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Math Help - integration :quick question

  1. #1
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    integration :quick question

    what do I get when I integrate
    x^2.5 dx
    Please?
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    what do I get when I integrate
    x^2.5 dx
    Please?

    Integration without bounds is equivalent to solving for an antiderivative.
    Antiderivative of functions of the form, f(x)=x^n \;where\; n\; is\; real, amounts to

    Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C
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  3. #3
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    Quote Originally Posted by vince View Post
    Integration without bounds is equivalent to solving for an antiderivative.
    Antiderivative of functions of the form, f(x)=x^n where n is real, amounts to

    Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C
    Sorry but I don't have a clue what your on about!!!!!!!!
    Could someone please show me the answer this part is delaying me
    Help please!
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  4. #4
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    Quote Originally Posted by wolfhound View Post
    Sorry but I don't have a clue what your on about!!!!!!!!
    Could someone please show me the answer this part is delaying me
    Help please!

    dude, \int x^{2.5}dx = \frac{x^{2.5+1}}{2.5+1}+C...clear now?
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  5. #5
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    My problem is what do I raise x^2.5 to? (x^?/?)
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  6. #6
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    so its x to the power of 3.5 divided by 3.5 + C ?
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  7. #7
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    \frac{x^{3.5}}{3.5} + C
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  8. #8
    Ted
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    Quote Originally Posted by vince View Post
    Integration without bounds is equivalent to solving for an antiderivative.
    Antiderivative of functions of the form, f(x)=x^n \;where\; n\; is\; real, amounts to

    Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C
    n \neq -1.
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  9. #9
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    Quote Originally Posted by Ted View Post
    n \neq -1.

    but of course, making sure if anyone was paying attention.
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  10. #10
    Ted
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    But you said where n is real.
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