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Thread: evaluate the definite integral

  1. #1
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    evaluate the definite integral

    can i take the the 2 so it looks like this

    = √2(1/√x)dx??
    Attached Thumbnails Attached Thumbnails evaluate the definite integral-screen-shot-2012-12-14-1.41.04-pm.png  
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    Re: evaluate the definite integral

    Quote Originally Posted by asilvester635 View Post
    can i take the the 2 so it looks like this
    = √2(1/√x)dx??

    What is the derivative of $\displaystyle \sqrt{2x}~?$
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    Re: evaluate the definite integral

    it will be x^-1/2
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    Re: evaluate the definite integral

    Quote Originally Posted by asilvester635 View Post
    it will be x^-1/2
    No it is not. If you cannot do derivatives, the integrals are impossible for you.
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    Re: evaluate the definite integral

    do i rewrite this?
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    Re: evaluate the definite integral

    Quote Originally Posted by asilvester635 View Post
    do i rewrite this?
    Look at this webpage.
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    Re: evaluate the definite integral

    The answer to the original question is yes, $\displaystyle \int_1^8\sqrt\frac{2}{x}\ dx = \sqrt{2}\int_1^8\sqrt\frac{1}{x}\ dx$.

    Plato has a point - you need some proficiency with derivatives before you try to do integrals.

    - Hollywood
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    Re: evaluate the definite integral

    Hello, asilvester635!

    $\displaystyle 18.\;\int^8_1\!\sqrt{\frac{2}{x}}\,dx$
    Can i take the the 2 so it looks like this? .$\displaystyle \sqrt{2}\left(\frac{1}{\sqrt{x}}\right)$ . Yes!

    $\displaystyle \sqrt{\frac{2}{x}} \;=\;\frac{\sqrt{2}}{\sqrt{x}} \;=\;\sqrt{2}\cdot\frac{1}{\sqrt{x}} \;=\;2x^{-\frac{1}{2}} $


    Therefore: .$\displaystyle \int^8_1\!\sqrt{\frac{2}{x}}\,dx \;=\;\sqrt{2}\!\int^8_1 x^{-\frac{1}{2}}\,dx$

    Got it?
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    Re: evaluate the definite integral

    Quote Originally Posted by asilvester635 View Post
    can i take the the 2 so it looks like this

    = √2(1/√x)dx??
    You can find the answer to this definite integral like this:

    $\displaystyle \begin{align*}\int^8_1\!\sqrt{\frac{2}{x}}\,dx =& 2\sqrt{2}\sqrt{x}\Big{]}^{8}_{1} \\ =& 2\sqrt{2} \sqrt{8} - 2\sqrt{2}\sqrt{1} \\ =& 8 - 2\sqrt{2}\end{align*}$

    $\displaystyle \therefore \int^8_1\!\sqrt{\frac{2}{x}}\,dx = 8 - 2\sqrt{2}$

    You can check the answer at:

    int (2)^(1/2) 1/(x)^(1/2) dx x=1 to 8 - Wolfram|Alpha
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