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 59 mins 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 59 mins 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
55 mins 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 59 mins 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 59 mins ago
BolboaBolboa
391516
asked 59 mins ago
BolboaBolboa
391516
asked 59 mins ago
BolboaBolboa
391516
asked 59 mins ago
BolboaBolboa
391516
asked 59 mins 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
55 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
55 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
55 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
55 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 55 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
54 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 mins ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 54 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
$endgroup$
add a comment |
Your Answer
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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 55 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
54 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 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 55 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
54 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 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 55 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 55 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 55 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 55 mins ago
Kavi Rama MurthyKavi Rama Murthy
73.6k53170
answered 55 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
54 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 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
54 mins ago
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 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
54 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
54 mins ago
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 mins ago
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
50 mins ago
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered 54 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 $-e^-r^2over 2$
answered 54 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 $-e^-r^2over 2$
answered 54 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 $-e^-r^2over 2$
answered 54 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
edited 52 mins ago
answered 54 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 54 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
answered 54 mins ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,836212
1,836212
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
55 mins ago