Sketch the region and find the area enclosed by the curves.

x = 1/y which is y=1/x (right?)

x=0

y=1

y=e ( this is the one I'm having trouble graphing/figuring out.)

Any help please? Thanks :D

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- Jun 1st 2010, 10:26 AMaction812Sketch the region and find the area.
Sketch the region and find the area enclosed by the curves.

x = 1/y which is y=1/x (right?)

x=0

y=1

y=e ( this is the one I'm having trouble graphing/figuring out.)

Any help please? Thanks :D - Jun 1st 2010, 10:44 AMUnknown008
e is the natural constant, and if you type it in your calculator, you'll have the value of e as 2.718281828

So, you graph y = 2.7 - Jun 1st 2010, 10:46 AMgoroner
it would be something like this :

- Jun 1st 2010, 10:47 AMgoroner
- Jun 1st 2010, 10:48 AMskeeter
- Jun 1st 2010, 10:48 AMaction812
Duh! Thanks Unknown, I can't believe I didn't put 2 and 2 together. Gonna work on it now...

And goroner, you didn't post anything?

EDIT:: Nevermind, I see it now. - Jun 1st 2010, 10:48 AMbigliSolution
Use $\displaystyle \int_1^e\int_0^\frac{1}{y}\! \,dx\,dy=1$.

- Jun 1st 2010, 10:49 AMgoroner
i post just below sry i misclicked the button and i replayed new one so dumb but you got it check out

- Jun 1st 2010, 10:50 AMUnknown008
In fact, you are being asked to find:

$\displaystyle \int^e_1 \frac{1}{y} dy$

EDIT: This post would have been after goroner's first post in this thread, but you all posted while I was typing the LaTeX :p - Jun 1st 2010, 10:59 AMaction812
I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys.

- Jun 1st 2010, 11:01 AMskeeter
- Jun 1st 2010, 11:04 AMUnknown008
Yes, integrating, you get ln e - ln 1 = 1 - 0 = 1 units^2 :)

EDIT: Yes, if you use a graph instead of the integration, you would get an approximation.