Doubt in integration by partsWhy doesn't this work for integrating $x^2e^-ax^2$ by parts?Volume of a solid involving integration by parts.Integration by parts of $fracln(1+sqrtx)sqrtx$Deriving the integration by parts formulaIntegration with probability density functionintegration by parts with $frac2sqrt2pi int^infty_0 z^2e^frac-z^22dz$Trying to understand integration by partsIntegration by Parts - $int fracx(x+1)^2 dx$Why do we ignore constant produced by integration of derivative when we derive integration by parts formula?Why doesn't this work for integrating $x^2e^-ax^2$ by parts?graphical meaning of integration by parts of this function
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Doubt in integration by parts
Why doesn't this work for integrating $x^2e^-ax^2$ by parts?Volume of a solid involving integration by parts.Integration by parts of $fracln(1+sqrtx)sqrtx$Deriving the integration by parts formulaIntegration with probability density functionintegration by parts with $frac2sqrt2pi int^infty_0 z^2e^frac-z^22dz$Trying to understand integration by partsIntegration by Parts - $int fracx(x+1)^2 dx$Why do we ignore constant produced by integration of derivative when we derive integration by parts formula?Why doesn't this work for integrating $x^2e^-ax^2$ by parts?graphical meaning of integration by parts of this function
$begingroup$
The integration by parts formula (in NCERT) is given as:$$
int f(x) g(x) = f(x)int g(x)- int f'(x)int g(x)dx$$
Can it be successfully applied to $$int _-infty^infty x^2 e^-alpha x^2 dx=x^2int _-infty^infty e^-alpha x^2-int _-infty^infty 2xint ^infty_-inftye^-alpha x^2dx$$$$
=x^2 sqrtpi/alpha-(x^2 sqrtpi/alpha)|^infty_-infty$$ I could get the required answer by using $$int ^infty_-infty udv= uv|^infty_-infty -int ^infty_-infty vdu.$$ I want to know why the ncert formula fails,if i take $f(x)= x^2$ and use the limits.Please tell me where im making a mistake.
integration
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The integration by parts formula (in NCERT) is given as:$$
int f(x) g(x) = f(x)int g(x)- int f'(x)int g(x)dx$$
Can it be successfully applied to $$int _-infty^infty x^2 e^-alpha x^2 dx=x^2int _-infty^infty e^-alpha x^2-int _-infty^infty 2xint ^infty_-inftye^-alpha x^2dx$$$$
=x^2 sqrtpi/alpha-(x^2 sqrtpi/alpha)|^infty_-infty$$ I could get the required answer by using $$int ^infty_-infty udv= uv|^infty_-infty -int ^infty_-infty vdu.$$ I want to know why the ncert formula fails,if i take $f(x)= x^2$ and use the limits.Please tell me where im making a mistake.
integration
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
1
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago
add a comment |
$begingroup$
The integration by parts formula (in NCERT) is given as:$$
int f(x) g(x) = f(x)int g(x)- int f'(x)int g(x)dx$$
Can it be successfully applied to $$int _-infty^infty x^2 e^-alpha x^2 dx=x^2int _-infty^infty e^-alpha x^2-int _-infty^infty 2xint ^infty_-inftye^-alpha x^2dx$$$$
=x^2 sqrtpi/alpha-(x^2 sqrtpi/alpha)|^infty_-infty$$ I could get the required answer by using $$int ^infty_-infty udv= uv|^infty_-infty -int ^infty_-infty vdu.$$ I want to know why the ncert formula fails,if i take $f(x)= x^2$ and use the limits.Please tell me where im making a mistake.
integration
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The integration by parts formula (in NCERT) is given as:$$
int f(x) g(x) = f(x)int g(x)- int f'(x)int g(x)dx$$
Can it be successfully applied to $$int _-infty^infty x^2 e^-alpha x^2 dx=x^2int _-infty^infty e^-alpha x^2-int _-infty^infty 2xint ^infty_-inftye^-alpha x^2dx$$$$
=x^2 sqrtpi/alpha-(x^2 sqrtpi/alpha)|^infty_-infty$$ I could get the required answer by using $$int ^infty_-infty udv= uv|^infty_-infty -int ^infty_-infty vdu.$$ I want to know why the ncert formula fails,if i take $f(x)= x^2$ and use the limits.Please tell me where im making a mistake.
integration
integration
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Vortex
asked 2 days ago
VortexVortex
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
VortexVortex
62
asked 2 days ago
VortexVortex
62
62
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vortex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
1
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
1
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
1
1
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the formula with $g(x)=xe^-alpha x^2$ and $f(x)=x$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
add a comment |
$begingroup$
$$beginbmatrixf \ mathrm dgendbmatrix=beginbmatrixx\xe^-alpha x^2endbmatriximpliesint x^2e^-alpha x^2mathrm dx=-dfracxmathrme^-alpha x^22alpha+dfrac12alphaint e^-alpha x^2mathrm dx\ int_-infty^+infty x^2e^-alpha x^2mathrm dx=dfracsqrtpioperatornameerfleft(sqrtaxright)4a^3/2-dfracxmathrme^-ax^22aBiggl|_-infty^+infty=dfracsqrtpi2alpha^3/2
$$
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
$endgroup$
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
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
$begingroup$
You can use the formula with $g(x)=xe^-alpha x^2$ and $f(x)=x$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
add a comment |
$begingroup$
You can use the formula with $g(x)=xe^-alpha x^2$ and $f(x)=x$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
add a comment |
$begingroup$
You can use the formula with $g(x)=xe^-alpha x^2$ and $f(x)=x$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
You can use the formula with $g(x)=xe^-alpha x^2$ and $f(x)=x$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
66.6k53067
add a comment |
add a comment |
$begingroup$
$$beginbmatrixf \ mathrm dgendbmatrix=beginbmatrixx\xe^-alpha x^2endbmatriximpliesint x^2e^-alpha x^2mathrm dx=-dfracxmathrme^-alpha x^22alpha+dfrac12alphaint e^-alpha x^2mathrm dx\ int_-infty^+infty x^2e^-alpha x^2mathrm dx=dfracsqrtpioperatornameerfleft(sqrtaxright)4a^3/2-dfracxmathrme^-ax^22aBiggl|_-infty^+infty=dfracsqrtpi2alpha^3/2
$$
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
$endgroup$
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
$$beginbmatrixf \ mathrm dgendbmatrix=beginbmatrixx\xe^-alpha x^2endbmatriximpliesint x^2e^-alpha x^2mathrm dx=-dfracxmathrme^-alpha x^22alpha+dfrac12alphaint e^-alpha x^2mathrm dx\ int_-infty^+infty x^2e^-alpha x^2mathrm dx=dfracsqrtpioperatornameerfleft(sqrtaxright)4a^3/2-dfracxmathrme^-ax^22aBiggl|_-infty^+infty=dfracsqrtpi2alpha^3/2
$$
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
$endgroup$
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
$$beginbmatrixf \ mathrm dgendbmatrix=beginbmatrixx\xe^-alpha x^2endbmatriximpliesint x^2e^-alpha x^2mathrm dx=-dfracxmathrme^-alpha x^22alpha+dfrac12alphaint e^-alpha x^2mathrm dx\ int_-infty^+infty x^2e^-alpha x^2mathrm dx=dfracsqrtpioperatornameerfleft(sqrtaxright)4a^3/2-dfracxmathrme^-ax^22aBiggl|_-infty^+infty=dfracsqrtpi2alpha^3/2
$$
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
$endgroup$
$$beginbmatrixf \ mathrm dgendbmatrix=beginbmatrixx\xe^-alpha x^2endbmatriximpliesint x^2e^-alpha x^2mathrm dx=-dfracxmathrme^-alpha x^22alpha+dfrac12alphaint e^-alpha x^2mathrm dx\ int_-infty^+infty x^2e^-alpha x^2mathrm dx=dfracsqrtpioperatornameerfleft(sqrtaxright)4a^3/2-dfracxmathrme^-ax^22aBiggl|_-infty^+infty=dfracsqrtpi2alpha^3/2
$$
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
answered 2 days ago
Paras KhoslaParas Khosla
1,828219
1,828219
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
add a comment |
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
sir can u show me how the evaluation goes if i use f(x)=x^2.I wanted to know why ncert method fails if i do like this!.
$endgroup$
– Vortex
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
$begingroup$
Have a look at this related question: math.stackexchange.com/questions/3089298/…
$endgroup$
– Paras Khosla
yesterday
add a comment |
Vortex is a new contributor. Be nice, and check out our Code of Conduct.
Vortex is a new contributor. Be nice, and check out our Code of Conduct.
Vortex is a new contributor. Be nice, and check out our Code of Conduct.
Vortex is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
2 days ago
1
$begingroup$
In particular, to apply that formula, you should choose $f,g$ so that differentiation of $f$ is easy, and integration of $g$ is easy. Otherwise you replace one integral by a (possibly harder) integral. The formula does not "fail" when you take $f(x) =x^2$. It yields something that is correct, but not helpful.
$endgroup$
– GEdgar
2 days ago