Find the area S of the region

  • #1
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.

http://www.webassign.net/www28/symImages/2/b/4def6fad2db11c7dd01c3a136b1fba.gif


Then find the area S of the region.


1. How do i know whether to integrate with respect to x or y?


2. is 2/x the same thing as x^-2 ?? (if not how do i rewrite it) ?


help.

i graphed it and didn't know what to do with the x = 7.
 
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Answers and Replies

  • #2
Office_Shredder
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x=7 is going to be a vertical line, all points of the form (7,y).

You can integrate with respect to either x or y, try writing both integrals out and see which one looks easier

[tex] \frac{2}{x} =/= x^{-2}[/tex] I don't know why you would think that... [tex]\frac{2}{x} = 2* \frac{1}{x}[/tex] If you're unsure of how fractions and exponentials work this is going to be a difficult problem to solve
 
  • #3
HallsofIvy
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I would start by determining where the two graphs intersect. Where is
[tex]y= \frac{2}{x}= \frac{2}{x^2}[/tex]?

Now, if y is a number between those two y values, for what x values is y= 1/x or y= 1/x2?
If x is number between the two x-values of the intersection points, what are y= 1/x and y= 1/x2?

Which of those is easier to integrate?
 
  • #4
i set the = and i think i got (0,1) is that right?

and im not following the second part, sorry.
 
  • #5
HallsofIvy
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i set the = and i think i got (0,1) is that right?

and im not following the second part, sorry.
No, that's not right. How did you do it? Did you put x= 0 and x= 1 back into the equation to check?
 

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