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Integration of a func raised to a fraction

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to solve the following integral:

    { x * ((1-x)/3)^(1/7) dx

    where { represents the integral symbol b/c I don't know how to type that.

    3. The attempt at a solution

    I've tried integration by parts and don't get anywhere. None of the integration techiques I know seem to work and nothing I've found online says what to do when you have this kind of a situation.

    Any help would be greatly appreciated. I'm lost. Thanks everyone!!

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 25, 2009 #2

    LeonhardEuler

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    Can you think of a substitution that would make the (1/7) power part easy to integrate?
     
  4. Aug 25, 2009 #3
    I don't think he's coming back (as that was his only post), but I'm relatively new to calc and would like help solving this. Anyone willing to help?
     
  5. Aug 25, 2009 #4
    Thanks for the response,

    Well, I tried to substitute that inside the parenthesis, but there's that blasted extra x outside of it to deal with. In integrating by parts, I made x the integration term(v) and the other part the differential term, but the result is just another integral only more complicated.
     
  6. Aug 25, 2009 #5

    VietDao29

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    No, you don't use Integration by Parts in this problem.

    To type maths beautifully, one should learn how to https://www.physicsforums.com/showthread.php?t=8997". And you can begin to dig in the awesome world of LaTeX today.

    So, your integral is:

    [tex]\int x \left( \frac{1 - x}{3} \right) ^ {\frac{1}{7}} dx[/tex], right?

    The first step is to take all constants out of it:

    [tex]\frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx[/tex]

    The product of (1 - x)1 / 7 and x is a little bit complicated to integrate, right? Because you just cannot expand the part (1 - x)1 / 7.

    If the problem was: [tex]\int x \left( 1 - x \right) ^ {2} dx[/tex], then it would be much easier, because we can expand (1 - x)2 = 1 - 2x + x2. And the whole integral would become: [tex]\int (x ^ 3 - 2 x ^ 2 + x) dx[/tex], which should be like, a piece of cake.

    So, back to the problem, we then think about a simple u-substitution, which eliminates the needs to expand (1 - x)1 / 7. Let's see if you can figure out what u should be. :)
     
    Last edited by a moderator: Apr 24, 2017
  7. Aug 25, 2009 #6
    Would it be u=1/7?
     
  8. Aug 25, 2009 #7
    Letting u equal a constant implies du = 0 and the integral is trivial. The difficulty with the integrand was the (1 - x) being raised to a fractional power. If the base were simpler....


    --Elucidus
     
  9. Aug 26, 2009 #8
    So right now we have [tex]
    \frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx
    [/tex] correct?
    i guess we could distribute the x in front of the parentheses so we'll have[tex]
    \frac{1}{\sqrt[7]{3}} \int \left( x - x^2 \right) ^ {\frac{1}{7}} dx
    [/tex]? then simplify that and pull the negative sign out front? but then there's no substitution...

    the only other idea would be to make a substitution so that u=(1-x)?
     
  10. Aug 26, 2009 #9
    Bring all constants outside then integrate. Use integration by parts. u=x, du=dx,
    dv=(1-x)^1/7 dx, v=-7/8(1-x)^8/7. Any corrections if I integrated it wrong.
     
  11. Aug 26, 2009 #10

    Cyosis

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    This is not correct, you cannot distribute unequal powers. Note that [itex]x=(x^7)^{\frac{1}{7}}[/itex]. Now that they are equal powers you can distribute, which would yield [itex]x(1-x)^{\frac{1}{7}}=(x^7-x^8)^{\frac{1}{7}}[/itex]. This doesn't make things any easier.
    As hinted before use [itex]u=1-x[/itex].
     
  12. Aug 26, 2009 #11
    To go from [itex]dv=(1-x)^{1/7}dx \text{ to }v=-\frac{7}{8}(1-x)^{8/7}[/itex] requires integration by substitution (in fact, using the same substitution that would solve it directly). The integral can be done more simply than by integration by parts - although "by parts" is valid.

    --Elucidus
     
  13. Aug 26, 2009 #12
    I just can't figure it out. I've never formally taken a calculus class, just taught myself out of a cheap self teach book from barnes and noble. I substituted so that u = (1-x) and now have [tex]\frac{1}{\sqrt[7]{3}} \int x \left( u\right) ^ {\frac{1}{7}} dx[/tex]. And, once again i'm stuck.
     
  14. Aug 26, 2009 #13

    LeonhardEuler

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    If u= 1- x, then what is 'x'? You need to make the substitution everywhere there is an 'x'. This means you must also integrate du instead of dx.
     
  15. Aug 26, 2009 #14

    Cyosis

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    As pointed out you need to express every x in terms of u, including the differential. If you substitute u=f(x), then du=f'(x)dx. In your case f(x)=1-x. Can you continue from here?
     
  16. Aug 26, 2009 #15
    u= 1-x
    du= -1
    correct?
     
  17. Aug 26, 2009 #16

    Cyosis

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    No, if f(x)=1-x and du=f'(x)dx, then f'(x)=? and du=?
     
  18. Aug 26, 2009 #17
    The derivative of 1-x isnt -1?
     
  19. Aug 26, 2009 #18

    Cyosis

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    Yes that part is correct, however incomplete. You want to express the differential, du in terms of the differential dx.
     
  20. Aug 26, 2009 #19
    du/dx = -1? i have no clue why i'm so lost right now.
     
  21. Aug 26, 2009 #20

    Cyosis

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    It's not much more than plug and play. Two differentials are related as [itex]du=\frac{du}{dx}dx[/itex]. This is the exact same as the one in my qu0te, just with a different notation. Use these properties to find du in terms of x.
     
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