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Math Help - integratetion #3

  1. #1
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    integratetion #3

    Hi:
    I think I have gone horribly wrong here:
    Could someone take a peek please?

    Attached Thumbnails Attached Thumbnails integratetion #3-q3.gif  
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  2. #2
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    Let u = 5x^4 + 5.

    Then:

    \int 3x^3(5x^4 + 5)^3 \cdot dx =

    \int 3 \cdot \frac{20}{20} x^3(5x^4 + 5)^3 \cdot dx =

    \int \frac{3}{20} \cdot 20x^3(5x^4 + 5)^3 \cdot dx =

    \int \frac{3}{20} \cdot u^3 \cdot du
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  3. #3
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    Hi Icemanfan:
    I'm new to integration: Can I ask how you got 20/20?
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  4. #4
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    Quote Originally Posted by stealthmaths View Post
    Hi Icemanfan:
    I'm new to integration: Can I ask how you got 20/20?
    You can multiply by one and the expression is the same. I needed a factor of 20 in the numerator since du = 20x^3 \cdot dx.
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  5. #5
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    How's this?

    Attached Thumbnails Attached Thumbnails integratetion #3-q3.1.gif  
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  6. #6
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    sorry icemanfan but i couldn't see what your trying to say with that.
    anyway i think i made a mistake above by leaving out the root^3
    i have adjusted if someone could check please

    <img>http://www.mathhelpforum.com/math-help/attachments/calculus/15753d1267660768-integratetion-3-q3.2.gif</img>
    Attached Thumbnails Attached Thumbnails integratetion #3-q3.2.gif  
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  7. #7
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    messed up that image.
    here it is

    Last edited by stealthmaths; March 3rd 2010 at 03:17 PM.
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  8. #8
    Member VitaX's Avatar
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    I think I have gone horribly wrong here:
    Could someone take a peek please?

    3 \int x^3(5x^4 + 5)^3 dx

    u = 5x^4 + 5

    du=20x^3 dx \rightarrow \frac{1}{20}du=x^3 dx

    3\left(\frac{1}{20}\right) \int u^3 du

    \frac{3}{20}\left[\frac{u^4}{4}\right] + C
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  9. #9
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    Hi:
    So have I got the final answer here?



    If so, I'm a little confused. Is this integration by parts?


    Because the way that this is calculated doesn't go with the formula I've been taught and method.

    If this is the correct final answer, can somebody explain please?
    Attached Thumbnails Attached Thumbnails integratetion #3-q3.3.gif   integratetion #3-q3.3parts.gif  
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  10. #10
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    So have I got the final answer here?



    If so, I'm a little confused. Is this integration by parts?


    Because the way that this is calculated doesn't go with the formula I've been taught and method.

    If this is the correct final answer, can somebody explain please?
    Your answer's right
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  11. #11
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    Great!
    Thanks Vita X.
    Thanks to everyone who contributed.

    Any tips on why I got my method mixed up here?
    I think the part I'm confused with is du=20x^3dx and 1/20du=x^3
    It seems I am getting confused with integral products.
    Last edited by stealthmaths; March 3rd 2010 at 10:32 PM. Reason: extended explanation
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  12. #12
    Member VitaX's Avatar
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    Quote Originally Posted by stealthmaths View Post
    Great!
    Thanks Vita X.
    Thanks to everyone who contributed.

    Any tips on why I got my method mixed up here?
    I think the part I'm confused with is du=20x^3dx and 1/20du=x^3
    It seems I am getting confused with integral products.
    du=20x^3 dx is just the derivative of the u substitution. 1/20 du = x^3 dx is getting it so that 1/20 du will be substituted back into the integral so that no x variables are left over, leaving a nice and easy integral.
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  13. #13
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    Yes. It does seem like a really simple, clean and easy method. So can I look at this method as a kind of shorthand to 'integration by parts'?
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  14. #14
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    Quote Originally Posted by stealthmaths View Post
    Yes. It does seem like a really simple, clean and easy method. So can I look at this method as a kind of shorthand to 'integration by parts'?
    No, integration by parts would only be viable if U substitution doesn't work out. Trial and error really. If no u sub works, go to integration by parts in an integral dealing with a composition of 2 functions.
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  15. #15
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    Thank you
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