Nov 23 at 3:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Prove that the integral
$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1} le sqrt{2016/2015}$
calculus inequality
edited Nov 22 at 20:43
Mason
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
Prove that the integral
$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1} le sqrt{2016/2015}$
calculus inequality
calculus inequality
edited Nov 22 at 20:43
Mason
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
edited Nov 22 at 20:43
Mason
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
edited Nov 22 at 20:43
Mason
1,8881527
edited Nov 22 at 20:43
Mason
1,8881527
edited Nov 22 at 20:43
Mason
1,8881527
1,8881527
asked Nov 22 at 20:22
Andrei Gabor
556
asked Nov 22 at 20:22
Andrei Gabor
556
asked Nov 22 at 20:22
Andrei Gabor
556
556
closed as off-topic by T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797 Nov 23 at 3:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797 Nov 23 at 3:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Davide Giraudo, Leucippus, Cesareo, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20
add a comment |
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx{=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_{-1}^0 frac{sqrt{1+x^{2014}}}{2015^x+1}dx\=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_0^1 frac{sqrt{1+(-x)^{2014}}}{2015^{-x}+1}dx\=int_{0}^1 sqrt{1+x^{2014}}dx\le sqrt{int_{0}^1 1+x^{2014}dx}\=sqrt{2016over 2015}}$$where the inequality comes from Cauchy-Schwarz inequality for integrals with $f(x)=sqrt{1+x^{2014}}$ and $g(x)=1$.
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx{=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_{-1}^0 frac{sqrt{1+x^{2014}}}{2015^x+1}dx\=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_0^1 frac{sqrt{1+(-x)^{2014}}}{2015^{-x}+1}dx\=int_{0}^1 sqrt{1+x^{2014}}dx\le sqrt{int_{0}^1 1+x^{2014}dx}\=sqrt{2016over 2015}}$$where the inequality comes from Cauchy-Schwarz inequality for integrals with $f(x)=sqrt{1+x^{2014}}$ and $g(x)=1$.
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
add a comment |
up vote
3
down vote
accepted
$$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx{=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_{-1}^0 frac{sqrt{1+x^{2014}}}{2015^x+1}dx\=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_0^1 frac{sqrt{1+(-x)^{2014}}}{2015^{-x}+1}dx\=int_{0}^1 sqrt{1+x^{2014}}dx\le sqrt{int_{0}^1 1+x^{2014}dx}\=sqrt{2016over 2015}}$$where the inequality comes from Cauchy-Schwarz inequality for integrals with $f(x)=sqrt{1+x^{2014}}$ and $g(x)=1$.
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx{=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_{-1}^0 frac{sqrt{1+x^{2014}}}{2015^x+1}dx\=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_0^1 frac{sqrt{1+(-x)^{2014}}}{2015^{-x}+1}dx\=int_{0}^1 sqrt{1+x^{2014}}dx\le sqrt{int_{0}^1 1+x^{2014}dx}\=sqrt{2016over 2015}}$$where the inequality comes from Cauchy-Schwarz inequality for integrals with $f(x)=sqrt{1+x^{2014}}$ and $g(x)=1$.
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
$$int_{-1}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx{=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_{-1}^0 frac{sqrt{1+x^{2014}}}{2015^x+1}dx\=int_{0}^1 frac{sqrt{1+x^{2014}}}{2015^x+1}dx+int_0^1 frac{sqrt{1+(-x)^{2014}}}{2015^{-x}+1}dx\=int_{0}^1 sqrt{1+x^{2014}}dx\le sqrt{int_{0}^1 1+x^{2014}dx}\=sqrt{2016over 2015}}$$where the inequality comes from Cauchy-Schwarz inequality for integrals with $f(x)=sqrt{1+x^{2014}}$ and $g(x)=1$.
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
answered Nov 22 at 22:32
Mostafa Ayaz
13.6k3836
13.6k3836
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
add a comment |
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
So in the second integral you changed x to -x.the the integral will be from 1 to 0 and to make it from 0 to 1 don't you have to put a - in front of it?
– Andrei Gabor
Nov 22 at 22:46
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
The minus sign gets cancelled with that of differential variable because$$int_{-1}^0f(x)dx=int_{1}^0f(-u)d(-u)=-int_{1}^0f(-u)du=int_0^1f(-u)du$$
– Mostafa Ayaz
Nov 23 at 7:31
add a comment |
It's x^2014 but it doesn't compile right
– Andrei Gabor
Nov 22 at 20:23
Claim seems true. desmos.com/calculator/vqvwnh6dbp
– Mason
Nov 22 at 20:28
Can you tell us where the problem is from? It looks like it's some exercise from 2014, 2015, or 2016... Is it homework? Is it from a book? Context usually helps in this type of thing.
– Mason
Nov 22 at 20:32
Use x^{2014} to make it appear right
– Rhys Hughes
Nov 22 at 20:43
It's from a 2014 local romanian olympiad .
– Andrei Gabor
Nov 22 at 22:20