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Area between two curves

  1. May 20, 2015 #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.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 20, 2015 #2
    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.
     
  4. May 20, 2015 #3

    Mark44

    Staff: Mentor

    Minor point, but the above is NOT an equation -- there's no =, which is the hallmark of an equation.
    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: May 7, 2017
  5. May 21, 2015 #4

    Mark44

    Staff: Mentor

    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.
     
  6. May 21, 2015 #5

    phion

    User Avatar
    Gold Member

    I would also try a trigonometric substitution.
     
  7. May 21, 2015 #6
    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##? )
     
  8. May 21, 2015 #7

    phion

    User Avatar
    Gold Member

    I just plugged the integral into Mathematica and it appears to be kind of a nightmare to solve.
     
  9. May 21, 2015 #8
    Mathematica? What does it do?
     
  10. May 21, 2015 #9

    phion

    User Avatar
    Gold Member

  11. May 21, 2015 #10
    Nightmare!!!!!!!! I agree. :))
     
  12. May 21, 2015 #11
    Some integrals have special(and shorter) methods. That just needs recognising the integral type. Otherwise they can be nightmares!!!!!
     
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