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?
What is the steepest angle that a canal can be traversable without locks?
Springs with some finite mass
aging parents with no investments
Can distinct morphisms between curves induce the same morphism on singular cohomology?
Inflated grade on resume at previous job, might former employer tell new employer?
What tool would a Roman-age civilization have to grind silver and other metals into dust?
Why don't Unix/Linux systems traverse through directories until they find the required version of a linked library?
The difference between dialogue marks
Time travel alters history but people keep saying nothing's changed
Patience, young "Padovan"
Any good smartcontract for "business calendar" oracles?
How to answer pointed "are you quitting" questioning when I don't want them to suspect
Why could you hear an Amstrad CPC working?
Is flight data recorder erased after every flight?
Confusion about non-derivable continuous functions
Pristine Bit Checking
Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints
Is there a name of the flying bionic bird?
How was Skylab's orbit inclination chosen?
Is domain driven design an anti-SQL pattern?
Are there any other methods to apply to solving simultaneous equations?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
Output the Arecibo Message
Why is Grand Jury testimony secret?
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...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
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...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
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
1 hour ago
add a comment |
$begingroup$
I am attempting to solve the integral of the following...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
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...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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 1 hour 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
1 hour ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 1 hour 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 1 hour 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
1 hour ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour 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 1 hour 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
1 hour ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour 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 1 hour 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 1 hour ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 1 hour ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 1 hour ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 1 hour 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
1 hour ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour 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
1 hour ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour 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
1 hour 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
1 hour ago
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour ago
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
1 hour ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 1 hour 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 $-e^-r^2over 2$
answered 1 hour 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 $-e^-r^2over 2$
answered 1 hour ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 1 hour ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
edited 1 hour ago
answered 1 hour ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 1 hour ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 1 hour 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
1 hour ago