Integrate $int_D 2e^{-x^2}$ with $D = lbrace (x,y) in Bbb{R}^2 | 0le x le 1, 0 le y le x rbrace$











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I was given the following task but I don't know how to keep going.
The task is:




Integrate $int_D 2e^{-x^2}$ with $D = lbrace (x,y) in Bbb{R}^2 |
> 0le x le 1, 0 le y le x rbrace$




I started with the following integral:



$$int_0^1 int_0^x 2cdot e^{-x^2}dy;dx$$
$$=int_0^1 left[2cdot e^{-x^2} right]_0^xdx$$
$$=int_0^1 xcdot2cdot e^{-x^2}dx$$



I am stuck at evaluating this integral.
I assumed that $erf(x)$ will appear.



The thing is that my university published the solution:



$$int_D 2e^{-x^2} = 1 - dfrac{1}{e}$$



I would be very happy if someone could explain to me, what I did wrong.



Greetings,
Finn






integration







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I was given the following task but I don't know how to keep going.
    The task is:




    Integrate $int_D 2e^{-x^2}$ with $D = lbrace (x,y) in Bbb{R}^2 |
    > 0le x le 1, 0 le y le x rbrace$




    I started with the following integral:



    $$int_0^1 int_0^x 2cdot e^{-x^2}dy;dx$$
    $$=int_0^1 left[2cdot e^{-x^2} right]_0^xdx$$
    $$=int_0^1 xcdot2cdot e^{-x^2}dx$$



    I am stuck at evaluating this integral.
    I assumed that $erf(x)$ will appear.



    The thing is that my university published the solution:



    $$int_D 2e^{-x^2} = 1 - dfrac{1}{e}$$



    I would be very happy if someone could explain to me, what I did wrong.



    Greetings,
    Finn






    integration







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I was given the following task but I don't know how to keep going.
      The task is:




      Integrate $int_D 2e^{-x^2}$ with $D = lbrace (x,y) in Bbb{R}^2 |
      > 0le x le 1, 0 le y le x rbrace$




      I started with the following integral:



      $$int_0^1 int_0^x 2cdot e^{-x^2}dy;dx$$
      $$=int_0^1 left[2cdot e^{-x^2} right]_0^xdx$$
      $$=int_0^1 xcdot2cdot e^{-x^2}dx$$



      I am stuck at evaluating this integral.
      I assumed that $erf(x)$ will appear.



      The thing is that my university published the solution:



      $$int_D 2e^{-x^2} = 1 - dfrac{1}{e}$$



      I would be very happy if someone could explain to me, what I did wrong.



      Greetings,
      Finn






      integration







      share|cite|improve this question















      I was given the following task but I don't know how to keep going.
      The task is:




      Integrate $int_D 2e^{-x^2}$ with $D = lbrace (x,y) in Bbb{R}^2 |
      > 0le x le 1, 0 le y le x rbrace$




      I started with the following integral:



      $$int_0^1 int_0^x 2cdot e^{-x^2}dy;dx$$
      $$=int_0^1 left[2cdot e^{-x^2} right]_0^xdx$$
      $$=int_0^1 xcdot2cdot e^{-x^2}dx$$



      I am stuck at evaluating this integral.
      I assumed that $erf(x)$ will appear.



      The thing is that my university published the solution:



      $$int_D 2e^{-x^2} = 1 - dfrac{1}{e}$$



      I would be very happy if someone could explain to me, what I did wrong.



      Greetings,
      Finn






      integration




      integration






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      share|cite|improve this question













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      edited Nov 22 at 12:57

























      asked Nov 22 at 12:51









      Finn Eggers

      364213




      364213






















          1 Answer
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          down vote



          accepted










          In $I=int 2xe^{x^2}dx$,
          let $y=x^2$.



          Then $dy = 2xdx$ so
          $I=int e^ydy$.






          share|cite|improve this answer

















          • 1




            Oh.. Wow I feel pretty dumb now. Thanks a lot!
            – Finn Eggers
            Nov 22 at 12:58










          • You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
            – marty cohen
            Nov 22 at 18:45











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          In $I=int 2xe^{x^2}dx$,
          let $y=x^2$.



          Then $dy = 2xdx$ so
          $I=int e^ydy$.






          share|cite|improve this answer

















          • 1




            Oh.. Wow I feel pretty dumb now. Thanks a lot!
            – Finn Eggers
            Nov 22 at 12:58










          • You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
            – marty cohen
            Nov 22 at 18:45















          up vote
          1
          down vote



          accepted










          In $I=int 2xe^{x^2}dx$,
          let $y=x^2$.



          Then $dy = 2xdx$ so
          $I=int e^ydy$.






          share|cite|improve this answer

















          • 1




            Oh.. Wow I feel pretty dumb now. Thanks a lot!
            – Finn Eggers
            Nov 22 at 12:58










          • You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
            – marty cohen
            Nov 22 at 18:45













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          In $I=int 2xe^{x^2}dx$,
          let $y=x^2$.



          Then $dy = 2xdx$ so
          $I=int e^ydy$.






          share|cite|improve this answer












          In $I=int 2xe^{x^2}dx$,
          let $y=x^2$.



          Then $dy = 2xdx$ so
          $I=int e^ydy$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 12:58









          marty cohen

          72k547126




          72k547126








          • 1




            Oh.. Wow I feel pretty dumb now. Thanks a lot!
            – Finn Eggers
            Nov 22 at 12:58










          • You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
            – marty cohen
            Nov 22 at 18:45














          • 1




            Oh.. Wow I feel pretty dumb now. Thanks a lot!
            – Finn Eggers
            Nov 22 at 12:58










          • You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
            – marty cohen
            Nov 22 at 18:45








          1




          1




          Oh.. Wow I feel pretty dumb now. Thanks a lot!
          – Finn Eggers
          Nov 22 at 12:58




          Oh.. Wow I feel pretty dumb now. Thanks a lot!
          – Finn Eggers
          Nov 22 at 12:58












          You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
          – marty cohen
          Nov 22 at 18:45




          You are welcome. It's just the standard special case of $int f'(x)e^{f(x)}dx = int e^ydy = e^y = e^{f(x)}$ with $y = f(x)$.
          – marty cohen
          Nov 22 at 18:45


















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