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Math Help - Integral with Sq Root of u on bottom question

  1. #1
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    Integral with Sq Root of u on bottom question

    Going through my book I've found this problem which I can't seem to find the answer to.

    Directions: Evaluate the definite integral of the algebraic function.
    --------------------------
    Problem: Inegral symbol (the bottom of the symbol there is a 1 and at the top a 4)

    u-2/ u^(1/2) .


    ----------------------


    Can anyone help me in answering this? This probably seems like a simple and dumb question, but it seems that my algebra may be a bit weak.

    Thanks
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  2. #2
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    You may wish to note this:

     <br />
\frac{u-2}{\sqrt{u}}\;=\;\sqrt{u} - \frac{2}{\sqrt{u}}\;=\; u^{1/2} - 2u^{-1/2}<br />

    Looking easier, yet?
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  3. #3
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    \int \frac{u-2}{u^{1/2}} du = \int u^{1/2}du - 2\int u^{-1/2}du.

    Should be enough for you to find the answer.
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  4. #4
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    I can see what both of you did, and thank you. My last question would be what exactly do I do to bring the u^(1/2) to the numerator?

    AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still.

    Thanks to both of you again
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  5. #5
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    Quote Originally Posted by Shadoowned View Post
    My last question would be what exactly do I do to bring the u^(1/2) to the numerator?

    AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still.
    I'm not exactly clear on what you mean by "bringing it to the numerator," but \frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2} by the basic rules of exponents. So \frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.

    The integral portion follows from the basic rule

    \int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.

    Make sense?
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  6. #6
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    Quote Originally Posted by AlephZero View Post
    I'm not exactly clear on what you mean by "bringing it to the numerator," but \frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2} by the basic rules of exponents. So \frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.

    The integral portion follows from the basic rule

    \int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.

    Make sense?
    It all clicked after reading that! Thank you so much
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  7. #7
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    Technical note: Is \frac{u}{\sqrt{u}} = \sqrt{u}?

    Really? Truly? Absolutely in every way?

    It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?
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  8. #8
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    Quote Originally Posted by TKHunny View Post
    Technical note: Is \frac{u}{\sqrt{u}} = \sqrt{u}?

    Really? Truly? Absolutely in every way?

    It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?
    Well, obviously not if the the denominator were 0. In the same sense that \frac{x^2- 4}{x- 2}= x+ 2 as long as x\ne 2.

    If the lower limit of the integral were 0, that is, if it were \int_0^4 \frac{u}{\sqrt{u}}du, the integrand is not defined at u= 0 and so the integral is an improper integral. Your point is a good one but we can do the following:

    Replace \int_0^4 \frac{u}{\sqrt{u}}du with \lim_{\epsilon\rightarrow 0}^4 u^{1/2} du. We can do that because u is not 0 there. That gives \left[\frac{2}{3}u^{3/2}\right]_\epsilon^4= \frac{2}{3}\left(8- \epsilon^{3/2}\right). And that goes to \frac{16}{3} as \epsilon goes to 0, the same as if we had must integrated \int_0^4 u^{1/2} du
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  9. #9
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    I was just wondering if anyone ever had introduced Shadoowned to Domain considerations. Ploughing through without validity is bad business. One should at least think about it, I think.
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