Prove that if $g(x)leq g(y)$ then $F(x)leq F(y),$ where $F(x)=int^{g(x)}_{a}f(t)dt$












1












$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24


















1












$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24
















1












1








1


1



$begingroup$


Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification







share|cite|improve this question











$endgroup$




Can you help me check if this is correct?




Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$
$$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$




PROOF 1



Since $g:[a,b]to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)geq 0,;forall;xin [a,b]$
begin{align} F'(x)&=dfrac{d}{dx}int^{g(x)}_{a}f(t)dt \&=fleft(g(x)right)g'(x) geq 0,;forall;xin [a,b]end{align}
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$



PROOF 2



Let $x,yin[a,b]$ be fixed such that $xleq y$. Since $g$ is nondecreasing, then $g(x)leq g(y).$ Thus, $[a,g(x)]subseteq [a,g(y)]$ and
$$F(x)=int^{g(x)}_{a}f(t)dtleq int^{g(y)}_{a}f(t)dt =F(y)$$
Hence, $$F(x)=int^{g(x)}_{a}f(t)dt$$
is nondecreasing on $[a,b].$






real-analysis calculus analysis proof-verification




real-analysis calculus analysis proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 3:27







Omojola Micheal

















asked Jan 14 at 21:04









Omojola MichealOmojola Micheal

1,999424




1,999424








  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24
















  • 2




    $begingroup$
    Looks OK, but the best you can conclude is that $F$ is nondecreasing.
    $endgroup$
    – Umberto P.
    Jan 14 at 21:09










  • $begingroup$
    Thanks for your feedback! I'll edit!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:10








  • 1




    $begingroup$
    No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
    $endgroup$
    – Arthur
    Jan 14 at 21:14












  • $begingroup$
    @Arthur: Hmm... I see you what you've stated!
    $endgroup$
    – Omojola Micheal
    Jan 14 at 21:18






  • 1




    $begingroup$
    @UmbertoP. I think the constant function is monotonically increasing as well.
    $endgroup$
    – Botond
    Jan 14 at 21:24










2




2




$begingroup$
Looks OK, but the best you can conclude is that $F$ is nondecreasing.
$endgroup$
– Umberto P.
Jan 14 at 21:09




$begingroup$
Looks OK, but the best you can conclude is that $F$ is nondecreasing.
$endgroup$
– Umberto P.
Jan 14 at 21:09












$begingroup$
Thanks for your feedback! I'll edit!
$endgroup$
– Omojola Micheal
Jan 14 at 21:10






$begingroup$
Thanks for your feedback! I'll edit!
$endgroup$
– Omojola Micheal
Jan 14 at 21:10






1




1




$begingroup$
No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
$endgroup$
– Arthur
Jan 14 at 21:14






$begingroup$
No matter how many times I hear it, nondecreasing will sound to me like "not monotonically decreasing". Intellectually, I know that's not what it means. But intuitively that's what I think when I see the word. "Increasing" is a better term in my opinion.
$endgroup$
– Arthur
Jan 14 at 21:14














$begingroup$
@Arthur: Hmm... I see you what you've stated!
$endgroup$
– Omojola Micheal
Jan 14 at 21:18




$begingroup$
@Arthur: Hmm... I see you what you've stated!
$endgroup$
– Omojola Micheal
Jan 14 at 21:18




1




1




$begingroup$
@UmbertoP. I think the constant function is monotonically increasing as well.
$endgroup$
– Botond
Jan 14 at 21:24






$begingroup$
@UmbertoP. I think the constant function is monotonically increasing as well.
$endgroup$
– Botond
Jan 14 at 21:24












1 Answer
1






active

oldest

votes


















2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07











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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07
















2












$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07














2












2








2





$begingroup$

This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}






share|cite|improve this answer











$endgroup$



This is correct; A little more general:




Claim. Let $[a,b] subsetmathbb{R}$ be a interval. Let $f: [a,b]rightarrow[0,infty[$ be a Riemann-integrable function such that $xle y implies f(x) le f(y) (text{for all } x,yin[a,b])$. Let $g: [a,b]rightarrow[a,b]$ be an arbitrary function that satisfies $xle y implies g(x) le g(y) (text{for all } x,yin[a,b]).\text{Then } displaystyle F(x) := int_a^{g(x)} f(t) mathrm{d}t text{ (for } xin[a,b]text{) is "nondecreasing".}$




Proof. Let $x,yin[a,b]$ with $xle y$. Then (by properties of the Riemann-Integral),
begin{equation*}
F(y) := int_a^{g(y)} f(t) mathrm{d}t = int_a^{g(x)} f(t) mathrm{d}t + int_{g(x)}^{g(y)} f(t) mathrm{d}t underbracegeq_{text{see (*)}} int_a^{g(x)} f(t) mathrm{d}t = F(x). quadsquare
end{equation*}



ADDENDUM
begin{align}tag{*}
& text{Note that } int_{g(x)}^{g(y)} f(t) mathrm{d}t
overbracegeq^{text{property of Riemann-Integral}} (g(y)-g(x)) cdotinf_{xin[a,b]} f(x)
geq 0,\
& text{since f }geq0text{ on }[a,b]text{ and }g(y)geq g(x)text{ i.e. } g(y)-g(x)geq0.
end{align}







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share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 21:59

























answered Jan 14 at 21:47









Maximilian JanischMaximilian Janisch

45612




45612












  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07


















  • $begingroup$
    Maximilian Janisch: This is nice! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 14 at 22:07
















$begingroup$
Maximilian Janisch: This is nice! (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 22:07




$begingroup$
Maximilian Janisch: This is nice! (+1)
$endgroup$
– Omojola Micheal
Jan 14 at 22:07


















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

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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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