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Thread: solving an indefinite integral

  1. #1
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    solving an indefinite integral

    \int 1/2*sqrt(x) dx = sqrt(x) - Why?

    I cannot understand how the author got this? Could you explain ...
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    MHF Contributor chisigma's Avatar
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    Given a real [or even complex...] $\displaystyle \alpha$ is...

    $\displaystyle \frac{d}{dx} x^{\alpha} = \alpha\ x^{\alpha-1}$ (1)

    ... and from (1) $\displaystyle \forall \alpha \ne -1$ You derive...

    $\displaystyle \int x^{\alpha}\ dx = \frac{x^{\alpha+1}}{\alpha+1} + c $ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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    WHAT? nothing understandable
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    Quote Originally Posted by problady View Post
    \int 1/2*sqrt(x) dx = sqrt(x) - Why?
    I cannot understand how the author got this? Could you explain ...
    Do you understand that $\displaystyle \frac{{d\left( {\sqrt x } \right)}}{{dx}} = \frac{1}{{2\sqrt x }}~?$

    If not, then a good place for you to start is a review of basic rules of derivatives.
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