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Integration by subsitution (give constant value), why?

  1. Sep 22, 2011 #1
    [itex]\int(x^{2}-5)^{2}x dx[/itex]

    By substiution:
    1. [itex]u = x^{2}-5[/itex]

    2. [itex]du = 2x dx[/itex]

    3. [itex]\frac{du}{2x}= dx[/itex]

    4. [itex]\int u^{2}x \frac{du}{2x}[/itex]

    5. [itex]\int u^{2} \frac{1}{2} du[/itex]

    6. [itex] \frac{1}{3} u^{3} \frac{1}{2}[/itex]

    7. [itex] \frac{1}{6} u^{3}[/itex]

    8. [itex] \frac{1}{6} (x^{2}-5)^{3}[/itex]

    9. [itex] \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125][/itex]


    By normal integration factorize the from beginning

    from: [itex]\int(x^{2}-5)^{2}x dx[/itex]


    to [itex]\int [x^{4} - 10x^{2} + 25 ] x dx[/itex]

    then: [itex]\int x^{5} - 10x^{3} + 25x dx[/itex]

    and finally : [itex] \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2}] + C[/itex]


    Probably this is a easy one, i been looking on internet, but had hard time to find the right keywords for an explanation...

    Question is: the one give me some kind of constant and the other i just add one.. Which one is correct? I mean both gives same result except of one provide a "real" constant value.
     
    Last edited: Sep 22, 2011
  2. jcsd
  3. Sep 22, 2011 #2
    A constant should be added in the 6th step of method one.The answers tally.
     
  4. Sep 22, 2011 #3

    gb7nash

    User Avatar
    Homework Helper

    You blundered on step 6. When integrating, you must always add an arbitrary constant after integrating. The cool thing though, is that any constant you get at the end is correct (unless more information is given). There's an infinite number of answers you can get. As long as your constant is a real number, you're fine. That's why we just add a constant C to be general.
     
  5. Sep 22, 2011 #4
    Aha yes my fault but this doesnt explain why i get 125 value from the first method

    since unfactorise [itex](x^{2} - 5)^{3} = x^{6} - 15x^{4} + 75x^{2} - 125[/itex]
     
  6. Sep 22, 2011 #5
    So why does both methods differs -125/6 ?
     
  7. Sep 22, 2011 #6
    Sorry I will correct this into new post

    By substiution:
    1. [itex]u = x^{2}-5[/itex]

    2. [itex]du = 2x dx[/itex]

    3. [itex]\frac{du}{2x}= dx[/itex]

    4. [itex]\int u^{2}x \frac{du}{2x}[/itex]

    5. [itex]\int u^{2} \frac{1}{2} du[/itex]

    6. [itex] \frac{1}{3} u^{3} \frac{1}{2} + C[/itex]

    7. [itex] \frac{1}{6} u^{3} +C[/itex]

    8. [itex] \frac{1}{6} (x^{2}-5)^{3} +C[/itex]

    9. [itex] \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125] +C[/itex]

    So comparing substitution method and normal unfactor in beginning give me different answer

    [itex] \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2} +125] +C \neq \frac{1}{6} [x^{6} - 15x^{4} + 75x^{2}] + C[/itex]

    Where a value of 126/6 appears (a constant), so the substitution method gives me "2" constants? but 126/6 + C = C ?

    Or did i make any mistake somewhere? I mean I know C can be anything sure, but shoulnt C be same for both methods?
     
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