$int_{mathbb R^n}e^{{-|x|}^n}dx$












4












$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08
















4












$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08














4












4








4





$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$




I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 16:00









gt6989b

34k22455




34k22455










asked Jan 16 at 15:57









Oria GruberOria Gruber

6,44332460




6,44332460












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08


















  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08
















$begingroup$
Make rather a polar coordinate change?
$endgroup$
– Mindlack
Jan 16 at 16:08




$begingroup$
Make rather a polar coordinate change?
$endgroup$
– Mindlack
Jan 16 at 16:08












$begingroup$
I don't think that's the intended way. I might be wrong.
$endgroup$
– Oria Gruber
Jan 16 at 16:08




$begingroup$
I don't think that's the intended way. I might be wrong.
$endgroup$
– Oria Gruber
Jan 16 at 16:08










1 Answer
1






active

oldest

votes


















3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04











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%2f3075902%2fint-mathbb-rne-xndx%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









3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04
















3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04














3












3








3





$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$



I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 16:21

























answered Jan 16 at 16:09









SongSong

12.2k630




12.2k630












  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04


















  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04
















$begingroup$
Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
$endgroup$
– Oria Gruber
Jan 16 at 16:13




$begingroup$
Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
$endgroup$
– Oria Gruber
Jan 16 at 16:13












$begingroup$
@OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
$endgroup$
– Song
Jan 16 at 16:14






$begingroup$
@OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
$endgroup$
– Song
Jan 16 at 16:14














$begingroup$
@OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
$endgroup$
– mathworker21
Jan 16 at 16:19




$begingroup$
@OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
$endgroup$
– mathworker21
Jan 16 at 16:19












$begingroup$
@mathworker21 Yes! That is what exactly I did. Thanks for clarification.
$endgroup$
– Song
Jan 16 at 16:20




$begingroup$
@mathworker21 Yes! That is what exactly I did. Thanks for clarification.
$endgroup$
– Song
Jan 16 at 16:20












$begingroup$
May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
$endgroup$
– Oria Gruber
Jan 16 at 17:04




$begingroup$
May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
$endgroup$
– Oria Gruber
Jan 16 at 17:04


















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%2f3075902%2fint-mathbb-rne-xndx%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

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more