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

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

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

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

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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

or

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

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[ 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.

[/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

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

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so as the question in the original post said,,it should be the region under y=x(as i knew) and therefore which gives exp(-1)

ahh now i see that is above the line,,,but

why do we start with dydx and where does y= come from in

my only thoughts as to why we start dydx can be that the region is bounded by x= and this needs to be the last integral calculated with the upper limit 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 but how were the limits changed and what procedure would youuse to get the rhs without the lhs

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

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