$\displaystyle \int^{\infty}_{0} \int^{x}_{0}\frac {e^{-{\frac {x}{y}-y}}}{y} dy dx = \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx$

in ref to http://www.mathhelpforum.com/math-he...ob-153251.html i dont think it is

Results 1 to 5 of 5

- Aug 11th 2010, 12:37 AM #1
## is this possible?

$\displaystyle \int^{\infty}_{0} \int^{x}_{0}\frac {e^{-{\frac {x}{y}-y}}}{y} dy dx = \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx$

in ref to http://www.mathhelpforum.com/math-he...ob-153251.html i dont think it is

- Aug 11th 2010, 03:33 AM #2

- Joined
- Apr 2005
- Posts
- 19,714
- Thanks
- 3003

You are right. They are not the same The first integral takes x going from 0 to infinity and, for each x, y going from 0 to x. That integral is over the region in the first quadrant bounded by the x-axis, y= 0, and by the line y= x, that is, the area in the first quadrant

**below**the line y= x. The second integral take y going from 0 to infinity and, for each y, x going from 0 to y- the region in the first quadrant bounded by the y-axis, x= 0, and by the line y= x. That is the region in the first quadrant**above**the line y= x.

What would be correct is either

$\displaystyle \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy$

or

$\displaystyle \int_0^\infty\int_x^\infty f(x,y)dydx= \int_0^\infty\int_0^y f(x,y) dxdy$

- Aug 11th 2010, 03:50 AM #3

- Joined
- Aug 2010
- Posts
- 20

[ going from 0 to infinity and,

**for each y**, x going from 0 to y- the region in the first quadrant bounded by the y-axis, x= 0, and by the line y= x. That is the region in the first quadrant**above**the line y= x.

$\displaystyle \int_0^\infty\int_x^\infty f(x,y)dydx= \int_0^\infty\int_0^y f(x,y) dxdy$[/QUOTE]

which region is this then? and do you not mean for each x?

bare with me a second,,i believe youve answered my question giving exp(-1) as the answer as opposed to 1-exp(-1) which gets me somewhere,,just a sec

- Aug 11th 2010, 03:53 AM #4

- Joined
- Aug 2010
- Posts
- 20

so as the question in the original post said,,it should be the region under y=x(as i knew) and therefore$\displaystyle \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy$ which gives exp(-1)

ahh now i see that $\displaystyle \int_0^\infty\int_0^y f(x,y) dxdy$ is above the line,,,but

why do we start with dydx and where does y=$\displaystyle \infty$ come from in $\displaystyle \int_0^\infty\int_y^\infty f(x,y)dxdy$

my only thoughts as to why we start dydx can be that the region is bounded by x=$\displaystyle \infty$ and this needs to be the last integral calculated with the upper limit $\displaystyle \infty$ to give a numeric answer although i cant convince myself this is a valid way to tackle the question

ok cool so the answer is $\displaystyle \int_0^\infty\int_0^x f(x,y)dydx= \int_0^\infty\int_y^\infty f(x,y)dxdy$ but how were the limits changed and what procedure would youuse to get the rhs without the lhs

- Aug 11th 2010, 04:20 AM #5

- Joined
- Aug 2010
- Posts
- 20

YouTube - ‪Double Integrals - Changing Order of Integration - Full Ex.‬‎

ok its cool i think i got it now,,,i can start dxdy if i wanted or dydx doesnt matter the answer is exp(-1)