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
asked Nov 22 at 12:51
Finn Eggers
364213
add a comment |
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
asked Nov 22 at 12:51
Finn Eggers
364213
add a comment |
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
asked Nov 22 at 12:51
Finn Eggers
364213
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
asked Nov 22 at 12:51
Finn Eggers
364213
asked Nov 22 at 12:51
Finn Eggers
364213
edited Nov 22 at 12:57
asked Nov 22 at 12:51
Finn Eggers
364213
asked Nov 22 at 12:51
Finn Eggers
364213
asked Nov 22 at 12:51
Finn Eggers
364213
364213
add a comment |
add a comment |
1 Answer
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1
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accepted
In $I=int 2xe^{x^2}dx$,
let $y=x^2$.
Then $dy = 2xdx$ so
$I=int e^ydy$.
answered Nov 22 at 12:58
marty cohen
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In $I=int 2xe^{x^2}dx$,
let $y=x^2$.
Then $dy = 2xdx$ so
$I=int e^ydy$.
answered Nov 22 at 12:58
marty cohen
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
add a comment |
up vote
1
down vote
accepted
In $I=int 2xe^{x^2}dx$,
let $y=x^2$.
Then $dy = 2xdx$ so
$I=int e^ydy$.
answered Nov 22 at 12:58
marty cohen
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
add a comment |
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
$I=int e^ydy$.
answered Nov 22 at 12:58
marty cohen
72k547126
In $I=int 2xe^{x^2}dx$,
let $y=x^2$.
Then $dy = 2xdx$ so
$I=int e^ydy$.
answered Nov 22 at 12:58
marty cohen
72k547126
answered Nov 22 at 12:58
marty cohen
72k547126
answered Nov 22 at 12:58
marty cohen
72k547126
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
add a comment |
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
add a comment |
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