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Ntegration on my maths book

  1. Jun 3, 2006 #1
    Hi. I read a few things about integration on my maths book and googled it up. I think i got the idea of it, but there's something that's still confusing me:

    I found this formula but i can't see how it can be true

    [tex] \int u^n du = \frac{u^{n+1}}{n+1} + C [/tex]

    however, as far as i remember:

    (u^n)' = n*u'*u^(n-1)

    How can this be right. where did that (u') go???

    For example: plz tell me how to solve this one:

    [tex] \int (1-x^2)^\frac{1}{2} dx [/tex]

    using the formula i'd say:

    [tex] \frac{2(1-x^2)^\frac{3}{2}}{3} [/tex]

    but this can't be right.:frown:
  2. jcsd
  3. Jun 3, 2006 #2


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    u' comes in when you differentiate with respect to a variable other than u. For example, if u=u(x), then:

    [tex]\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx} [/tex]

    If you differentiate with respect to u (in other words, take u=x in the above equation), then du/du=1, and so:

    [tex]\frac{d}{du}(u^n) = n u^{n-1} [/tex]

    Applying this to the indefinite integral above you get back the integrand as expected.

    As for your second question, this is more complicated because u=1-x2, so du/dx is not 1 as it was before. The final answer will involve the arcsin function. If you're interested, this would go like:

    [tex]\int \sqrt{1-x^2} dx = \int \frac{1-x^2}{\sqrt{1-x^2}} dx[/tex]
    [tex]= \int \frac{dx}{\sqrt{1-x^2}} - \int \frac{x^2 dx}{\sqrt{1-x^2}} = \sin^{-1} x - \left( (x)(-\sqrt{1-x^2}) - \int (-\sqrt{1-x^2}) dx \right)[/tex]

    [tex]\int \sqrt{1-x^2} dx = \frac{1}{2} (\sin^{-1} x + x\sqrt{1-x^2})[/tex]
  4. Jul 2, 2006 #3

    I think this is one nice example

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  5. Jul 2, 2006 #4


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    is NOT correct. In fact, it really doesn't makes sense because it doesn't say with respect to what variable the derivative is done.

    What is true is that
    [tex]\frac{du^n}{dx}= n u^{n-1}\frac{du}{dx}[/tex]
    In particular, if u= x,
    [tex]\frac{du}{dx}= 1[/tex]
    [tex]\frac{du^n}{du}= n u^{n-1}[/tex]

    That also means that
    [tex]\frac{1}{n+1}\frac{du^{n+1}}{du}= \frac{1}{n+1}(n+1)x^{n+1-1}= u^n[/tex]
    [tex]\int u^n du= \frac{1}{n+1}u^{n+1}[/tex]
  6. Jul 6, 2006 #5
    It's simply...the abstractization of the antiderivative written in a very fancy way.

    assume your problem asks for integration of 3x^2. you know answer is x^3.

    So look. 3x^2 integrated => 3x ^ (2+1) divided by (2+1) => 3x^3 divided by 3 or...x^3.
  7. Jul 7, 2006 #6


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    Hopefully, you know the answer is x^3+ C!:biggrin:
  8. Jul 7, 2006 #7
    not +C! but plain +C w00t!
    But yeah, good point. Just for the record, in first few months of calculus I forgot every dx and every +C I encountered and should have written...all until differentials. Then...God, they mattered :D
  9. Jul 8, 2006 #8


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  10. Jul 12, 2006 #9
    Maybe i used the wrong word, I'm not a native english speaker...The thing where universal variables replace numbers...basically a model that you can use for your numbers if you can figure out what goes where. At times I found it to be more difficult than just...say it in baby terms.

    Oh you know what I mean! I was trying to sound smart :(
  11. Jul 12, 2006 #10


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    Robokapp: "Abstraction" is a perfectly good word and works in place of "Abstractization". Unfortunately, in English, it is easy to create new, overly long, words by adding "endings". Don't use long words where shorter ones will do!

    By the way, your English is far better than my (put pretty much any language you wish here).
  12. Jul 12, 2006 #11

    Abstract is the root word...Abstraction is the noun formed by derivation...Abstractization is the noun reflecting "Abstraction done by someone on something"...a noun coming from the verb...which I think goes "to abstractize". Or is it "to abstract"?

    Well that's how i thought it out I think. I don't start mubling these things before i write a word...it just comes out.

    I'll remember it for the future hehe. Thx!
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