# Area between two curves

#### liquidheineken

Find the Area between the two functions.

http://www4a.wolframalpha.com/Calculate/MSP/MSP16151chba72gif4040fg0000686h2hea7f943994?MSPStoreType=image/gif&s=54&w=381.&h=306.&cdf=Coordinates&cdf=Tooltips [Broken]

I know the bounds are from x=[-1,1] which gives me the equation...
∫ 2/(1+x4) - x2 dx

I just can't figure out how to integrate 2/(1+x4). Substitution won't work.

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#### mooncrater

This type of integration is a special case.
To do it:
Do you see that 2 in the numerator?
1)Convert that into $1+x^2+1-x^2$.
2)separate The two fractions.... I. E.
$\frac {1-x^2}{1+x^4}$+$\frac {1+x^2}{1+x^4}$
3)devide numerator and denominator by $x^2$.
4) know that the differential of $x+\frac {1}{x}$ is $1-\frac {1}{x^2}$.
I think these hints are sufficient.

#### Mark44

Mentor
Find the Area between the two functions.

http://www4a.wolframalpha.com/Calculate/MSP/MSP16151chba72gif4040fg0000686h2hea7f943994?MSPStoreType=image/gif&s=54&w=381.&h=306.&cdf=Coordinates&cdf=Tooltips [Broken]

I know the bounds are from x=[-1,1] which gives me the equation...
∫ 2/(1+x4) - x2 dx
Minor point, but the above is NOT an equation -- there's no =, which is the hallmark of an equation.
liquidheineken said:
I just can't figure out how to integrate 2/(1+x4). Substitution won't work.
I would be inclined to try a trig substitution.

BTW, when you post a problem, do not delete the homework template. The three parts are required.

Last edited by a moderator:

#### Mark44

Mentor
This type of integration is a special case.
To do it:
Do you see that 2 in the numerator?
1)Convert that into $1+x^2+1-x^2$.
2)separate The two fractions.... I. E.
$\frac {1-x^2}{1+x^4}$+$\frac {1+x^2}{1+x^4}$
3)devide numerator and denominator by $x^2$.
4) know that the differential of $x+\frac {1}{x}$ is $1-\frac {1}{x^2}$.
This is true, but I don't see how it is relevant to this problem. If you divide numerator and denominator of step 2 by x2, you get $\frac{1/x^2 - 1}{1/x^2 + x^2} + \frac{1/x^2 + 1}{1/x^2 + x^2}$. I don't see how this is an improvement.
mooncrater said:
I think these hints are sufficient.

#### phion

Gold Member
I would also try a trigonometric substitution.

#### mooncrater

This is true, but I don't see how it is relevant to this problem. If you divide numerator and denominator of step 2 by x2, you get $\frac{1/x^2 - 1}{1/x^2 + x^2} + \frac{1/x^2 + 1}{1/x^2 + x^2}$. I don't see how this is an improvement.
You can write $x^2+1/x^2$ as $[x+1/x]^2-2$ as well as $[x-1/x]^2+2$. There fore when you take $t=x+1/x$ you get $dt=1-1/x^2$ and when you take $n=x-1/x$ then you get $dn=1+1/x^2$......
These substitutions can be used to calculate the integral.
Well.. I also would like to see the trigonometric substitution method.
(Are you talking about substituting $x^2=tan\theta$? )

#### phion

Gold Member
I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.

#### mooncrater

I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.
Mathematica? What does it do?

#### phion

Gold Member #### mooncrater Nightmare!!!!!!!! I agree. #### mooncrater

Some integrals have special(and shorter) methods. That just needs recognising the integral type. Otherwise they can be nightmares!!!!!

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