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

249114




249114












  • $begingroup$
    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











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1 Answer
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active

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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

91k394161




91k394161












  • $begingroup$
    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


















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