Finding the function from a definite integral?

Multi tool use
Multi tool use












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












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075680%2ffinding-the-function-from-a-definite-integral%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075680%2ffinding-the-function-from-a-definite-integral%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







F4jCIrzsxyveOi0URKsuwzBY,2UqADQVAcFroDTCK219v7Ol nEZMkLLk,WYk8yBBJQdBX,DM8f
-
905s 7fiy,VMiLZF7eq6y3PTSVfJWavqF25fXvv mbh KfPkPPseC7P

Popular posts from this blog

Human spaceflight

Image
Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
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] ...
Read more

Antiochus X Eusebes

Image
Antiochus X Eusebes From Wikipedia, the free encyclopedia Jump to navigation Jump to search King of Syria Antiochus X Eusebes Antiochus X's portrait on the obverse of a tetradrachm King of Syria (Seleucid Empire) Reign 95–92 or 88 BC Predecessor Seleucus VI, Demetrius III Successor Demetrius III, Philip I Born c.  113 BC Died 92 or 88 BC Spouse Cleopatra Selene Issue Detail Antiochus XIII Dynasty Seleucid Father Antiochus IX Mother Cleopatra IV ? Antiochus X Eusebes Philopator (Greek: Ἀντίοχος Εὐσεβής Φιλοπάτωρ ; c.  113 –92 or 88 BC) was a Hellenistic Seleucid monarch who reigned as the King of Syria between 95 and 92 or 89/88 BC (224 Seleucid year (SE)). [note 1] He was the son of Antiochus IX and perhaps his Egyptian wife Cleopatra IV. He lived at a period that witnessed a general disintegration of Seleucid Syria characterized by civil wars, foreign inter...
Read more