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If $int_2^x^2xcdot f(s) ds=2x^5-x^4+2$, then find $f(9)$

Integral $ int_-pi/2^pi/2 frac12007^x+1cdot fracsin^2008xsin^2008x+cos^2008xdx $Calculus find extreme values of integraluse fundamental theorem of calculus to find a function $f(x)$ and a number $a$Evaluate $ int_0^pi/4left(cos 2x right)^11/2cdot cos x;dx $Is integrating both sides of an equation useful?Initial value problem without explicit constant findingDefinite integral of exponential of nested functionFind a function using definite integral from the second fundamental theorem of calculusKinematics equations with calculus what does dv mean?Method to solve this integral with algebraic and trigonometric terms

0

1

$begingroup$

$$int_2^x^2xcdot f(s)ds=2x^5-x^4+2quad(x>0)Rightarrow f(9)=?$$

I would have taken the derivative of both sides of the equation if the integral was not definite. In this case, I don't have a valid solution.

How can we solve this problem?

calculus definite-integrals

share|cite|improve this question

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

$endgroup$

• 
  
  
  

  
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  $begingroup$
  Where did you come across this problem? That integral notation is... not good.
  $endgroup$
  – Arthur
  Mar 12 at 9:00
  

  
  
  
  


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  Do you mean $$f(x) int_2^x^2 x mathrm dx$$ on the LHS?
  $endgroup$
  – George Coote
  Mar 12 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  Sorry, I'm not good at MathJax. I edited my question
  $endgroup$
  – Eldar Rahimli
  Mar 12 at 9:03
  
  

  
  
  
  


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  'strictly they should be using a different variable' Exactly. So I would say $int_2^x^2tf(t)dt = 2x^5-x^4+2$. The phrase 'being "different" $x$s' should never be a thing one has to specify.
  $endgroup$
  – Arthur
  Mar 12 at 9:09
  
  

  
  
  
  


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  @GeorgeCoote I think that is highly unusual and, most probably, a mistake. The integration variable must be different from the one(s) on the integral's limits, otherwise the problem could be considered ill posed, imo.
  $endgroup$
  – DonAntonio
  Mar 12 at 9:11
  

  
  
  
  

 | 
show 8 more comments

0

1

$begingroup$

$$int_2^x^2xcdot f(s)ds=2x^5-x^4+2quad(x>0)Rightarrow f(9)=?$$

I would have taken the derivative of both sides of the equation if the integral was not definite. In this case, I don't have a valid solution.

How can we solve this problem?

calculus definite-integrals

share|cite|improve this question

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Where did you come across this problem? That integral notation is... not good.
  $endgroup$
  – Arthur
  Mar 12 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $$f(x) int_2^x^2 x mathrm dx$$ on the LHS?
  $endgroup$
  – George Coote
  Mar 12 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I'm not good at MathJax. I edited my question
  $endgroup$
  – Eldar Rahimli
  Mar 12 at 9:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  'strictly they should be using a different variable' Exactly. So I would say $int_2^x^2tf(t)dt = 2x^5-x^4+2$. The phrase 'being "different" $x$s' should never be a thing one has to specify.
  $endgroup$
  – Arthur
  Mar 12 at 9:09
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeCoote I think that is highly unusual and, most probably, a mistake. The integration variable must be different from the one(s) on the integral's limits, otherwise the problem could be considered ill posed, imo.
  $endgroup$
  – DonAntonio
  Mar 12 at 9:11
  

  
  
  
  

 | 
show 8 more comments

$begingroup$

$$int_2^x^2xcdot f(s)ds=2x^5-x^4+2quad(x>0)Rightarrow f(9)=?$$

I would have taken the derivative of both sides of the equation if the integral was not definite. In this case, I don't have a valid solution.

How can we solve this problem?

calculus definite-integrals

share|cite|improve this question

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

$endgroup$

share|cite|improve this question

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

share|cite|improve this question

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

edited Mar 12 at 10:41

Saad

20k92352

edited Mar 12 at 10:41

Saad

20k92352

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

asked Mar 12 at 8:58

Eldar RahimliEldar Rahimli

36310

3 Answers
3

active

oldest

votes

5

$begingroup$

This solution was posted before the question was edited.

Use Chain Rule. The derivative of LHS is $x^2f(x^2) (2x)$. Put $x=3$ after differentiating.

For the revised version first divide by $x$, then apply Chain Rule and put $x=3$.

share|cite|improve this answer

edited Mar 12 at 10:00

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

add a comment | 

2

$begingroup$

Realize that the $x$ in $dx$ (and in $xf(x)$) is a dummy variable. It means that we can replace it with anything without anything changing - so LHS reads:

$$textLHS(x) = intlimits_2^x^2 t f(t) dt$$

It should be now clear - if you take the derivative with respect to $x$ you need only to take care in the $x$ dependence in the upper limit. If you don't know how to deal with this kind of derivatives then first set $x^2 = y$, take derivative with respect to $y$ and figure out how derivative w.r.t. $y$ relates to one w.r.t. $x$.

share|cite|improve this answer

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

$endgroup$

add a comment | 

2

$begingroup$

There does not exist such $f$ since otherwise$$
0 = int_2^(sqrt2)^2 y f(y) ,mathrmdy = 2(sqrt2)^5 - (sqrt2)^4 + 2 = 8sqrt2 - 2 ≠ 0,
$$
a contradiction.

share|cite|improve this answer

answered Mar 12 at 9:18

SaadSaad

20k92352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur What? As given, the question implies that the equality is true for any $;x;$ ...so this answer is correct. Now, if there are some constraints then they must be explicitly stated.
  $endgroup$
  – DonAntonio
  Mar 12 at 9:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur As clarified in the comments below the question, the unambiguous notation is $displaystyleint_2^x^2t·f(t),mathrm dt=2x^5-x^4+2$. And even if it were to mean $displaystyleint_2^x^2x·f(t),mathrm dt=2x^5-x^4+2$, the deduction above still leads to a contradiction.
  $endgroup$
  – Saad
  Mar 12 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can agree with those interpretations but not as it is stated.
  $endgroup$
  – Maria Mazur
  Mar 12 at 9:45
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

This solution was posted before the question was edited.

Use Chain Rule. The derivative of LHS is $x^2f(x^2) (2x)$. Put $x=3$ after differentiating.

For the revised version first divide by $x$, then apply Chain Rule and put $x=3$.

share|cite|improve this answer

edited Mar 12 at 10:00

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

add a comment | 

5

$begingroup$

This solution was posted before the question was edited.

Use Chain Rule. The derivative of LHS is $x^2f(x^2) (2x)$. Put $x=3$ after differentiating.

For the revised version first divide by $x$, then apply Chain Rule and put $x=3$.

share|cite|improve this answer

edited Mar 12 at 10:00

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

add a comment | 

$begingroup$

This solution was posted before the question was edited.

Use Chain Rule. The derivative of LHS is $x^2f(x^2) (2x)$. Put $x=3$ after differentiating.

For the revised version first divide by $x$, then apply Chain Rule and put $x=3$.

share|cite|improve this answer

edited Mar 12 at 10:00

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

$endgroup$

share|cite|improve this answer

edited Mar 12 at 10:00

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

answered Mar 12 at 9:00

Kavi Rama MurthyKavi Rama Murthy

67.6k53067

2

$begingroup$

Realize that the $x$ in $dx$ (and in $xf(x)$) is a dummy variable. It means that we can replace it with anything without anything changing - so LHS reads:

$$textLHS(x) = intlimits_2^x^2 t f(t) dt$$

It should be now clear - if you take the derivative with respect to $x$ you need only to take care in the $x$ dependence in the upper limit. If you don't know how to deal with this kind of derivatives then first set $x^2 = y$, take derivative with respect to $y$ and figure out how derivative w.r.t. $y$ relates to one w.r.t. $x$.

share|cite|improve this answer

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

$endgroup$

add a comment | 

2

$begingroup$

Realize that the $x$ in $dx$ (and in $xf(x)$) is a dummy variable. It means that we can replace it with anything without anything changing - so LHS reads:

$$textLHS(x) = intlimits_2^x^2 t f(t) dt$$

It should be now clear - if you take the derivative with respect to $x$ you need only to take care in the $x$ dependence in the upper limit. If you don't know how to deal with this kind of derivatives then first set $x^2 = y$, take derivative with respect to $y$ and figure out how derivative w.r.t. $y$ relates to one w.r.t. $x$.

share|cite|improve this answer

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

$endgroup$

add a comment | 

$begingroup$

Realize that the $x$ in $dx$ (and in $xf(x)$) is a dummy variable. It means that we can replace it with anything without anything changing - so LHS reads:

$$textLHS(x) = intlimits_2^x^2 t f(t) dt$$

It should be now clear - if you take the derivative with respect to $x$ you need only to take care in the $x$ dependence in the upper limit. If you don't know how to deal with this kind of derivatives then first set $x^2 = y$, take derivative with respect to $y$ and figure out how derivative w.r.t. $y$ relates to one w.r.t. $x$.

share|cite|improve this answer

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

answered Mar 12 at 9:10

Piotr BenedysiukPiotr Benedysiuk

1,344519

2

$begingroup$

There does not exist such $f$ since otherwise$$
0 = int_2^(sqrt2)^2 y f(y) ,mathrmdy = 2(sqrt2)^5 - (sqrt2)^4 + 2 = 8sqrt2 - 2 ≠ 0,
$$
a contradiction.

share|cite|improve this answer

answered Mar 12 at 9:18

SaadSaad

20k92352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur What? As given, the question implies that the equality is true for any $;x;$ ...so this answer is correct. Now, if there are some constraints then they must be explicitly stated.
  $endgroup$
  – DonAntonio
  Mar 12 at 9:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur As clarified in the comments below the question, the unambiguous notation is $displaystyleint_2^x^2t·f(t),mathrm dt=2x^5-x^4+2$. And even if it were to mean $displaystyleint_2^x^2x·f(t),mathrm dt=2x^5-x^4+2$, the deduction above still leads to a contradiction.
  $endgroup$
  – Saad
  Mar 12 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can agree with those interpretations but not as it is stated.
  $endgroup$
  – Maria Mazur
  Mar 12 at 9:45
  
  

  
  
  
  

add a comment | 

2

$begingroup$

There does not exist such $f$ since otherwise$$
0 = int_2^(sqrt2)^2 y f(y) ,mathrmdy = 2(sqrt2)^5 - (sqrt2)^4 + 2 = 8sqrt2 - 2 ≠ 0,
$$
a contradiction.

share|cite|improve this answer

answered Mar 12 at 9:18

SaadSaad

20k92352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur What? As given, the question implies that the equality is true for any $;x;$ ...so this answer is correct. Now, if there are some constraints then they must be explicitly stated.
  $endgroup$
  – DonAntonio
  Mar 12 at 9:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MariaMazur As clarified in the comments below the question, the unambiguous notation is $displaystyleint_2^x^2t·f(t),mathrm dt=2x^5-x^4+2$. And even if it were to mean $displaystyleint_2^x^2x·f(t),mathrm dt=2x^5-x^4+2$, the deduction above still leads to a contradiction.
  $endgroup$
  – Saad
  Mar 12 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can agree with those interpretations but not as it is stated.
  $endgroup$
  – Maria Mazur
  Mar 12 at 9:45
  
  

  
  
  
  

add a comment | 

$begingroup$

There does not exist such $f$ since otherwise$$
0 = int_2^(sqrt2)^2 y f(y) ,mathrmdy = 2(sqrt2)^5 - (sqrt2)^4 + 2 = 8sqrt2 - 2 ≠ 0,
$$
a contradiction.

share|cite|improve this answer

answered Mar 12 at 9:18

SaadSaad

20k92352

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:18

SaadSaad

20k92352

answered Mar 12 at 9:18

SaadSaad

20k92352

answered Mar 12 at 9:18

SaadSaad

20k92352