# Thread: integrate to find area

1. ## 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

2. I am not sure with what you are getting at. Can you just post the question from the book?

3. 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.

4. 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

$\int_1^5 \frac{1}{x^2} dx = [\frac{x^{-2+1}}{-2+1}]_1^5 = [\frac{-1}{x}]_1^5 = [\frac{-1}{5} - \frac{-1}{1}] = \frac{-1}{5} + 1 = \frac{4}{5}$

5. Integrate like you normal would. Look above harish21 did it.

6. this is probably a really stupid question but how did you get -1/x.

i like ur sig btw.

7. 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.

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