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Math Help - Remain Sum (What am I doing for this)

  1. #1
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    Remain Sum (What am I doing for this)

    I haven't got to the lesson in class about this, but I'd like to jump start before I learn anything in class about it.

    The Question:

    Consider the Integral

    <br />
\int_{2}^{6}(\frac{4}{x} +5)

    (a) Find the Riemann sum for this integral using right endpoints and n=4.
    (b) Find the Riemann sum for this same integral, using left endpoints and n=4
    To do this, am I taking the anti-derivative to get f(x) then solving for part a and b?

    The Anti-derivative by the way is:
    4ln(\left|x \right|)+5x

    Lastly, this has nothing to do with the problem, but the thread its self. What tags should I put on this to help someone else in the future?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Zanderist View Post
    I haven't got to the lesson in class about this, but I'd like to jump start before I learn anything in class about it.

    The Question:

    Consider the Integral

    <br />
\int_{2}^{6}(\frac{4}{x} +5)

    (a) Find the Riemann sum for this integral using right endpoints and n=4.
    (b) Find the Riemann sum for this same integral, using left endpoints and n=4

    To do this, am I taking the anti-derivative to get f(x) then solving for part a and b?
    No, I don't think that this is what you are asked to do.

    The Riemann sum for this integral using right endpoints would be something like

    \sum\limits_{k=1}^n f\left(a+k\cdot\frac{b-a}{n}\right)\cdot \frac{b-a}{n} =\sum\limits_{k=1}^n f\left(2+k\cdot\frac{6-2}{n}\right)\cdot \frac{6-2}{n}

    and using left endpoints it would be

    \sum\limits_{k=1}^n f\left(a+(k-1)\cdot\frac{b-a}{n}\right)\cdot \frac{b-a}{n}=\sum\limits_{k=1}^n f\left(2+(k-1)\cdot\frac{6-2}{n}\right)\cdot \frac{6-2}{n}

    where f(x) := \frac{4}{x}+5.

    Just search your textbook, your lecture notes, or Wikipedia for "Riemann sum", if the above formulas seem completely foreign to you.

    Now plug in the given value of n.
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