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?
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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 2 days ago
BolboaBolboa
408616
$endgroup$
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 2 days ago
BolboaBolboa
408616
$endgroup$
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
2 days ago
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday
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 2 days ago
BolboaBolboa
408616
$endgroup$
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 2 days ago
BolboaBolboa
408616
asked 2 days ago
BolboaBolboa
408616
asked 2 days ago
BolboaBolboa
408616
asked 2 days ago
BolboaBolboa
408616
asked 2 days ago
BolboaBolboa
408616
408616
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
2 days ago
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday
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
2 days ago
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday
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
2 days 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
2 days ago
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday
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 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
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
2 days ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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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 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
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
2 days ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
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
2 days ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
74.1k53270
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
2 days ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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
2 days ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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
2 days 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
2 days ago
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
2 days 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 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
$endgroup$
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 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
$endgroup$
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 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
$endgroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
edited 2 days ago
answered 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
answered 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
answered 2 days ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,953212
1,953212
add a comment |
add a comment |
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-calculus, integration, limits
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
2 days ago
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
yesterday