Find the area S of the region

[Princeofdark]

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.

 

[Office_Shredder]

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

$$\frac{2}{x} =/= x^{-2}$$ I don't know why you would think that... $$\frac{2}{x} = 2* \frac{1}{x}$$ If you're unsure of how fractions and exponentials work this is going to be a difficult problem to solve

 

[HallsofIvy]

I would start by determining where the two graphs intersect. Where is
$$y= \frac{2}{x}= \frac{2}{x^2}$$?

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?

 

[Princeofdark]

i set the = and i think i got (0,1) is that right?

and I am not following the second part, sorry.

 

[HallsofIvy]

i set the = and i think i got (0,1) is that right?

and I am 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?