How to change the limits of integration The 2019 Stack Overflow Developer Survey Results Are...
Evaluating number of iteration with a certain map with While
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Any good smartcontract for "business calendar" oracles?
Why isn't airport relocation done gradually?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
Can I write a for loop that iterates over both collections and arrays?
The difference between dialogue marks
What do hard-Brexiteers want with respect to the Irish border?
Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?
How to answer pointed "are you quitting" questioning when I don't want them to suspect
If a poisoned arrow's piercing damage is reduced to 0, do you still get poisoned?
Carnot-Caratheodory metric
How was Skylab's orbit inclination chosen?
Where to refill my bottle in India?
Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints
Where does the "burst of radiance" from Holy Weapon originate?
Is domain driven design an anti-SQL pattern?
How come people say “Would of”?
It's possible to achieve negative score?
Why could you hear an Amstrad CPC working?
Idiomatic way to prevent slicing?
What tool would a Roman-age civilization have to grind silver and other metals into dust?
Should I use my personal or workplace e-mail when registering to external websites for work purpose?
How to deal with fear of taking dependencies
How to change the limits of integration
The 2019 Stack Overflow Developer Survey Results Are InIntegration limits of the double integral after conversion to the polar coordinatesConvergent or Divergent using LimitsHow do I solve a double integral with an absolute value?How are limits of integration changed?Converting limits of integrationHow to set the limits for Jacobian IntegrationLimits of integration in multivariable integrals during change of variablesChange of limits of integrationWhat theorem(s) is(are) used to change between the various improper integrals.how does an integral becoming negative effect limits of integration?
$begingroup$
I am attempting to solve the integral of the following...
$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$
So I do the following step...
$$=2 piint_{0}^{infty}e^{-r^2}rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked 1 hour ago
BolboaBolboa
391516
$endgroup$
add a comment |
$begingroup$
I am attempting to solve the integral of the following...
$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$
So I do the following step...
$$=2 piint_{0}^{infty}e^{-r^2}rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked 1 hour ago
BolboaBolboa
391516
$endgroup$
1
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago
add a comment |
$begingroup$
I am attempting to solve the integral of the following...
$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$
So I do the following step...
$$=2 piint_{0}^{infty}e^{-r^2}rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked 1 hour ago
BolboaBolboa
391516
$endgroup$
I am attempting to solve the integral of the following...
$$int_{0}^{2 pi}int_{0}^{infty}e^{-r^2}rdrTheta $$
So I do the following step...
$$=2 piint_{0}^{infty}e^{-r^2}rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_{- infty}^{0}frac{1}{2}e^{s}ds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
calculus integration limits
asked 1 hour ago
BolboaBolboa
391516
asked 1 hour ago
BolboaBolboa
391516
asked 1 hour ago
BolboaBolboa
391516
asked 1 hour ago
BolboaBolboa
391516
asked 1 hour ago
BolboaBolboa
391516
391516
1
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago
add a comment |
1
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago
1
1
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181696%2fhow-to-change-the-limits-of-integration%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
$begingroup$
$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
add a comment |
$begingroup$
$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
add a comment |
$begingroup$
$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
$endgroup$
$s=-r^{2}$ gives $ds=-2rdr$ so $dr =-frac 1 {2r} ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 57 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
73.6k53170
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
add a comment |
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
1
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
55 mins ago
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
52 mins ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
Do you really need substitution. We already know the antiderivative of $re^{-r^2}$ and it is $-e^{-r^2}over 2$
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
edited 54 mins ago
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 56 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
1,836212
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181696%2fhow-to-change-the-limits-of-integration%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
$$s = -r^{2} implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
57 mins ago