How to change the limits of integration The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integration 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 In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integration 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 Apr 9 at 23:40
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 Apr 9 at 23:40
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
Apr 9 at 23:44
$begingroup$
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
Apr 10 at 4:30
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 Apr 9 at 23:40
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 Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
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
Apr 9 at 23:44
$begingroup$
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
Apr 10 at 4:30
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
Apr 9 at 23:44
$begingroup$
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
Apr 10 at 4:30
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
Apr 9 at 23:44
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
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
Apr 10 at 4:30
$begingroup$
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
Apr 10 at 4:30
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 Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
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
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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 Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
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
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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 Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
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
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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 Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
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
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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
Apr 9 at 23:45
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
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 Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
$endgroup$
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 Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
$endgroup$
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 Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
$endgroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
edited Apr 9 at 23:47
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,220212
2,220212
add a comment |
add a comment |
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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
Apr 9 at 23:44
$begingroup$
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
Apr 10 at 4:30