Definite Integral in u and u' Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integrating Factors Via MinimizationDefinite integral involving e and lnCan every definite integral be computed symbolically?Finding a definite integral by residue integration?Question about a definite integralSolving a definite integralTo determine a definite integralSolving definite integralEvaluation of given definite integralDefinite integral over singularityIntegration by parts for definite integral
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Definite Integral in u and u'
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integrating Factors Via MinimizationDefinite integral involving e and lnCan every definite integral be computed symbolically?Finding a definite integral by residue integration?Question about a definite integralSolving a definite integralTo determine a definite integralSolving definite integralEvaluation of given definite integralDefinite integral over singularityIntegration by parts for definite integral
$begingroup$
Suppose I want to solve $int_-1^1 u(x)^ 2 dx$. I should be able to write this as $int_-1^1 u^ 2 frac1u' du$.
Since $int_-1^1 u' du$ is presumably just $u(1) - u(-1)$, can I also find the definite integral of an integrand in $u$ and $u'$, such as $int_-1^1 u^ 2 frac1u' du$? How would I proceed in doing so?
calculus integration functions derivatives definite-integrals
asked Mar 26 at 16:22
user10478user10478
499413
$endgroup$
add a comment |
$begingroup$
Suppose I want to solve $int_-1^1 u(x)^ 2 dx$. I should be able to write this as $int_-1^1 u^ 2 frac1u' du$.
Since $int_-1^1 u' du$ is presumably just $u(1) - u(-1)$, can I also find the definite integral of an integrand in $u$ and $u'$, such as $int_-1^1 u^ 2 frac1u' du$? How would I proceed in doing so?
calculus integration functions derivatives definite-integrals
asked Mar 26 at 16:22
user10478user10478
499413
$endgroup$
$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48
add a comment |
$begingroup$
Suppose I want to solve $int_-1^1 u(x)^ 2 dx$. I should be able to write this as $int_-1^1 u^ 2 frac1u' du$.
Since $int_-1^1 u' du$ is presumably just $u(1) - u(-1)$, can I also find the definite integral of an integrand in $u$ and $u'$, such as $int_-1^1 u^ 2 frac1u' du$? How would I proceed in doing so?
calculus integration functions derivatives definite-integrals
asked Mar 26 at 16:22
user10478user10478
499413
$endgroup$
Suppose I want to solve $int_-1^1 u(x)^ 2 dx$. I should be able to write this as $int_-1^1 u^ 2 frac1u' du$.
Since $int_-1^1 u' du$ is presumably just $u(1) - u(-1)$, can I also find the definite integral of an integrand in $u$ and $u'$, such as $int_-1^1 u^ 2 frac1u' du$? How would I proceed in doing so?
calculus integration functions derivatives definite-integrals
calculus integration functions derivatives definite-integrals
asked Mar 26 at 16:22
user10478user10478
499413
asked Mar 26 at 16:22
user10478user10478
499413
asked Mar 26 at 16:22
user10478user10478
499413
asked Mar 26 at 16:22
user10478user10478
499413
asked Mar 26 at 16:22
user10478user10478
499413
499413
$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48
add a comment |
$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48
$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48
add a comment |
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$begingroup$
you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$.
$endgroup$
– Yizhar Amir
Mar 26 at 16:43
$begingroup$
Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid.
$endgroup$
– herb steinberg
Mar 26 at 17:32
$begingroup$
@YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $fracdudu$ to get $fracdxdu du implies frac1u' du$.
$endgroup$
– user10478
Mar 27 at 3:48