Integral of $int_0^infty {x^2}{e^{-3x}},dx$ [closed]
$begingroup$
$$int_0^infty {x^2}{ e^{-3x}},dx$$
What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.
improper-integrals
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
$endgroup$
closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$int_0^infty {x^2}{ e^{-3x}},dx$$
What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.
improper-integrals
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
$endgroup$
closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17
add a comment |
$begingroup$
$$int_0^infty {x^2}{ e^{-3x}},dx$$
What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.
improper-integrals
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
$endgroup$
$$int_0^infty {x^2}{ e^{-3x}},dx$$
What I attempted was integration by parts twice, but I end with $-frac{2}{27}$. That's obviously wrong, should be positive. I am also unsure whether or not $$lim_{tto infty} frac{t^2}{e^{3t}}$$ converges to 0 or not.
improper-integrals
improper-integrals
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
edited Dec 12 '18 at 5:10
Key Flex
8,28261233
8,28261233
asked Dec 12 '18 at 5:04
user623028user623028
235
asked Dec 12 '18 at 5:04
user623028user623028
235
asked Dec 12 '18 at 5:04
user623028user623028
235
235
closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Lord_Farin, user91500, José Carlos Santos, Abcd Dec 27 '18 at 15:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, user91500, José Carlos Santos, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17
add a comment |
1
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17
1
1
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Given $$int_0^infty x^2e^{-3x} dx$$
First compute without boundaries.
By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
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@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
add a comment |
$begingroup$
The answer is $boxed{frac{2}{27}}$
More generally, it can be shown that
$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$
Here, we have $n = 2$ and $a = 3$, so the result follows.
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
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add a comment |
$begingroup$
using the substitution $u = 3x$ the integral becomes:
$$
I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
= frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
= frac{Gamma(3)}{27} \
= frac2{27}
$$
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given $$int_0^infty x^2e^{-3x} dx$$
First compute without boundaries.
By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
add a comment |
$begingroup$
Given $$int_0^infty x^2e^{-3x} dx$$
First compute without boundaries.
By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
add a comment |
$begingroup$
Given $$int_0^infty x^2e^{-3x} dx$$
First compute without boundaries.
By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
$endgroup$
Given $$int_0^infty x^2e^{-3x} dx$$
First compute without boundaries.
By applying u-substitution: $u=-3x$, we get $$=-dfrac{1}{27}int e^uu^2 du$$
By applying Integration by parts $u=u^2,v^{prime}=e^u$, we get
$$-dfrac{1}{27}(u^2e^u-int2ue^u du)=-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})$$
Now apply the boundaries $$left[-dfrac{1}{27}(9e^{-3x}x^2+6e^{-3x}+2e^{-3x})right]_0^infty=0-left(-dfrac{2}{27}right)=dfrac{2}{27}$$
By L hospitals rule
$lim_{x to infty} e^{-3x}x^2 = limfrac{x^2}{e^3} =limfrac{2x}{3e^3}= limfrac{2}{9e^3}=0$
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
edited Dec 12 '18 at 13:48
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
answered Dec 12 '18 at 5:10
Key FlexKey Flex
8,28261233
8,28261233
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
add a comment |
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
When you apply the boundaries, why doesn't that first term diverge? 9e^{3x}x^2 ?
$endgroup$
– user623028
Dec 12 '18 at 5:16
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
@BigArsole: The first term is $9e^{-3x}x^2$ which converges to $0$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 5:20
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
$begingroup$
Oh did you make a typo, because that's not written? It's fine though thanks.
$endgroup$
– user623028
Dec 12 '18 at 5:45
add a comment |
$begingroup$
The answer is $boxed{frac{2}{27}}$
More generally, it can be shown that
$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$
Here, we have $n = 2$ and $a = 3$, so the result follows.
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
add a comment |
$begingroup$
The answer is $boxed{frac{2}{27}}$
More generally, it can be shown that
$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$
Here, we have $n = 2$ and $a = 3$, so the result follows.
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
add a comment |
$begingroup$
The answer is $boxed{frac{2}{27}}$
More generally, it can be shown that
$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$
Here, we have $n = 2$ and $a = 3$, so the result follows.
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
The answer is $boxed{frac{2}{27}}$
More generally, it can be shown that
$$int_{0}^{infty} x^{n}e^{-ax} mathop{dx} = frac{n!}{a^{n + 1}}.$$
Here, we have $n = 2$ and $a = 3$, so the result follows.
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
answered Dec 12 '18 at 5:46
Ekesh KumarEkesh Kumar
1,00428
1,00428
add a comment |
add a comment |
$begingroup$
using the substitution $u = 3x$ the integral becomes:
$$
I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
= frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
= frac{Gamma(3)}{27} \
= frac2{27}
$$
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
$endgroup$
add a comment |
$begingroup$
using the substitution $u = 3x$ the integral becomes:
$$
I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
= frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
= frac{Gamma(3)}{27} \
= frac2{27}
$$
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
$endgroup$
add a comment |
$begingroup$
using the substitution $u = 3x$ the integral becomes:
$$
I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
= frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
= frac{Gamma(3)}{27} \
= frac2{27}
$$
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
$endgroup$
using the substitution $u = 3x$ the integral becomes:
$$
I = int_0^{infty} bigg(frac{u}3bigg)^2 e^{-u} dbigg(frac{u}3bigg) \
= frac1{27} int_0^{infty} u^{3-1} e^{-u} du \
= frac{Gamma(3)}{27} \
= frac2{27}
$$
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
answered Dec 12 '18 at 5:48
David HoldenDavid Holden
14.9k21224
14.9k21224
add a comment |
add a comment |
1
$begingroup$
That does tend to zero: exponentials beat powers.
$endgroup$
– Lord Shark the Unknown
Dec 12 '18 at 5:07
$begingroup$
See math.stackexchange.com/questions/3015570/…
$endgroup$
– lab bhattacharjee
Dec 12 '18 at 5:17