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
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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
it would be something like this :
Yfrog Image : yfrog.com/3mgraphtj
Originally Posted by action812 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 $\displaystyle y = e$ is just a horizontal line ... you do know the value of e, correct? $\displaystyle A = \int_1^e \frac{1}{y} \, dy $
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.
Use $\displaystyle \int_1^e\int_0^\frac{1}{y}\! \,dx\,dy=1$.
i post just below sry i misclicked the button and i replayed new one so dumb but you got it check out
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
I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys.
Originally Posted by action812 I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys. not approximately ... exactly one.
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.
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