How to change the limits of integration The 2019 Stack Overflow Developer Survey Results Are...
What do the Banks children have against barley water?
Can't find the latex code for the ⍎ (down tack jot) symbol
In microwave frequencies, do you use a circulator when you need a (near) perfect diode?
Confusion about non-derivable continuous functions
Access elements in std::string where positon of string is greater than its size
Is there a name of the flying bionic bird?
Does it makes sense to buy a new cycle to learn riding?
"To split hairs" vs "To be pedantic"
I looked up a future colleague on linkedin before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?
Lethal sonic weapons
JSON.serialize: is it possible to suppress null values of a map?
"What time...?" or "At what time...?" - what is more grammatically correct?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
What tool would a Roman-age civilization have to grind silver and other metals into dust?
Feasability of miniature nuclear reactors for humanoid cyborgs
Patience, young "Padovan"
Is flight data recorder erased after every flight?
Can distinct morphisms between curves induce the same morphism on singular cohomology?
How can I fix this gap between bookcases I made?
Does duplicating a spell with wish count as casting that spell?
Is three citations per paragraph excessive for undergraduate research paper?
Are there any other methods to apply to solving simultaneous equations?
Why can Shazam do this?
Inflated grade on resume at previous job, might former employer tell new employer?
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 48 mins ago
BolboaBolboa
386516
$endgroup$
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 48 mins ago
BolboaBolboa
386516
$endgroup$
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
44 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 48 mins ago
BolboaBolboa
386516
$endgroup$
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 48 mins ago
BolboaBolboa
386516
asked 48 mins ago
BolboaBolboa
386516
asked 48 mins ago
BolboaBolboa
386516
asked 48 mins ago
BolboaBolboa
386516
asked 48 mins ago
BolboaBolboa
386516
386516
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
44 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
44 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
44 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
44 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 44 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
43 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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 43 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 44 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
43 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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 44 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
43 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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 44 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 44 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 44 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 44 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 44 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
43 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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
43 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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
43 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
43 mins ago
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 mins ago
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
39 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 43 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 43 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 43 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 43 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
edited 41 mins ago
answered 43 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 43 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 43 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
44 mins ago