integrate to find area

• Apr 8th 2010, 05:49 PM
FinalFantasy9291
integrate to find area
i am doing area between curves. now i im on like the third step. i did the top minus bottom and now my integral is this.

integral going from 1 to 5 (1/x^2) dx...

now the next step i see is that that equals [-1/x]with integral 1 to 5. how do i get from the first integral to the second?

thx
• Apr 8th 2010, 05:53 PM
dwsmith
I am not sure with what you are getting at. Can you just post the question from the book?
• Apr 8th 2010, 05:56 PM
FinalFantasy9291
okay it says sketch the region bounded by the graphs of the equasions, and determien the area of the region.

the problem is y=1/x^2, y=0, x=1, x=5.

so that means the integrals are going from 1 to 5. i know how to sketch the graph. and then so i got

integral from 1 to 5 (1/x^2) dx because there are vertical slices. not sure how to integrate after that to find the area. lol sorry if im being confusing.
• Apr 8th 2010, 05:58 PM
harish21
Quote:

Originally Posted by FinalFantasy9291
i am doing area between curves. now i im on like the third step. i did the top minus bottom and now my integral is this.

integral going from 1 to 5 (1/x^2) dx...

now the next step i see is that that equals [-1/x]with integral 1 to 5. how do i get from the first integral to the second?

thx

$\int_1^5 \frac{1}{x^2} dx = [\frac{x^{-2+1}}{-2+1}]_1^5 = [\frac{-1}{x}]_1^5 = [\frac{-1}{5} - \frac{-1}{1}] = \frac{-1}{5} + 1 = \frac{4}{5}$
• Apr 8th 2010, 06:01 PM
dwsmith
Integrate like you normal would. Look above harish21 did it.
• Apr 8th 2010, 06:04 PM
FinalFantasy9291
this is probably a really stupid question but how did you get -1/x.

i like ur sig btw.
• Apr 8th 2010, 06:10 PM
harish21
Quote:

Originally Posted by FinalFantasy9291
this is probably a really stupid question but how did you get -1/x.

i like ur sig btw.

You need to have a good look at your book and learn the basic integration skills.

$\int \frac{1}{x^2} dx = \int x^{-2} dx= \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = - {x^{-1}} = \frac{-1}{x}$