Area between two curves

  • #1
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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Answers and Replies

  • #2
217
18
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.
 
  • #3
34,172
5,786
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:
  • #4
34,172
5,786
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.
 
  • #5
phion
Gold Member
175
39
I would also try a trigonometric substitution.
 
  • #6
217
18
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##? )
 
  • #7
phion
Gold Member
175
39
I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.
 
  • #8
217
18
I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.
Mathematica? What does it do?
 
  • #9
phion
Gold Member
175
39
Wolfram_Alpha_integrate_2_1_x_4_x_2_from_1.gif
 
  • #11
217
18
Some integrals have special(and shorter) methods. That just needs recognising the integral type. Otherwise they can be nightmares!!!!!
 

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