Find the area between two curves - Help finding the limits of integration - Again

  • #1
264
0

Homework Statement



The curves are:

[tex] y = \frac{x^{4}}{x^{2}+1} [/tex]

and

[tex] y = \frac{1}{x^{2}+1} [/tex]


The Attempt at a Solution



So again I assume that:

[tex] \frac {x^{4}}{x^{2}+1} = \frac {1}{x^{2}+1} [/tex]

and then cross multiply:

[tex] (x^{2}+1) = x^{4}(x^{2}+1) [/tex]

not really sure at this point if i should distribute the x^4 but if i do it looks like so:

[tex] (x^{2}+1) = (x^{6}+x^{4}) [/tex]

so:

[tex] (x^{2}+1)-(x^{6}+x^{4}) = 0 [/tex]

and I am not really sure what to do at this point, I do have a polynomial if I do the subtraction which is:

[tex] -x^{6}-x^{4}+x^{2}+1 = 0 [/tex]

but I don't know how to factor it....

thanks guys!
 

Answers and Replies

  • #2
34,010
5,660

Homework Statement



The curves are:

[tex] y = \frac{x^{4}}{x^{2}+1} [/tex]

and

[tex] y = \frac{1}{x^{2}+1} [/tex]


The Attempt at a Solution



So again I assume that:

[tex] \frac {x^{4}}{x^{2}+1} = \frac {1}{x^{2}+1} [/tex]

and then cross multiply:
Don't cross multiply. Multiply both sides by x^2 + 1.
[tex] (x^{2}+1) = x^{4}(x^{2}+1) [/tex]

not really sure at this point if i should distribute the x^4 but if i do it looks like so:

[tex] (x^{2}+1) = (x^{6}+x^{4}) [/tex]

so:

[tex] (x^{2}+1)-(x^{6}+x^{4}) = 0 [/tex]

and I am not really sure what to do at this point, I do have a polynomial if I do the subtraction which is:

[tex] -x^{6}-x^{4}+x^{2}+1 = 0 [/tex]

but I don't know how to factor it....

thanks guys!
 
  • #3
264
0
thanks man, worked out, but I am having a WAY hard time trying to do this problem so I will post the actual problem another thread. thanks for the help here though.
 

Related Threads on Find the area between two curves - Help finding the limits of integration - Again

  • Last Post
Replies
1
Views
987
Replies
1
Views
1K
Replies
2
Views
2K
Replies
4
Views
3K
  • Last Post
Replies
9
Views
5K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
10
Views
688
  • Last Post
Replies
4
Views
913
  • Last Post
Replies
4
Views
4K
Replies
4
Views
4K
Top