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Finding the constrain of integral

ML-inequality for real integralsEvaluating the integral $int_0^infty frac{sin x} x ,mathrm dx = frac pi 2$?Proving the inequality of an integral applied to a sumFind values so that integral is a bounded operatorProblem in series expansionSolving the improper integral $1/(x^a+y^b)$Convergence of the integral: $I_{alpha }=int _0^{infty }left(frac{e^{-alpha x}}{sqrt{x}}right):dx$definition of Fourier transform questionsHow to calculate $iint_D{} (x^2 + y^2) , dxdy$ with $D = {(frac{x}{2})^2 + (frac{y}{3})^2 leq 1}$?Integral transform with reciprocal complex exponential functions?permute double integral and limit

2

1

$begingroup$

$ f_n(x) = left(4x-x^2right)^n$

I'm trying to prove that $int _0^4:f_n(x),dx < 4^{n+1}$

When I'm doing a general integral to $n > 1$, I'm getting an expression that equals to zero when $X = 4$ or $X = 0$.

This is the expression:

$$frac{left(4x-x^2right)^{n+1}}{(4-2X)(n+1)} $$

What am I missing? Because it doesn't not make sense that every integral with $n > 1$ is equal to zero

analysis

share|cite|improve this question

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

$endgroup$

add a comment | 

2

1

$begingroup$

$ f_n(x) = left(4x-x^2right)^n$

I'm trying to prove that $int _0^4:f_n(x),dx < 4^{n+1}$

When I'm doing a general integral to $n > 1$, I'm getting an expression that equals to zero when $X = 4$ or $X = 0$.

This is the expression:

$$frac{left(4x-x^2right)^{n+1}}{(4-2X)(n+1)} $$

What am I missing? Because it doesn't not make sense that every integral with $n > 1$ is equal to zero

analysis

share|cite|improve this question

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

$endgroup$

add a comment | 

$begingroup$

$ f_n(x) = left(4x-x^2right)^n$

I'm trying to prove that $int _0^4:f_n(x),dx < 4^{n+1}$

When I'm doing a general integral to $n > 1$, I'm getting an expression that equals to zero when $X = 4$ or $X = 0$.

This is the expression:

$$frac{left(4x-x^2right)^{n+1}}{(4-2X)(n+1)} $$

What am I missing? Because it doesn't not make sense that every integral with $n > 1$ is equal to zero

analysis

share|cite|improve this question

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

$endgroup$

share|cite|improve this question

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

share|cite|improve this question

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

edited 8 hours ago

Bernard

127k743120

edited 8 hours ago

Bernard

127k743120

asked 8 hours ago

DanielDaniel

111

asked 8 hours ago

DanielDaniel

111

3 Answers
3

active

oldest

votes

4

$begingroup$

Hint:

$4x-x^2$ is a quadratic polynomial , with a negative leading coefficient. Furthermore, it vanishes for $x=0$ and $x=4$. Can you find it (absolute) maximum?

Once you've found this maximum, you can use the mean value inequality for integrals.

share|cite|improve this answer

answered 7 hours ago

BernardBernard

127k743120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, you can estimate this by the maximum! very nice approach. I like this argument. This is probably how the question was meant to be done ^^
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You cannot just blindly apply the reverse power rule. Remember the chain rule? You have a function within another function.

Consider the case when n=2

$ int_0^4 (4x + x^2)^2 dx = int_0^4 16x^2 + 8x^3 + x^4 dx$

share|cite|improve this answer

answered 7 hours ago

agentnolaagentnola

389

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would it be better for him to use the binomial expansion, you think?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That would be an excellent approach. I was also thinking that you could attempt to find a pattern and then use induction to prove it.
  $endgroup$
  – agentnola
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I started working out the binomial theorem approach, you can switch the order of integration and summation and then integrate the power of $x$, this should not be too hard, although it will get a bit messy: $$int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now reduce everything to some power of $x$, move the integral inside and then we hope to get something pretty?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Below is a suggestion for a very tedious route, but I wanted to post it anyway. This is probably not how the question is meant to be done.

Observe how by the binomial theorem we have that:
$$I=int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
We make this look a bit prettier:
$$I=int _{x=0}^4: sum_{k=0}^n binom{n}{k} 4^{k} cdot (-1)^{n-k}x^{2n-k},dx$$

We may switch the order of integration and summation since this concerns polynomials (what I mean is that they are of finite order), we compute:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{k} cdot (-1)^{n-k}4^{2n-k+1}-0$$
We collect some terms:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{2n+1} cdot (-1)^{n-k}$$
Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be:
$$ I =frac{sqrt {pi} cdot n! cdot 2^{2n+1}}{(n+frac{1}{2})!}=frac{sqrt pi cdot n!}{(n+frac{1}{2})!} cdot 4^{n + frac{1}{2}}$$
We then need to prove by induction that $frac{sqrt pi cdot n!}{(n+frac{1}{2})!} > sqrt{4}$ for positive $n$.

Yes. This is a lot of hard work. You could also just take the approach that was posted by Bernard and observe that $4x-x^2$ has a global maximum on the interval $[0,4]$ (the begin and endpoint are zeros of this polynomial), this maximum by symmetry occurs at $x=2$. The function has maximum value $8-4=4$. Observe then by the estimation lemma that $M =max_{x in [0,4]}((4x-x^2)^n)=4^n$, and the length of our path is $4$. We get:
$$ int _{x=0}^4 f_n dx leq M L = 4^n cdot 4 = 4^{n+1} $$

Also see: ML-inequality for real integrals

share|cite|improve this answer

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

$endgroup$

add a comment | 

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4

$begingroup$

Hint:

$4x-x^2$ is a quadratic polynomial , with a negative leading coefficient. Furthermore, it vanishes for $x=0$ and $x=4$. Can you find it (absolute) maximum?

Once you've found this maximum, you can use the mean value inequality for integrals.

share|cite|improve this answer

answered 7 hours ago

BernardBernard

127k743120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, you can estimate this by the maximum! very nice approach. I like this argument. This is probably how the question was meant to be done ^^
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Hint:

$4x-x^2$ is a quadratic polynomial , with a negative leading coefficient. Furthermore, it vanishes for $x=0$ and $x=4$. Can you find it (absolute) maximum?

Once you've found this maximum, you can use the mean value inequality for integrals.

share|cite|improve this answer

answered 7 hours ago

BernardBernard

127k743120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, you can estimate this by the maximum! very nice approach. I like this argument. This is probably how the question was meant to be done ^^
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

$4x-x^2$ is a quadratic polynomial , with a negative leading coefficient. Furthermore, it vanishes for $x=0$ and $x=4$. Can you find it (absolute) maximum?

Once you've found this maximum, you can use the mean value inequality for integrals.

share|cite|improve this answer

answered 7 hours ago

BernardBernard

127k743120

$endgroup$

share|cite|improve this answer

answered 7 hours ago

BernardBernard

127k743120

answered 7 hours ago

BernardBernard

127k743120

answered 7 hours ago

BernardBernard

127k743120

1

$begingroup$

You cannot just blindly apply the reverse power rule. Remember the chain rule? You have a function within another function.

Consider the case when n=2

$ int_0^4 (4x + x^2)^2 dx = int_0^4 16x^2 + 8x^3 + x^4 dx$

share|cite|improve this answer

answered 7 hours ago

agentnolaagentnola

389

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would it be better for him to use the binomial expansion, you think?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That would be an excellent approach. I was also thinking that you could attempt to find a pattern and then use induction to prove it.
  $endgroup$
  – agentnola
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I started working out the binomial theorem approach, you can switch the order of integration and summation and then integrate the power of $x$, this should not be too hard, although it will get a bit messy: $$int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now reduce everything to some power of $x$, move the integral inside and then we hope to get something pretty?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

You cannot just blindly apply the reverse power rule. Remember the chain rule? You have a function within another function.

Consider the case when n=2

$ int_0^4 (4x + x^2)^2 dx = int_0^4 16x^2 + 8x^3 + x^4 dx$

share|cite|improve this answer

answered 7 hours ago

agentnolaagentnola

389

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Would it be better for him to use the binomial expansion, you think?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That would be an excellent approach. I was also thinking that you could attempt to find a pattern and then use induction to prove it.
  $endgroup$
  – agentnola
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I started working out the binomial theorem approach, you can switch the order of integration and summation and then integrate the power of $x$, this should not be too hard, although it will get a bit messy: $$int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
  $endgroup$
  – Wesley Strik
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now reduce everything to some power of $x$, move the integral inside and then we hope to get something pretty?
  $endgroup$
  – Wesley Strik
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You cannot just blindly apply the reverse power rule. Remember the chain rule? You have a function within another function.

Consider the case when n=2

$ int_0^4 (4x + x^2)^2 dx = int_0^4 16x^2 + 8x^3 + x^4 dx$

share|cite|improve this answer

answered 7 hours ago

agentnolaagentnola

389

$endgroup$

share|cite|improve this answer

answered 7 hours ago

agentnolaagentnola

389

answered 7 hours ago

agentnolaagentnola

389

answered 7 hours ago

agentnolaagentnola

389

0

$begingroup$

Below is a suggestion for a very tedious route, but I wanted to post it anyway. This is probably not how the question is meant to be done.

Observe how by the binomial theorem we have that:
$$I=int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
We make this look a bit prettier:
$$I=int _{x=0}^4: sum_{k=0}^n binom{n}{k} 4^{k} cdot (-1)^{n-k}x^{2n-k},dx$$

We may switch the order of integration and summation since this concerns polynomials (what I mean is that they are of finite order), we compute:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{k} cdot (-1)^{n-k}4^{2n-k+1}-0$$
We collect some terms:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{2n+1} cdot (-1)^{n-k}$$
Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be:
$$ I =frac{sqrt {pi} cdot n! cdot 2^{2n+1}}{(n+frac{1}{2})!}=frac{sqrt pi cdot n!}{(n+frac{1}{2})!} cdot 4^{n + frac{1}{2}}$$
We then need to prove by induction that $frac{sqrt pi cdot n!}{(n+frac{1}{2})!} > sqrt{4}$ for positive $n$.

Yes. This is a lot of hard work. You could also just take the approach that was posted by Bernard and observe that $4x-x^2$ has a global maximum on the interval $[0,4]$ (the begin and endpoint are zeros of this polynomial), this maximum by symmetry occurs at $x=2$. The function has maximum value $8-4=4$. Observe then by the estimation lemma that $M =max_{x in [0,4]}((4x-x^2)^n)=4^n$, and the length of our path is $4$. We get:
$$ int _{x=0}^4 f_n dx leq M L = 4^n cdot 4 = 4^{n+1} $$

Also see: ML-inequality for real integrals

share|cite|improve this answer

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

$endgroup$

add a comment | 

0

$begingroup$

Below is a suggestion for a very tedious route, but I wanted to post it anyway. This is probably not how the question is meant to be done.

Observe how by the binomial theorem we have that:
$$I=int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
We make this look a bit prettier:
$$I=int _{x=0}^4: sum_{k=0}^n binom{n}{k} 4^{k} cdot (-1)^{n-k}x^{2n-k},dx$$

We may switch the order of integration and summation since this concerns polynomials (what I mean is that they are of finite order), we compute:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{k} cdot (-1)^{n-k}4^{2n-k+1}-0$$
We collect some terms:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{2n+1} cdot (-1)^{n-k}$$
Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be:
$$ I =frac{sqrt {pi} cdot n! cdot 2^{2n+1}}{(n+frac{1}{2})!}=frac{sqrt pi cdot n!}{(n+frac{1}{2})!} cdot 4^{n + frac{1}{2}}$$
We then need to prove by induction that $frac{sqrt pi cdot n!}{(n+frac{1}{2})!} > sqrt{4}$ for positive $n$.

Yes. This is a lot of hard work. You could also just take the approach that was posted by Bernard and observe that $4x-x^2$ has a global maximum on the interval $[0,4]$ (the begin and endpoint are zeros of this polynomial), this maximum by symmetry occurs at $x=2$. The function has maximum value $8-4=4$. Observe then by the estimation lemma that $M =max_{x in [0,4]}((4x-x^2)^n)=4^n$, and the length of our path is $4$. We get:
$$ int _{x=0}^4 f_n dx leq M L = 4^n cdot 4 = 4^{n+1} $$

Also see: ML-inequality for real integrals

share|cite|improve this answer

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

$endgroup$

add a comment | 

$begingroup$

Below is a suggestion for a very tedious route, but I wanted to post it anyway. This is probably not how the question is meant to be done.

Observe how by the binomial theorem we have that:
$$I=int _{x=0}^4:f_n(x),dx =int _{x=0}^4: (4x-x^2 )^n,dx =int _{x=0}^4: sum_{k=0}^n binom{n}{k} (4x)^{k} cdot (-x^2)^{n-k},dx$$
We make this look a bit prettier:
$$I=int _{x=0}^4: sum_{k=0}^n binom{n}{k} 4^{k} cdot (-1)^{n-k}x^{2n-k},dx$$

We may switch the order of integration and summation since this concerns polynomials (what I mean is that they are of finite order), we compute:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{k} cdot (-1)^{n-k}4^{2n-k+1}-0$$
We collect some terms:
$$I=sum_{k=0}^n binom{n}{k} frac{1}{2n-k+1} 4^{2n+1} cdot (-1)^{n-k}$$
Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be:
$$ I =frac{sqrt {pi} cdot n! cdot 2^{2n+1}}{(n+frac{1}{2})!}=frac{sqrt pi cdot n!}{(n+frac{1}{2})!} cdot 4^{n + frac{1}{2}}$$
We then need to prove by induction that $frac{sqrt pi cdot n!}{(n+frac{1}{2})!} > sqrt{4}$ for positive $n$.

Yes. This is a lot of hard work. You could also just take the approach that was posted by Bernard and observe that $4x-x^2$ has a global maximum on the interval $[0,4]$ (the begin and endpoint are zeros of this polynomial), this maximum by symmetry occurs at $x=2$. The function has maximum value $8-4=4$. Observe then by the estimation lemma that $M =max_{x in [0,4]}((4x-x^2)^n)=4^n$, and the length of our path is $4$. We get:
$$ int _{x=0}^4 f_n dx leq M L = 4^n cdot 4 = 4^{n+1} $$

Also see: ML-inequality for real integrals

share|cite|improve this answer

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

answered 7 hours ago

Wesley StrikWesley Strik

2,310424

answered 7 hours ago

Wesley StrikWesley Strik

2,310424