# Math Help - Sketch the region and find the area.

1. ## Sketch 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.)

2. 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

3. it would be something like this :

4. 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.)

$y = e$ is just a horizontal line ... you do know the value of e, correct?

$
A = \int_1^e \frac{1}{y} \, dy
$

5. 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.

6. ## Solution

Use $\int_1^e\int_0^\frac{1}{y}\! \,dx\,dy=1$.

7. i post just below sry i misclicked the button and i replayed new one so dumb but you got it check out

8. In fact, you are being asked to find:

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

EDIT: This post would have been after goroner's first post in this thread, but you all posted while I was typing the LaTeX

9. I got the area to be approximately 1. Can anyone confirm this? Thanks for all the help guys.

10. 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.

11. 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.