Finding the function from a definite integral?












0












$begingroup$


If I had an integral, as shown below,
$$dfrac{int_{0}^{1} a(x) cdot x^{2} ; dx}{int_{0}^{1} a(x); dx}=dfrac{1}{2}$$
How would I figure out the function $a(x)$, is there way or do I have to make a guess and test it?






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
    $endgroup$
    – Jack D'Aurizio
    Jan 16 at 12:39






  • 1




    $begingroup$
    Every non-zero constant function will do, also $a(x) = x$ from polynomials
    $endgroup$
    – Jakobian
    Jan 16 at 12:49






  • 1




    $begingroup$
    @Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
    $endgroup$
    – W. Zhu
    Jan 16 at 12:55
















0












$begingroup$


If I had an integral, as shown below,
$$dfrac{int_{0}^{1} a(x) cdot x^{2} ; dx}{int_{0}^{1} a(x); dx}=dfrac{1}{2}$$
How would I figure out the function $a(x)$, is there way or do I have to make a guess and test it?






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
    $endgroup$
    – Jack D'Aurizio
    Jan 16 at 12:39






  • 1




    $begingroup$
    Every non-zero constant function will do, also $a(x) = x$ from polynomials
    $endgroup$
    – Jakobian
    Jan 16 at 12:49






  • 1




    $begingroup$
    @Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
    $endgroup$
    – W. Zhu
    Jan 16 at 12:55














0












0








0


0



$begingroup$


If I had an integral, as shown below,
$$dfrac{int_{0}^{1} a(x) cdot x^{2} ; dx}{int_{0}^{1} a(x); dx}=dfrac{1}{2}$$
How would I figure out the function $a(x)$, is there way or do I have to make a guess and test it?






calculus







share|cite|improve this question









$endgroup$




If I had an integral, as shown below,
$$dfrac{int_{0}^{1} a(x) cdot x^{2} ; dx}{int_{0}^{1} a(x); dx}=dfrac{1}{2}$$
How would I figure out the function $a(x)$, is there way or do I have to make a guess and test it?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 12:37









D. QaD. Qa

1656




1656








  • 1




    $begingroup$
    The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
    $endgroup$
    – Jack D'Aurizio
    Jan 16 at 12:39






  • 1




    $begingroup$
    Every non-zero constant function will do, also $a(x) = x$ from polynomials
    $endgroup$
    – Jakobian
    Jan 16 at 12:49






  • 1




    $begingroup$
    @Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
    $endgroup$
    – W. Zhu
    Jan 16 at 12:55














  • 1




    $begingroup$
    The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
    $endgroup$
    – Jack D'Aurizio
    Jan 16 at 12:39






  • 1




    $begingroup$
    Every non-zero constant function will do, also $a(x) = x$ from polynomials
    $endgroup$
    – Jakobian
    Jan 16 at 12:49






  • 1




    $begingroup$
    @Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
    $endgroup$
    – W. Zhu
    Jan 16 at 12:55








1




1




$begingroup$
The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
$endgroup$
– Jack D'Aurizio
Jan 16 at 12:39




$begingroup$
The function is simply not fixed. This is more or less like asking given $f(1)=2019$, what is $f(x)$?
$endgroup$
– Jack D'Aurizio
Jan 16 at 12:39




1




1




$begingroup$
Every non-zero constant function will do, also $a(x) = x$ from polynomials
$endgroup$
– Jakobian
Jan 16 at 12:49




$begingroup$
Every non-zero constant function will do, also $a(x) = x$ from polynomials
$endgroup$
– Jakobian
Jan 16 at 12:49




1




1




$begingroup$
@Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
$endgroup$
– W. Zhu
Jan 16 at 12:55




$begingroup$
@Jakobian $a(x)=c$ will give $frac{frac c3}{c}=frac13$.
$endgroup$
– W. Zhu
Jan 16 at 12:55










2 Answers
2






active

oldest

votes


















1












$begingroup$

There are too many choices. At some point you're going to have to make a choice arbitrarily, or use information from the deeper context of the question (things you haven't told us about) to choose.



We can see $a(x) = 1$ (or any constant) won't work. The average value of $x^2$ for $x in [0,1]$ is only $frac13.$ We need to "weight the average" to count the larger values of $x^2$ more heavily so that the "average" comes out to $frac12$ instead.
The function $a(x)$ is the "weight".



One possible approach is to assume $a(x) = x^k$ for some real number $k$. Work out the ratio of the integrals, which will be some expression in $k$.
You want that expression to equal $frac12.$ Assume it does, and solve for $k.$



There are many other possibilities, for example if you don't mind an integrand with finitely many jump discontinuities then $a(x) = lfloor x + krfloor$ could do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
    $endgroup$
    – D. Qa
    Jan 16 at 13:26






  • 1




    $begingroup$
    The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
    $endgroup$
    – David K
    Jan 16 at 13:33












  • $begingroup$
    a(x) can be x but as pointed out, there are lots of other possibilities too.
    $endgroup$
    – Paul
    Jan 16 at 14:03



















0












$begingroup$

One possibility is the function $a(x)=0$ for all $x$ but there might be others.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But then we divide by $0$
    $endgroup$
    – Jakobian
    Jan 16 at 12:43










  • $begingroup$
    You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
    $endgroup$
    – James
    Jan 16 at 12:44












  • $begingroup$
    $frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
    $endgroup$
    – W. Zhu
    Jan 16 at 13:03










  • $begingroup$
    I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
    $endgroup$
    – James
    Jan 16 at 13:06












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

There are too many choices. At some point you're going to have to make a choice arbitrarily, or use information from the deeper context of the question (things you haven't told us about) to choose.



We can see $a(x) = 1$ (or any constant) won't work. The average value of $x^2$ for $x in [0,1]$ is only $frac13.$ We need to "weight the average" to count the larger values of $x^2$ more heavily so that the "average" comes out to $frac12$ instead.
The function $a(x)$ is the "weight".



One possible approach is to assume $a(x) = x^k$ for some real number $k$. Work out the ratio of the integrals, which will be some expression in $k$.
You want that expression to equal $frac12.$ Assume it does, and solve for $k.$



There are many other possibilities, for example if you don't mind an integrand with finitely many jump discontinuities then $a(x) = lfloor x + krfloor$ could do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
    $endgroup$
    – D. Qa
    Jan 16 at 13:26






  • 1




    $begingroup$
    The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
    $endgroup$
    – David K
    Jan 16 at 13:33












  • $begingroup$
    a(x) can be x but as pointed out, there are lots of other possibilities too.
    $endgroup$
    – Paul
    Jan 16 at 14:03
















1












$begingroup$

There are too many choices. At some point you're going to have to make a choice arbitrarily, or use information from the deeper context of the question (things you haven't told us about) to choose.



We can see $a(x) = 1$ (or any constant) won't work. The average value of $x^2$ for $x in [0,1]$ is only $frac13.$ We need to "weight the average" to count the larger values of $x^2$ more heavily so that the "average" comes out to $frac12$ instead.
The function $a(x)$ is the "weight".



One possible approach is to assume $a(x) = x^k$ for some real number $k$. Work out the ratio of the integrals, which will be some expression in $k$.
You want that expression to equal $frac12.$ Assume it does, and solve for $k.$



There are many other possibilities, for example if you don't mind an integrand with finitely many jump discontinuities then $a(x) = lfloor x + krfloor$ could do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
    $endgroup$
    – D. Qa
    Jan 16 at 13:26






  • 1




    $begingroup$
    The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
    $endgroup$
    – David K
    Jan 16 at 13:33












  • $begingroup$
    a(x) can be x but as pointed out, there are lots of other possibilities too.
    $endgroup$
    – Paul
    Jan 16 at 14:03














1












1








1





$begingroup$

There are too many choices. At some point you're going to have to make a choice arbitrarily, or use information from the deeper context of the question (things you haven't told us about) to choose.



We can see $a(x) = 1$ (or any constant) won't work. The average value of $x^2$ for $x in [0,1]$ is only $frac13.$ We need to "weight the average" to count the larger values of $x^2$ more heavily so that the "average" comes out to $frac12$ instead.
The function $a(x)$ is the "weight".



One possible approach is to assume $a(x) = x^k$ for some real number $k$. Work out the ratio of the integrals, which will be some expression in $k$.
You want that expression to equal $frac12.$ Assume it does, and solve for $k.$



There are many other possibilities, for example if you don't mind an integrand with finitely many jump discontinuities then $a(x) = lfloor x + krfloor$ could do the job.






share|cite|improve this answer









$endgroup$



There are too many choices. At some point you're going to have to make a choice arbitrarily, or use information from the deeper context of the question (things you haven't told us about) to choose.



We can see $a(x) = 1$ (or any constant) won't work. The average value of $x^2$ for $x in [0,1]$ is only $frac13.$ We need to "weight the average" to count the larger values of $x^2$ more heavily so that the "average" comes out to $frac12$ instead.
The function $a(x)$ is the "weight".



One possible approach is to assume $a(x) = x^k$ for some real number $k$. Work out the ratio of the integrals, which will be some expression in $k$.
You want that expression to equal $frac12.$ Assume it does, and solve for $k.$



There are many other possibilities, for example if you don't mind an integrand with finitely many jump discontinuities then $a(x) = lfloor x + krfloor$ could do the job.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 12:58









David KDavid K

55.5k345121




55.5k345121












  • $begingroup$
    I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
    $endgroup$
    – D. Qa
    Jan 16 at 13:26






  • 1




    $begingroup$
    The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
    $endgroup$
    – David K
    Jan 16 at 13:33












  • $begingroup$
    a(x) can be x but as pointed out, there are lots of other possibilities too.
    $endgroup$
    – Paul
    Jan 16 at 14:03


















  • $begingroup$
    I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
    $endgroup$
    – D. Qa
    Jan 16 at 13:26






  • 1




    $begingroup$
    The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
    $endgroup$
    – David K
    Jan 16 at 13:33












  • $begingroup$
    a(x) can be x but as pointed out, there are lots of other possibilities too.
    $endgroup$
    – Paul
    Jan 16 at 14:03
















$begingroup$
I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
$endgroup$
– D. Qa
Jan 16 at 13:26




$begingroup$
I did assume $x^{k}$ and after solving, managed to get a value for $k$, which is $k=1$, therefore $a(x)=x$. That was very helpful, thank you so much!
$endgroup$
– D. Qa
Jan 16 at 13:26




1




1




$begingroup$
The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
$endgroup$
– David K
Jan 16 at 13:33






$begingroup$
The solution $a(x) = x$ was also given away in a comment under the question. But it seems more satisfying to derive it from a method that doesn't assume you have guessed the exact answer immediately.
$endgroup$
– David K
Jan 16 at 13:33














$begingroup$
a(x) can be x but as pointed out, there are lots of other possibilities too.
$endgroup$
– Paul
Jan 16 at 14:03




$begingroup$
a(x) can be x but as pointed out, there are lots of other possibilities too.
$endgroup$
– Paul
Jan 16 at 14:03











0












$begingroup$

One possibility is the function $a(x)=0$ for all $x$ but there might be others.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But then we divide by $0$
    $endgroup$
    – Jakobian
    Jan 16 at 12:43










  • $begingroup$
    You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
    $endgroup$
    – James
    Jan 16 at 12:44












  • $begingroup$
    $frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
    $endgroup$
    – W. Zhu
    Jan 16 at 13:03










  • $begingroup$
    I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
    $endgroup$
    – James
    Jan 16 at 13:06
















0












$begingroup$

One possibility is the function $a(x)=0$ for all $x$ but there might be others.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But then we divide by $0$
    $endgroup$
    – Jakobian
    Jan 16 at 12:43










  • $begingroup$
    You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
    $endgroup$
    – James
    Jan 16 at 12:44












  • $begingroup$
    $frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
    $endgroup$
    – W. Zhu
    Jan 16 at 13:03










  • $begingroup$
    I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
    $endgroup$
    – James
    Jan 16 at 13:06














0












0








0





$begingroup$

One possibility is the function $a(x)=0$ for all $x$ but there might be others.






share|cite|improve this answer









$endgroup$



One possibility is the function $a(x)=0$ for all $x$ but there might be others.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 12:43









JamesJames

2,112422




2,112422












  • $begingroup$
    But then we divide by $0$
    $endgroup$
    – Jakobian
    Jan 16 at 12:43










  • $begingroup$
    You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
    $endgroup$
    – James
    Jan 16 at 12:44












  • $begingroup$
    $frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
    $endgroup$
    – W. Zhu
    Jan 16 at 13:03










  • $begingroup$
    I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
    $endgroup$
    – James
    Jan 16 at 13:06


















  • $begingroup$
    But then we divide by $0$
    $endgroup$
    – Jakobian
    Jan 16 at 12:43










  • $begingroup$
    You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
    $endgroup$
    – James
    Jan 16 at 12:44












  • $begingroup$
    $frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
    $endgroup$
    – W. Zhu
    Jan 16 at 13:03










  • $begingroup$
    I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
    $endgroup$
    – James
    Jan 16 at 13:06
















$begingroup$
But then we divide by $0$
$endgroup$
– Jakobian
Jan 16 at 12:43




$begingroup$
But then we divide by $0$
$endgroup$
– Jakobian
Jan 16 at 12:43












$begingroup$
You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
$endgroup$
– James
Jan 16 at 12:44






$begingroup$
You can rewrite the formula by bringing the denominator on the right site. Then there are no issues with dividing by 0 and the solution is the same.
$endgroup$
– James
Jan 16 at 12:44














$begingroup$
$frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
$endgroup$
– W. Zhu
Jan 16 at 13:03




$begingroup$
$frac00$ is indeterminate. You can make $frac00=1$ and $frac00=2$ by "moving the denominator to the other side", but $1ne2$.
$endgroup$
– W. Zhu
Jan 16 at 13:03












$begingroup$
I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
$endgroup$
– James
Jan 16 at 13:06




$begingroup$
I know. But the equation in question is $int dots = frac{1}{2}int dots$ which can be solved without any problems.
$endgroup$
– James
Jan 16 at 13:06


















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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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File:DeusFollowingSea.jpg

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File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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