# Area Between two curves - can't figure it out

1. May 1, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

So the curves are:

$$y = \frac {x^{4}}{x^{2}+1}$$

and

$$y = \frac {1}{x^{2}+1}$$

3. The attempt at a solution

So first the limits of integration are:

$$\frac{x^{4}}{x^{2}+1} = \frac{1}{x^{2}+1}$$

$$(x^{2}+1)\frac{x^{4}}{x^{2}+1} = \frac{1}{x^{2}+1} (x^{2}+1)$$

$$x^{4} = 1$$

$$x^{4}-1 = 0$$

$$(x^{2}+1)(x^{2}-1)= 0$$

so

$$x^{2}+1 = 0 \rightarrow x^{2} =-1$$

and since:

$$\sqrt{-1}=i$$

I assume that it should be the negative square root of 1:

$$x= -\sqrt{1} = -1$$

and also:

$$x^{2}-1 = 0 \rightarrow x^{2}=1 \rightarrow x= \sqrt{1}=1$$

so the integral should go from -1 to 1

Also I can tell from the graph of the two functions that:

$$y = \frac {1}{x^{2}+1}$$

bounds

$$y = \frac {x^{4}}{x^{2}+1}$$

so the integral should be set up this way (I will use two integrals purely for clarity):

$$\int^{1}_{-1} (x^{2}+1)^{-1} dx - \int^{1}_{-1} \frac{x^{4}}{x^{2}+1} dx$$

I am fairly certain that the integral on the left can be evaluated using this anti-derivative:

$$\frac{1}{2x}ln(x^{2}+1)$$

which I got by somewhat trial and error as u substitution did me no good on that problem, but that's easy one, I can not figure out the right side for the life of me, here's my shot at it, but its not very good.

Let:

$$u = x^{2}+1$$

$$du = 2xdx$$

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

so the problem now looks like:

$$\int^{1}_{-1} \frac{1}{2}\frac{x^{4}}{(u)(x)} du$$

and the x will remove 1 exponent so:

$$\int^{1}_{-1} \frac{1}{2}\frac{x^{3}}{(u)} du$$

which can be put:

$$\int^{1}_{-1} \frac{1}{2}(x)^{3}(u)^{-1} du$$

and since

$$u = x^{2}+1 \rightarrow x = \sqrt{u-1}$$

so now its:

$$\int^{1}_{-1} \frac{1}{2}(\sqrt{u-1})^{3}(u)^{-1} du$$

and

$$(\sqrt{u-1})^{3} = (u-1)\sqrt{u-1} = u\sqrt{u-1} -\sqrt{u-1}$$

and put together:

$$\int^{1}_{-1} \frac{1}{2}\frac{u\sqrt{u-1} -\sqrt{u-1}}{u} du$$

which cancels 1 u to make it:

$$\int^{1}_{-1} \frac{1}{2}\sqrt{u-1} -\frac {\sqrt{u-1}}{u} du$$

and I am just not seeing this work out from here.... did I make this too hard or what?

I guess I should say that I have tried u = x^4 as well, and it does work out slightly better but I get stuck at the point of:

$$\int^{1}_{-1} \frac{1}{4} \frac{u}{(u+u^{3/4})}$$

If i go that way I can't figure out how to do:

$$u * (u+u^{3/4})^{-1}$$

if it wasn't the reciprocal of u+u^3/4 it would go....

Last edited: May 1, 2010
2. May 1, 2010

### The Chaz

The "left integral" (i.e. anti-derivative) isn't right. In your trial and error, don't forget the product rule! Your anti-derivative should include the inverse tangent.

As for the "right integral", I'm going to do the work offline and show you the result (mostly since you've done a ton of work already). Here's the gist of it:
POLYNOMIAL LONG DIVISION. You should get a couple of easy terms to integrate, then

1) a ln(x^2 + 1) anti-derivative, and/or
2) another inverse tangent (from a constant remainder in the division)

btw... you could just combine the integrals before the long division. Not sure if that will speed things up...
DUH. Yeah, that will speed things up a LOT!

Let's integrate -(x^4 -1)/(x^2 + 1)
First, rewrite as -(x^2 + 1)(x^2 - 1)/(x^2 + 1)
cancel, which is ok since that factor never equals zero...
-(x^2 - 1) = -x^2 + 1. EZ

Last edited: May 1, 2010
3. May 1, 2010

### Asphyxiated

Ok man so if we rewrite the integral as:

$$\int^{1}_{-1} \frac {x^4-1}{x^{2}+1} dx$$

which is:

$$\int^{1}_{-1} \frac {(x^{2}+1)(x^{2}-1)}{x^{2}+1} dx$$

where did you get the -(x^{2}+1) from? I don't see anything in there to make it -... anyway it goes like so:

$$\int^{1}_{-1} x^{2}-1 dx$$

and for my equation the anti-derivative is:

$$\frac{1}{3}x^{3}-x$$

or yours would be:

$$-\frac{1}{3}x^{3}-x$$

so now that we have integrated the two equations does that mean that F(b)-F(a) of the anti-derivative will give me the answer to the area of the for the problem? just checking because I am not sure how it will work out since we combined them before integrating

4. May 1, 2010

### The Chaz

Right! Check my edit, which is now correct.
The first line of your most current reply is missing the (-1), since it's actually
1 - x^4.

I've really gotta learn latex... maybe when school breaks for the summer ;)

5. May 1, 2010

### Asphyxiated

Ah right..... cause of the subtraction order...

$$\int^{1}_{-1} \frac{1-x^{4}}{x^{2}+1} dx$$

$$\int^{1}_{-1} \frac{-(x^{4}+1)}{x^{2}+1} dx$$

$$\int^{1}_{-1} \frac{-(x^{2}+1)(x^{2}-1)}{x^{2}+1} dx$$

$$\int^{1}_{-1} -(x^{2}-1)dx$$

which is:

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

then of course:

$$-\frac{1}{3}x^{3}+x$$

I wrote that out so you could confirm that that is the correct way to get there for the start. Also latex is really not all that hard to learn for bulletin purposes. It gets complicated if you are actually writing a paper in latex (which is actually incredibly helpful to do because it will automatically link your sources and create sections and a full bibliography for you just by using the right code).

for the bulletin board here it is really simple, for an equation just put the equation between{tex} {/tex} but with brackets instead of curls you know, not {} but []. the bottom line of my equation is just

{tex} \frac {1}{3} x^{3}+x {/tex}

the only thing that is different in that line than the line that parses is the curls on the tex statements at the beginning and end.

6. May 1, 2010

### The Chaz

You're golden! Now use the FTC and you'll get a number. When (in the course of time) you eventually see something like 1/(x^2 + 1), use a trig substitution. But I assume you'll learn that when the time comes.
cya

7. May 1, 2010

### Asphyxiated

Integrals with trig substitution is the next chapter :O, did you get the last part of my reply? I edited it after posting, I tried to give you a simple guideline to post latex equations on bulletin boards, it will help you out immensely because it is so much easier for people to read and understand.

8. May 1, 2010

### The Chaz

No I read it too quickly! Thanks for the tips.