If f is even integrable function on [0,a] prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx...












0












$begingroup$


let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $



The answer is given below:



enter image description here



But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??






real-analysis calculus integration analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is confusing about the third line?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 10:28
















0












$begingroup$


let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $



The answer is given below:



enter image description here



But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??






real-analysis calculus integration analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is confusing about the third line?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 10:28














0












0








0





$begingroup$


let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $



The answer is given below:



enter image description here



But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??






real-analysis calculus integration analysis







share|cite|improve this question









$endgroup$




let f be an even function and suppose that f is integrable on [0,a].prove that f is integrable on [-a,a] and that $int_{-a}^{0} f(x) dx = int_{0}^{a} f(x) dx $



The answer is given below:



enter image description here



But I have a difficulty understanding the third line in the solution, does it contain a typo i.e. t must be -t in the second equality??






real-analysis calculus integration analysis




real-analysis calculus integration analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 10:24









hopefullyhopefully

187113




187113












  • $begingroup$
    What is confusing about the third line?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 10:28


















  • $begingroup$
    What is confusing about the third line?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 10:28
















$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28




$begingroup$
What is confusing about the third line?
$endgroup$
– Michael Burr
Dec 3 '18 at 10:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.



This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.




A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:



$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we have to multiply the inequality by -1 in the second part
    $endgroup$
    – hopefully
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:31






  • 1




    $begingroup$
    @hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:37






  • 1




    $begingroup$
    @hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:20






  • 1




    $begingroup$
    @hopefully That sounds about right, yes
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:45











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%2f3023882%2fif-f-is-even-integrable-function-on-0-a-prove-that-f-is-integrable-on-a-a-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.



This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.




A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:



$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we have to multiply the inequality by -1 in the second part
    $endgroup$
    – hopefully
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:31






  • 1




    $begingroup$
    @hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:37






  • 1




    $begingroup$
    @hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:20






  • 1




    $begingroup$
    @hopefully That sounds about right, yes
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:45
















1












$begingroup$

No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.



This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.




A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:



$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    but we have to multiply the inequality by -1 in the second part
    $endgroup$
    – hopefully
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:31






  • 1




    $begingroup$
    @hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:37






  • 1




    $begingroup$
    @hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:20






  • 1




    $begingroup$
    @hopefully That sounds about right, yes
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:45














1












1








1





$begingroup$

No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.



This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.




A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:



$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.






share|cite|improve this answer











$endgroup$



No, the second inequality contains a $t$. This is because $f$ is even, therefore, $f(-t)=f(t)$.



This means that the values $f$ takes on the interval $[x_{k-1}, x_k]$ are the same that $f$ takes on the interval $[-x_k, -x_{k-1}]$.




A strict proof of the claim would indeed require multiplying the inequalities, in particular,a more rigorous equality would be written like so:



$$begin{align}sup{f(t)| -x_k leq t leq -x_{k-1}} &= sup{f(-t)|-x_kleq t leq -x_{k-1}}\& = sup{f(-t)|x_{k-1}leq -t leq x_k}\&=sup{f(tau)|x_{k-1}leq tauleq x_k}end{align}$$
In line one, I use only the fact that $f$ is odd. In line $2$, I just multiply the inequality by $-1$, and in line $3$, I introduce a new variable $tau=-t$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 10:32

























answered Dec 3 '18 at 10:29









5xum5xum

90.2k393161




90.2k393161












  • $begingroup$
    but we have to multiply the inequality by -1 in the second part
    $endgroup$
    – hopefully
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:31






  • 1




    $begingroup$
    @hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:37






  • 1




    $begingroup$
    @hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:20






  • 1




    $begingroup$
    @hopefully That sounds about right, yes
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:45


















  • $begingroup$
    but we have to multiply the inequality by -1 in the second part
    $endgroup$
    – hopefully
    Dec 3 '18 at 10:30






  • 1




    $begingroup$
    @hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:31






  • 1




    $begingroup$
    @hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
    $endgroup$
    – 5xum
    Dec 3 '18 at 10:37






  • 1




    $begingroup$
    @hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:20






  • 1




    $begingroup$
    @hopefully That sounds about right, yes
    $endgroup$
    – 5xum
    Dec 3 '18 at 11:45
















$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30




$begingroup$
but we have to multiply the inequality by -1 in the second part
$endgroup$
– hopefully
Dec 3 '18 at 10:30




1




1




$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31




$begingroup$
@hopefully You don't multiply the inequality. You use the fact that for every $tin[-x_k, -x_{k-1}]$, there exists some $tauin[x_{k-1}, x_k]$ such that $f(t)=f(tau)$.
$endgroup$
– 5xum
Dec 3 '18 at 10:31




1




1




$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37




$begingroup$
@hopefully Technically, you do multiply the inequality, but you also introduce $-t$ into the argument of $f$. I updated my answer to show exactly what happens.
$endgroup$
– 5xum
Dec 3 '18 at 10:37




1




1




$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20




$begingroup$
@hopefully Well you won't have $f(t)=f(-t)$ any more. To see what exactly the difference will be, I advise you try to do it, and see for yourself!
$endgroup$
– 5xum
Dec 3 '18 at 11:20




1




1




$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45




$begingroup$
@hopefully That sounds about right, yes
$endgroup$
– 5xum
Dec 3 '18 at 11:45


















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%2f3023882%2fif-f-is-even-integrable-function-on-0-a-prove-that-f-is-integrable-on-a-a-a%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







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more