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Thread: Evaluate the definite integral of the algebraic function

  1. #1
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    Evaluate the definite integral of the algebraic function

    If you integrate the square root of 2/x does it come out to just 2x because when i rewrite it its going to be the square root of 2x^-1
    Then i turn the square root into an exponent which looks like this 2x^-1+2.... -1 + 2 = 1.... So i have just 1 as an exponent.... now i just have 2x
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    Re: Evaluate the definite integral of the algebraic function

    integral of (1/sqrt x) = integral (x ^(-1/2)} = x^(-1/2 + 1)/(-1/2 + 1) = sqrt(x)/(1/2)
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    Re: Evaluate the definite integral of the algebraic function

    ok i understand how you got x^-1/2 part but how did you get (-1/2 + 1)/(-1/2 + 1)????
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    Re: Evaluate the definite integral of the algebraic function

    Quote Originally Posted by asilvester635 View Post
    ok i understand how you got x^-1/2 part but how did you get (-1/2 + 1)/(-1/2 + 1)????
    You ought to know the basics.

    $\displaystyle n \ne - 1,\quad \int {x^n dx} = \frac{{x^{n + 1} }}{{n + 1}}$
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    Re: Evaluate the definite integral of the algebraic function

    integral of x^n ={1/(n+1)}*{ x^{n+1)}. Put n = -1/2 you get (1/2)*sqrt x
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    Re: Evaluate the definite integral of the algebraic function

    i know the basics...... im just asking how you would rewrite √2/x??? isn't it √2x^-1????
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    Re: Evaluate the definite integral of the algebraic function

    Quote Originally Posted by asilvester635 View Post
    i know the basics...... im just asking how you would rewrite √2/x??? isn't it √2x^-1????

    "i know the basics".Do you really?

    Basic algebra: $\displaystyle \sqrt{\frac{2}{x}}=\sqrt{2}x^{-\frac{1}{2}}$
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    Re: Evaluate the definite integral of the algebraic function

    yes that what i got but i put -1 instead of putting -1/2.... NOW WHAT DO I DO with the square root?
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    Re: Evaluate the definite integral of the algebraic function

    Quote Originally Posted by asilvester635 View Post
    yes that what i got but i put -1 instead of putting -1/2.... NOW WHAT DO I DO with the square root?

    The final answer to the integral is $\displaystyle 2\sqrt{2}\sqrt{x}+C$

    Or we could write it as $\displaystyle \sqrt{8x}+C$
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    Re: Evaluate the definite integral of the algebraic function

    oh it makes sense now but how is 2√2√x equal to √8x?
    And since this is definite integral we can remove the + C right?
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    Re: Evaluate the definite integral of the algebraic function

    Quote Originally Posted by asilvester635 View Post
    oh it makes sense now but how is 2√2√x equal to √8x?
    And since this is definite integral we can remove the + C right?

    $\displaystyle 2\sqrt{2}=\sqrt{4}\sqrt{2}=\sqrt{8}$
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    Re: Evaluate the definite integral of the algebraic function

    Quote Originally Posted by asilvester635 View Post
    oh it makes sense now but how is 2√2√x equal to √8x?
    And since this is definite integral we can remove the + C right?
    Your problem seems to me like you need to do the indefinite integral. You should be able to put the limits in yourself. As for the other question: $\displaystyle 2 \sqrt{2} = 2^1 \cdot 2^{1/2} = 2^{3/2} = \left ( 2^3 \right )^{1/2} = \left ( 2^3 \right ) ^{1/2} = \sqrt{8}$

    You might need a review of multiplication of radicals, which is a pre-Calculus topic.

    -Dan

    @Plato: Sorry for the snipe. I thought you were gone.
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    Re: Evaluate the definite integral of the algebraic function

    $\displaystyle \int_1^8 \sqrt{\frac{2}{x}} \, dx$

    $\displaystyle \sqrt{2} \int_1^8 \frac{1}{\sqrt{x}} \, dx$

    $\displaystyle 2\sqrt{2} \int_1^8 \frac{1}{2\sqrt{x}} \, dx$

    $\displaystyle \frac{1}{2\sqrt{x}}$ is the derivative of $\displaystyle \sqrt{x}$ ...

    $\displaystyle 2\sqrt{2} \left[\sqrt{x}\right]_1^8$

    $\displaystyle 2\sqrt{2} \left[\sqrt{8} - \sqrt{1} \right]$

    $\displaystyle 2\sqrt{16} - 2\sqrt{2}$

    $\displaystyle 8 - 2\sqrt{2}$
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