area between two curves?

Question

I have a graph that looks like a bowtie

the graph says that y=2x/x^2+4 and x= -2 and x= 2

so with substitution I get
u = x^2+4
du= 2x dx

so that leaves us with
1/u(du)

Then we have to find new x values and they are
0^2+4=4
-2^2+4=8
2^2+4=8
0^2+4=4

This can't be right can it

this can be right

Answer 1

Your question doesn't match its title, as there is only one curve, not two. You want the area between this curve and the x axis?
Yes, you are right so far.

The primitive of 1/u is ln u, and so
The area between the curve and the x axis between -2 and 0 is
[ln u when u = 4] - [ln u when u = 8]
=ln 4 - ln 8
= ln(1/2) which is negative because the curve is below the x axis here. We only want the number of units of area, because if we retain the sign and add the two areas together we get zero, because the graph has point-symmetry (It's an odd function)
So we take absolute value of the integral, and since
ln(1/2) = - ln 2, the abs val is
ln 2

On the right-hand side the area is
[ln u where u = 8 (x=2)] - [ln u where u = 4 (x=0)]
= ln 8 - ln 4
= ln 2, same as before, confirming the symmetry of the graph.

Total area
= ln 2 + ln 2
= 2*ln 2 ... [or, if you like, ln(2*2)]
= ln (2^2)
= ln 4, which is roughly 1.4 square units.

PS Why don't you allow email? It makes it easy to discuss your question further with an answerer, and your email address doesn't get revealed -- the messages all pass through Yahoo Answers.

Answer 2

You are almost there. The graph is symmetrical so:
int[-2 to 2, 2x/(x^2+4)]=2*int[0 to 2, 2x/(x^2+4)]

let u=x^2+4, du=2xdx
x=0 => u=4, x=2 => u=8

so we have
int[4 to 8, 1/u] = ln|u| from 4 to 8 = ln8-ln4 = ln(8/4) = ln2

Answer 3

I don't understand your question. What are the 2 curves? One is y = 2x/(x^2 + 4). I suppose. And the other?