Question

can anyone help?
Find the area of the region between the curves y=1/x,y=1/x^2,x=2 ?

Answer 1

Are you referring to the area between the curves from x=1 to x=2? See the attached.

Answer 2

im not quite sure? thats all they gave us?

Answer 3

Referring to x >= 1, the upper graph is 1/x (red) and the lower graph is 1/x^2 (blue). If this is the area for which you want the integral, then you can get the area by the following:\[A = \int\limits_{1}^{2}(\frac{ 1 }{ x } - \frac{ 1 }{ x ^{2} }) dx\]It would seem from both the graph and algebraically that the bounds on "x" fro which you want the area are "x" from 1 to 2. I would think that since you were given x=2, the value of x=1 forms a natural boundary because then the graphs interpose. So, all you have to do is evaluate this integral. Are you able to do that or do you need further help? Also, is this graph helpful?

Answer 4

oh yes i get you know :) but i may need further help let me try integrate this

Answer 5

ok, I'll stick around. The integral shouldn't be too hard. I figured it out already and I can confirm your answer once you get one.

Answer 6

is it 0.193147?

Answer 7

Yes! Good job!

Answer 8

thanks :) how did you draw the graph if you didnt know what x was?

Answer 9

Good luck to you in all of your studies and thx for the recognition! @mags093

And you're welcome!

Answer 10

could you help me with this one too? Find the area of the region between the curves y=|x| y=X^2-2 ?

Answer 11

I didn't need to know "x" to draw the graph. And really, I didn't have to draw the graph at all. I knew that they intersected at x=1. Also, that grid in the graph is helpful with 2x2 squares giving 1 square. It's easy to eyeball that the area in question is about 1/5.

And sure, I'll help with this one too if you like.

Answer 12

ok thanks because it helped better when i could see the graph :) yes please if you wouldnt mind?

Answer 13

If that second equation is y = x^2 - 2, then your graph looks like:

Answer 14

how did you make that graph? i tried and it came out like this http://assets.openstudy.com/updates/attachments/511eb4ade4b03d9dd0c4c6ce-tcarroll010-1360968679818-graph_040.png

Answer 15

it didnt show up like that a second ago? :/

Answer 16

You can forget about the absolute value and just use symmetry to get the area as:\[A = 2 \times \int\limits_{0}^{2}[x - (x ^{2} - 2)] dx\]As for making that graph, I used meta-calculator.

Answer 17

http://www.meta-calculator.com/online/

Answer 18

i was doing that but from 0 to 1 instead so thats where i was going wrong? is it from 0 to 2 because of the -2?

Answer 19

thanks for the graphing calculator :)

Answer 20

The whole area in question really from x = -2 to +2, but I went from 0 to +2 and then doubled the area because of symmetry. I chose 2 as the right boundary because that's where the graphs intersect.

Answer 21

oh ok :)

Answer 22

You're welcome about the graphing tool. I used the old version where you have to put in a "*" character for multiplication.

Answer 23

ya we use that for writing multiplication too so i understood it :) thanks

Answer 24

Did you figure your integral yet? I can verify your answer if you work it out.

Answer 25

i think its 20/3?

Answer 26

You got it right! Good job!

Answer 27

So, nice working with you and I hope you have a good rest of your day.

Answer 28

yay thanks :) same to you! too bad its night time here haha

Answer 29

It's late afternoon, but I feel a nap coming on!

Answer 30

have a good one :)

Answer 31

u2, bye now. @mags093