Substitution of x in Integral












0












$begingroup$


Hi I was wondering how to do the following:



I have this integral:
$$int _ { - L / 2 } ^ { + L / 2 } w' ( x ), d x$$



I know would like to normalize x with respect to L:
$$hat { x } = frac { x } { L }$$



How do I substitute correctly ?



I know that:
$$w' ( x ) = frac { 1 } { L } w' (hat { x })$$



but what about the $dx$?



Thanks!



EDIT:
From the chain rule:
$$
frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
$$



see here: Change of variables in differential equation?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:54












  • $begingroup$
    @k.stm Thanks ! , Uhm, no I am not 100% sure.
    $endgroup$
    – james
    Dec 14 '18 at 15:55










  • $begingroup$
    Well, what is $w$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:57










  • $begingroup$
    @james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
    $endgroup$
    – hyperkahler
    Dec 14 '18 at 15:57










  • $begingroup$
    @Arteom Please see my edit for the derivation
    $endgroup$
    – james
    Dec 14 '18 at 15:58
















0












$begingroup$


Hi I was wondering how to do the following:



I have this integral:
$$int _ { - L / 2 } ^ { + L / 2 } w' ( x ), d x$$



I know would like to normalize x with respect to L:
$$hat { x } = frac { x } { L }$$



How do I substitute correctly ?



I know that:
$$w' ( x ) = frac { 1 } { L } w' (hat { x })$$



but what about the $dx$?



Thanks!



EDIT:
From the chain rule:
$$
frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
$$



see here: Change of variables in differential equation?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:54












  • $begingroup$
    @k.stm Thanks ! , Uhm, no I am not 100% sure.
    $endgroup$
    – james
    Dec 14 '18 at 15:55










  • $begingroup$
    Well, what is $w$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:57










  • $begingroup$
    @james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
    $endgroup$
    – hyperkahler
    Dec 14 '18 at 15:57










  • $begingroup$
    @Arteom Please see my edit for the derivation
    $endgroup$
    – james
    Dec 14 '18 at 15:58














0












0








0





$begingroup$


Hi I was wondering how to do the following:



I have this integral:
$$int _ { - L / 2 } ^ { + L / 2 } w' ( x ), d x$$



I know would like to normalize x with respect to L:
$$hat { x } = frac { x } { L }$$



How do I substitute correctly ?



I know that:
$$w' ( x ) = frac { 1 } { L } w' (hat { x })$$



but what about the $dx$?



Thanks!



EDIT:
From the chain rule:
$$
frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
$$



see here: Change of variables in differential equation?






integration







share|cite|improve this question











$endgroup$




Hi I was wondering how to do the following:



I have this integral:
$$int _ { - L / 2 } ^ { + L / 2 } w' ( x ), d x$$



I know would like to normalize x with respect to L:
$$hat { x } = frac { x } { L }$$



How do I substitute correctly ?



I know that:
$$w' ( x ) = frac { 1 } { L } w' (hat { x })$$



but what about the $dx$?



Thanks!



EDIT:
From the chain rule:
$$
frac{dw}{dx} = frac{dhat{x}}{dx} frac{dw}{dhat{x}} = frac{1}{L} frac{dw}{dhat{x}}
$$



see here: Change of variables in differential equation?






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 18:20









egreg

183k1486205




183k1486205










asked Dec 14 '18 at 15:50









jamesjames

15110




15110












  • $begingroup$
    The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:54












  • $begingroup$
    @k.stm Thanks ! , Uhm, no I am not 100% sure.
    $endgroup$
    – james
    Dec 14 '18 at 15:55










  • $begingroup$
    Well, what is $w$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:57










  • $begingroup$
    @james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
    $endgroup$
    – hyperkahler
    Dec 14 '18 at 15:57










  • $begingroup$
    @Arteom Please see my edit for the derivation
    $endgroup$
    – james
    Dec 14 '18 at 15:58


















  • $begingroup$
    The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:54












  • $begingroup$
    @k.stm Thanks ! , Uhm, no I am not 100% sure.
    $endgroup$
    – james
    Dec 14 '18 at 15:55










  • $begingroup$
    Well, what is $w$?
    $endgroup$
    – k.stm
    Dec 14 '18 at 15:57










  • $begingroup$
    @james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
    $endgroup$
    – hyperkahler
    Dec 14 '18 at 15:57










  • $begingroup$
    @Arteom Please see my edit for the derivation
    $endgroup$
    – james
    Dec 14 '18 at 15:58
















$begingroup$
The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
$endgroup$
– k.stm
Dec 14 '18 at 15:54






$begingroup$
The differential $mathrm d x$ transforms linearly, that is $mathrm d x = mathrm d L hat x = L mathrm d hat x$. But are you sure about $w(x) = frac 1 L w(hat x)$?
$endgroup$
– k.stm
Dec 14 '18 at 15:54














$begingroup$
@k.stm Thanks ! , Uhm, no I am not 100% sure.
$endgroup$
– james
Dec 14 '18 at 15:55




$begingroup$
@k.stm Thanks ! , Uhm, no I am not 100% sure.
$endgroup$
– james
Dec 14 '18 at 15:55












$begingroup$
Well, what is $w$?
$endgroup$
– k.stm
Dec 14 '18 at 15:57




$begingroup$
Well, what is $w$?
$endgroup$
– k.stm
Dec 14 '18 at 15:57












$begingroup$
@james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
$endgroup$
– hyperkahler
Dec 14 '18 at 15:57




$begingroup$
@james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L hat{x})$
$endgroup$
– hyperkahler
Dec 14 '18 at 15:57












$begingroup$
@Arteom Please see my edit for the derivation
$endgroup$
– james
Dec 14 '18 at 15:58




$begingroup$
@Arteom Please see my edit for the derivation
$endgroup$
– james
Dec 14 '18 at 15:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

Note. You know that $displaystylefrac{d[w(x)]}{dx}=w'(x)implies d[w(x)]=w'(x)dx$



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$




Method $1$



$hat x=frac xLimplies x=Lhat x$



Since $x$ ranges from $-L/2to L/2, hat x=frac xL$ ranges from $-1/2to1/2$.



Substitute $x$ by $Lhat x$ wherever you find $x$ in the integrand.



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-1/2}^{1/2}w'(Lhat x)d(Lhat x)=Lint_{-1/2}^{1/2}w'(Lhat x)dhat xbecause d(Lhat x)=Ld(hat x)$



Method $2$



$displaystyle w'(x)=frac d{dx}[w(x)]=frac d{dx}[w(y)]$, where $y=Lhat x=x$



$displaystyle w'(x)=frac d{dx}[w(y)]=frac d{dy}[w(y)]times frac{dy}{dx}=w'(y)timesfrac{dy}{dx}=w'(y)timesfrac{dx}{dx}=w'(y)=w'(Lhat x)$



The error in your approach lies when you claim that $displaystylefrac{d[w(x)]}{dhat x}=w'(hat x)$. Since $displaystyle x=Lhat x,frac{d[w(x)]}{dhat x}=frac{d[w(Lhat x)]}{dhat x}$



By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $hat x$.



$displaystylefrac{d[w(Lhat x)]}{dhat x}=frac{d[w(Lhat x)]}{d(Lhat x)}timesfrac{d(Lhat x)}{dhat x}=Lw'(Lhat x)because w'(Lhat x)=frac d{d(Lhat x)}[w(Lhat x)]$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't you go further with $w(Lhat{x})$ ?
    $endgroup$
    – james
    Dec 14 '18 at 16:05










  • $begingroup$
    Not unless we have some more information about $w(x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:06










  • $begingroup$
    Please have a look at my new edit. Thanks !
    $endgroup$
    – james
    Dec 14 '18 at 16:09










  • $begingroup$
    Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
    $endgroup$
    – james
    Dec 14 '18 at 16:12










  • $begingroup$
    Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:12



















1












$begingroup$

If you substitute $hat{x}=x/L$, then $dhat{x}=L^{-1},dx$ and the integral becomes
$$
int_{-1/2}^{1/2} Lw'(Lhat{x}),dhat{x}
$$
Not really a progress, because you need to apply the chain rule to show that, if $u(hat{x})=w(Lhat{x})$, then
$$
u'(hat{x})=Lw'(Lhat{x})
$$
and so the integral can be rewritten as
$$
int_{-1/2}^{1/2} u'(hat{x}),dhat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2)
$$
which was directly clear from the beginning with the fundamental theorem of calculus.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ShubhamJohri Yes, you're right. Fixed.
    $endgroup$
    – egreg
    Dec 14 '18 at 21:02











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%2f3039529%2fsubstitution-of-x-in-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












$begingroup$

Note. You know that $displaystylefrac{d[w(x)]}{dx}=w'(x)implies d[w(x)]=w'(x)dx$



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$




Method $1$



$hat x=frac xLimplies x=Lhat x$



Since $x$ ranges from $-L/2to L/2, hat x=frac xL$ ranges from $-1/2to1/2$.



Substitute $x$ by $Lhat x$ wherever you find $x$ in the integrand.



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-1/2}^{1/2}w'(Lhat x)d(Lhat x)=Lint_{-1/2}^{1/2}w'(Lhat x)dhat xbecause d(Lhat x)=Ld(hat x)$



Method $2$



$displaystyle w'(x)=frac d{dx}[w(x)]=frac d{dx}[w(y)]$, where $y=Lhat x=x$



$displaystyle w'(x)=frac d{dx}[w(y)]=frac d{dy}[w(y)]times frac{dy}{dx}=w'(y)timesfrac{dy}{dx}=w'(y)timesfrac{dx}{dx}=w'(y)=w'(Lhat x)$



The error in your approach lies when you claim that $displaystylefrac{d[w(x)]}{dhat x}=w'(hat x)$. Since $displaystyle x=Lhat x,frac{d[w(x)]}{dhat x}=frac{d[w(Lhat x)]}{dhat x}$



By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $hat x$.



$displaystylefrac{d[w(Lhat x)]}{dhat x}=frac{d[w(Lhat x)]}{d(Lhat x)}timesfrac{d(Lhat x)}{dhat x}=Lw'(Lhat x)because w'(Lhat x)=frac d{d(Lhat x)}[w(Lhat x)]$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't you go further with $w(Lhat{x})$ ?
    $endgroup$
    – james
    Dec 14 '18 at 16:05










  • $begingroup$
    Not unless we have some more information about $w(x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:06










  • $begingroup$
    Please have a look at my new edit. Thanks !
    $endgroup$
    – james
    Dec 14 '18 at 16:09










  • $begingroup$
    Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
    $endgroup$
    – james
    Dec 14 '18 at 16:12










  • $begingroup$
    Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:12
















1












$begingroup$

Note. You know that $displaystylefrac{d[w(x)]}{dx}=w'(x)implies d[w(x)]=w'(x)dx$



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$




Method $1$



$hat x=frac xLimplies x=Lhat x$



Since $x$ ranges from $-L/2to L/2, hat x=frac xL$ ranges from $-1/2to1/2$.



Substitute $x$ by $Lhat x$ wherever you find $x$ in the integrand.



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-1/2}^{1/2}w'(Lhat x)d(Lhat x)=Lint_{-1/2}^{1/2}w'(Lhat x)dhat xbecause d(Lhat x)=Ld(hat x)$



Method $2$



$displaystyle w'(x)=frac d{dx}[w(x)]=frac d{dx}[w(y)]$, where $y=Lhat x=x$



$displaystyle w'(x)=frac d{dx}[w(y)]=frac d{dy}[w(y)]times frac{dy}{dx}=w'(y)timesfrac{dy}{dx}=w'(y)timesfrac{dx}{dx}=w'(y)=w'(Lhat x)$



The error in your approach lies when you claim that $displaystylefrac{d[w(x)]}{dhat x}=w'(hat x)$. Since $displaystyle x=Lhat x,frac{d[w(x)]}{dhat x}=frac{d[w(Lhat x)]}{dhat x}$



By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $hat x$.



$displaystylefrac{d[w(Lhat x)]}{dhat x}=frac{d[w(Lhat x)]}{d(Lhat x)}timesfrac{d(Lhat x)}{dhat x}=Lw'(Lhat x)because w'(Lhat x)=frac d{d(Lhat x)}[w(Lhat x)]$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't you go further with $w(Lhat{x})$ ?
    $endgroup$
    – james
    Dec 14 '18 at 16:05










  • $begingroup$
    Not unless we have some more information about $w(x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:06










  • $begingroup$
    Please have a look at my new edit. Thanks !
    $endgroup$
    – james
    Dec 14 '18 at 16:09










  • $begingroup$
    Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
    $endgroup$
    – james
    Dec 14 '18 at 16:12










  • $begingroup$
    Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:12














1












1








1





$begingroup$

Note. You know that $displaystylefrac{d[w(x)]}{dx}=w'(x)implies d[w(x)]=w'(x)dx$



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$




Method $1$



$hat x=frac xLimplies x=Lhat x$



Since $x$ ranges from $-L/2to L/2, hat x=frac xL$ ranges from $-1/2to1/2$.



Substitute $x$ by $Lhat x$ wherever you find $x$ in the integrand.



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-1/2}^{1/2}w'(Lhat x)d(Lhat x)=Lint_{-1/2}^{1/2}w'(Lhat x)dhat xbecause d(Lhat x)=Ld(hat x)$



Method $2$



$displaystyle w'(x)=frac d{dx}[w(x)]=frac d{dx}[w(y)]$, where $y=Lhat x=x$



$displaystyle w'(x)=frac d{dx}[w(y)]=frac d{dy}[w(y)]times frac{dy}{dx}=w'(y)timesfrac{dy}{dx}=w'(y)timesfrac{dx}{dx}=w'(y)=w'(Lhat x)$



The error in your approach lies when you claim that $displaystylefrac{d[w(x)]}{dhat x}=w'(hat x)$. Since $displaystyle x=Lhat x,frac{d[w(x)]}{dhat x}=frac{d[w(Lhat x)]}{dhat x}$



By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $hat x$.



$displaystylefrac{d[w(Lhat x)]}{dhat x}=frac{d[w(Lhat x)]}{d(Lhat x)}timesfrac{d(Lhat x)}{dhat x}=Lw'(Lhat x)because w'(Lhat x)=frac d{d(Lhat x)}[w(Lhat x)]$






share|cite|improve this answer











$endgroup$



Note. You know that $displaystylefrac{d[w(x)]}{dx}=w'(x)implies d[w(x)]=w'(x)dx$



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$




Method $1$



$hat x=frac xLimplies x=Lhat x$



Since $x$ ranges from $-L/2to L/2, hat x=frac xL$ ranges from $-1/2to1/2$.



Substitute $x$ by $Lhat x$ wherever you find $x$ in the integrand.



$displaystyleint_{-L/2}^{L/2}w'(x)dx=int_{-1/2}^{1/2}w'(Lhat x)d(Lhat x)=Lint_{-1/2}^{1/2}w'(Lhat x)dhat xbecause d(Lhat x)=Ld(hat x)$



Method $2$



$displaystyle w'(x)=frac d{dx}[w(x)]=frac d{dx}[w(y)]$, where $y=Lhat x=x$



$displaystyle w'(x)=frac d{dx}[w(y)]=frac d{dy}[w(y)]times frac{dy}{dx}=w'(y)timesfrac{dy}{dx}=w'(y)timesfrac{dx}{dx}=w'(y)=w'(Lhat x)$



The error in your approach lies when you claim that $displaystylefrac{d[w(x)]}{dhat x}=w'(hat x)$. Since $displaystyle x=Lhat x,frac{d[w(x)]}{dhat x}=frac{d[w(Lhat x)]}{dhat x}$



By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $hat x$.



$displaystylefrac{d[w(Lhat x)]}{dhat x}=frac{d[w(Lhat x)]}{d(Lhat x)}timesfrac{d(Lhat x)}{dhat x}=Lw'(Lhat x)because w'(Lhat x)=frac d{d(Lhat x)}[w(Lhat x)]$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 18:15

























answered Dec 14 '18 at 15:59









Shubham JohriShubham Johri

5,204718




5,204718












  • $begingroup$
    Can't you go further with $w(Lhat{x})$ ?
    $endgroup$
    – james
    Dec 14 '18 at 16:05










  • $begingroup$
    Not unless we have some more information about $w(x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:06










  • $begingroup$
    Please have a look at my new edit. Thanks !
    $endgroup$
    – james
    Dec 14 '18 at 16:09










  • $begingroup$
    Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
    $endgroup$
    – james
    Dec 14 '18 at 16:12










  • $begingroup$
    Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:12


















  • $begingroup$
    Can't you go further with $w(Lhat{x})$ ?
    $endgroup$
    – james
    Dec 14 '18 at 16:05










  • $begingroup$
    Not unless we have some more information about $w(x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:06










  • $begingroup$
    Please have a look at my new edit. Thanks !
    $endgroup$
    – james
    Dec 14 '18 at 16:09










  • $begingroup$
    Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
    $endgroup$
    – james
    Dec 14 '18 at 16:12










  • $begingroup$
    Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
    $endgroup$
    – Shubham Johri
    Dec 14 '18 at 16:12
















$begingroup$
Can't you go further with $w(Lhat{x})$ ?
$endgroup$
– james
Dec 14 '18 at 16:05




$begingroup$
Can't you go further with $w(Lhat{x})$ ?
$endgroup$
– james
Dec 14 '18 at 16:05












$begingroup$
Not unless we have some more information about $w(x)$
$endgroup$
– Shubham Johri
Dec 14 '18 at 16:06




$begingroup$
Not unless we have some more information about $w(x)$
$endgroup$
– Shubham Johri
Dec 14 '18 at 16:06












$begingroup$
Please have a look at my new edit. Thanks !
$endgroup$
– james
Dec 14 '18 at 16:09




$begingroup$
Please have a look at my new edit. Thanks !
$endgroup$
– james
Dec 14 '18 at 16:09












$begingroup$
Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
$endgroup$
– james
Dec 14 '18 at 16:12




$begingroup$
Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ?
$endgroup$
– james
Dec 14 '18 at 16:12












$begingroup$
Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
$endgroup$
– Shubham Johri
Dec 14 '18 at 16:12




$begingroup$
Note that $displaystylefrac{dw}{dhat x}ne w'(hat x)$
$endgroup$
– Shubham Johri
Dec 14 '18 at 16:12











1












$begingroup$

If you substitute $hat{x}=x/L$, then $dhat{x}=L^{-1},dx$ and the integral becomes
$$
int_{-1/2}^{1/2} Lw'(Lhat{x}),dhat{x}
$$
Not really a progress, because you need to apply the chain rule to show that, if $u(hat{x})=w(Lhat{x})$, then
$$
u'(hat{x})=Lw'(Lhat{x})
$$
and so the integral can be rewritten as
$$
int_{-1/2}^{1/2} u'(hat{x}),dhat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2)
$$
which was directly clear from the beginning with the fundamental theorem of calculus.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ShubhamJohri Yes, you're right. Fixed.
    $endgroup$
    – egreg
    Dec 14 '18 at 21:02
















1












$begingroup$

If you substitute $hat{x}=x/L$, then $dhat{x}=L^{-1},dx$ and the integral becomes
$$
int_{-1/2}^{1/2} Lw'(Lhat{x}),dhat{x}
$$
Not really a progress, because you need to apply the chain rule to show that, if $u(hat{x})=w(Lhat{x})$, then
$$
u'(hat{x})=Lw'(Lhat{x})
$$
and so the integral can be rewritten as
$$
int_{-1/2}^{1/2} u'(hat{x}),dhat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2)
$$
which was directly clear from the beginning with the fundamental theorem of calculus.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ShubhamJohri Yes, you're right. Fixed.
    $endgroup$
    – egreg
    Dec 14 '18 at 21:02














1












1








1





$begingroup$

If you substitute $hat{x}=x/L$, then $dhat{x}=L^{-1},dx$ and the integral becomes
$$
int_{-1/2}^{1/2} Lw'(Lhat{x}),dhat{x}
$$
Not really a progress, because you need to apply the chain rule to show that, if $u(hat{x})=w(Lhat{x})$, then
$$
u'(hat{x})=Lw'(Lhat{x})
$$
and so the integral can be rewritten as
$$
int_{-1/2}^{1/2} u'(hat{x}),dhat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2)
$$
which was directly clear from the beginning with the fundamental theorem of calculus.






share|cite|improve this answer











$endgroup$



If you substitute $hat{x}=x/L$, then $dhat{x}=L^{-1},dx$ and the integral becomes
$$
int_{-1/2}^{1/2} Lw'(Lhat{x}),dhat{x}
$$
Not really a progress, because you need to apply the chain rule to show that, if $u(hat{x})=w(Lhat{x})$, then
$$
u'(hat{x})=Lw'(Lhat{x})
$$
and so the integral can be rewritten as
$$
int_{-1/2}^{1/2} u'(hat{x}),dhat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2)
$$
which was directly clear from the beginning with the fundamental theorem of calculus.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 21:01

























answered Dec 14 '18 at 18:31









egregegreg

183k1486205




183k1486205












  • $begingroup$
    @ShubhamJohri Yes, you're right. Fixed.
    $endgroup$
    – egreg
    Dec 14 '18 at 21:02


















  • $begingroup$
    @ShubhamJohri Yes, you're right. Fixed.
    $endgroup$
    – egreg
    Dec 14 '18 at 21:02
















$begingroup$
@ShubhamJohri Yes, you're right. Fixed.
$endgroup$
– egreg
Dec 14 '18 at 21:02




$begingroup$
@ShubhamJohri Yes, you're right. Fixed.
$endgroup$
– egreg
Dec 14 '18 at 21:02


















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%2f3039529%2fsubstitution-of-x-in-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







-

Popular posts from this blog

Willebadessen

Image
Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
Read more

Residenzschloss Arolsen

Image
Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
Read more