Integral from 0 to infinity involving integration by parts

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I have the following:



$$int_0^infty x e^-x dx$$



How would I use integration of parts when I have the natural e in my problem?



Edit:Solved by the following...



I set $u=x$, $dv=e^-x$, $du=1dx$, and $v=-e^-x$



By parts, I came up with



$$-xe^-x - int -e^-x dx$$



From this I came up with $-xe^-x - e^-x$ from $0$ to $b$, which led to: $(-be^-b +e^-b) - (0-1)$.



Second $b$ term went to zero, used L'hopital's rule on the first one.



Integral equals $1$.



Apologies for not including more info in my original post, and I appreciate any and all feedback.




calculus integration




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  • 1




    Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
    – saulspatz
    Aug 7 at 16:08






  • 1




    Did you try anything when you know you have to integrate by parts?
    – StubbornAtom
    Aug 7 at 16:09






  • 2




    Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
    – Simply Beautiful Art
    Aug 7 at 16:10






  • 1




    I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
    – Simply Beautiful Art
    Aug 7 at 18:35






  • 1




    Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
    – saulspatz
    Aug 7 at 18:38















up vote
0
down vote

favorite












I have the following:



$$int_0^infty x e^-x dx$$



How would I use integration of parts when I have the natural e in my problem?



Edit:Solved by the following...



I set $u=x$, $dv=e^-x$, $du=1dx$, and $v=-e^-x$



By parts, I came up with



$$-xe^-x - int -e^-x dx$$



From this I came up with $-xe^-x - e^-x$ from $0$ to $b$, which led to: $(-be^-b +e^-b) - (0-1)$.



Second $b$ term went to zero, used L'hopital's rule on the first one.



Integral equals $1$.



Apologies for not including more info in my original post, and I appreciate any and all feedback.




calculus integration




share|cite|improve this question

















  • 1




    Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
    – saulspatz
    Aug 7 at 16:08






  • 1




    Did you try anything when you know you have to integrate by parts?
    – StubbornAtom
    Aug 7 at 16:09






  • 2




    Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
    – Simply Beautiful Art
    Aug 7 at 16:10






  • 1




    I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
    – Simply Beautiful Art
    Aug 7 at 18:35






  • 1




    Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
    – saulspatz
    Aug 7 at 18:38













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following:



$$int_0^infty x e^-x dx$$



How would I use integration of parts when I have the natural e in my problem?



Edit:Solved by the following...



I set $u=x$, $dv=e^-x$, $du=1dx$, and $v=-e^-x$



By parts, I came up with



$$-xe^-x - int -e^-x dx$$



From this I came up with $-xe^-x - e^-x$ from $0$ to $b$, which led to: $(-be^-b +e^-b) - (0-1)$.



Second $b$ term went to zero, used L'hopital's rule on the first one.



Integral equals $1$.



Apologies for not including more info in my original post, and I appreciate any and all feedback.




calculus integration




share|cite|improve this question













I have the following:



$$int_0^infty x e^-x dx$$



How would I use integration of parts when I have the natural e in my problem?



Edit:Solved by the following...



I set $u=x$, $dv=e^-x$, $du=1dx$, and $v=-e^-x$



By parts, I came up with



$$-xe^-x - int -e^-x dx$$



From this I came up with $-xe^-x - e^-x$ from $0$ to $b$, which led to: $(-be^-b +e^-b) - (0-1)$.



Second $b$ term went to zero, used L'hopital's rule on the first one.



Integral equals $1$.



Apologies for not including more info in my original post, and I appreciate any and all feedback.





calculus integration





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 7 at 18:32









Simply Beautiful Art

49.4k572172




49.4k572172









asked Aug 7 at 16:04









jackbenimbo

677




677







  • 1




    Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
    – saulspatz
    Aug 7 at 16:08






  • 1




    Did you try anything when you know you have to integrate by parts?
    – StubbornAtom
    Aug 7 at 16:09






  • 2




    Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
    – Simply Beautiful Art
    Aug 7 at 16:10






  • 1




    I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
    – Simply Beautiful Art
    Aug 7 at 18:35






  • 1




    Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
    – saulspatz
    Aug 7 at 18:38













  • 1




    Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
    – saulspatz
    Aug 7 at 16:08






  • 1




    Did you try anything when you know you have to integrate by parts?
    – StubbornAtom
    Aug 7 at 16:09






  • 2




    Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
    – Simply Beautiful Art
    Aug 7 at 16:10






  • 1




    I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
    – Simply Beautiful Art
    Aug 7 at 18:35






  • 1




    Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
    – saulspatz
    Aug 7 at 18:38








1




1




Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
– saulspatz
Aug 7 at 16:08




Well you have to choose the parts. What do want to integrate and what to you want to differentiate?
– saulspatz
Aug 7 at 16:08




1




1




Did you try anything when you know you have to integrate by parts?
– StubbornAtom
Aug 7 at 16:09




Did you try anything when you know you have to integrate by parts?
– StubbornAtom
Aug 7 at 16:09




2




2




Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
– Simply Beautiful Art
Aug 7 at 16:10




Hi and welcome! I see you are still learning some of the ropes concerning this site, so here's another: questions lacking context may become closed, especially questions of the form "Here's my problem" and not much else. For example, have you tried taking $x$ or $e^-x$ as your $u$ or $v$? If so, please try adding this in, as it'll also help us help you.
– Simply Beautiful Art
Aug 7 at 16:10




1




1




I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
– Simply Beautiful Art
Aug 7 at 18:35




I have also edited it to use MathJax. Note that ... acts like the parentheses in e^(-x) for the code, so you don't end up with things like $e^-x$ instead of $e^-x$.
– Simply Beautiful Art
Aug 7 at 18:35




1




1




Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
– saulspatz
Aug 7 at 18:38





Now it looks correct to me. Of course, if you just want to check that you got the right answer, you can plug it into Wolfram Alpha
– saulspatz
Aug 7 at 18:38











1 Answer
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Using integration by parts



$$
int_0^infty x e^-x , dx = left[ x (-e^-x) right]_0^infty - int_0^infty 1 (-e^-x) , dx = int_0^infty e^-x , dx = 1
$$






Using another technique that can sometimes be useful



First let
$$
f(t) := int_0^infty e^-xt , dx = frac1t
$$
Then,
$$
int_0^infty x e^-xt , dx = -f'(t) = frac1t^2
$$
and setting $t=1$ gives our integral,
$$
int_0^infty x e^-x , dx = frac11^2 = 1.
$$






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    1 Answer
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    down vote













    Using integration by parts



    $$
    int_0^infty x e^-x , dx = left[ x (-e^-x) right]_0^infty - int_0^infty 1 (-e^-x) , dx = int_0^infty e^-x , dx = 1
    $$






    Using another technique that can sometimes be useful



    First let
    $$
    f(t) := int_0^infty e^-xt , dx = frac1t
    $$
    Then,
    $$
    int_0^infty x e^-xt , dx = -f'(t) = frac1t^2
    $$
    and setting $t=1$ gives our integral,
    $$
    int_0^infty x e^-x , dx = frac11^2 = 1.
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      Using integration by parts



      $$
      int_0^infty x e^-x , dx = left[ x (-e^-x) right]_0^infty - int_0^infty 1 (-e^-x) , dx = int_0^infty e^-x , dx = 1
      $$






      Using another technique that can sometimes be useful



      First let
      $$
      f(t) := int_0^infty e^-xt , dx = frac1t
      $$
      Then,
      $$
      int_0^infty x e^-xt , dx = -f'(t) = frac1t^2
      $$
      and setting $t=1$ gives our integral,
      $$
      int_0^infty x e^-x , dx = frac11^2 = 1.
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Using integration by parts



        $$
        int_0^infty x e^-x , dx = left[ x (-e^-x) right]_0^infty - int_0^infty 1 (-e^-x) , dx = int_0^infty e^-x , dx = 1
        $$






        Using another technique that can sometimes be useful



        First let
        $$
        f(t) := int_0^infty e^-xt , dx = frac1t
        $$
        Then,
        $$
        int_0^infty x e^-xt , dx = -f'(t) = frac1t^2
        $$
        and setting $t=1$ gives our integral,
        $$
        int_0^infty x e^-x , dx = frac11^2 = 1.
        $$






        share|cite|improve this answer













        Using integration by parts



        $$
        int_0^infty x e^-x , dx = left[ x (-e^-x) right]_0^infty - int_0^infty 1 (-e^-x) , dx = int_0^infty e^-x , dx = 1
        $$






        Using another technique that can sometimes be useful



        First let
        $$
        f(t) := int_0^infty e^-xt , dx = frac1t
        $$
        Then,
        $$
        int_0^infty x e^-xt , dx = -f'(t) = frac1t^2
        $$
        and setting $t=1$ gives our integral,
        $$
        int_0^infty x e^-x , dx = frac11^2 = 1.
        $$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 7 at 19:33









        md2perpe

        6,07111022




        6,07111022






















             

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