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Thread: Limit process to find area of regions

  1. #1
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    Limit process to find area of regions

    I'm unsure of how to set up these types of problems.

    Here's a couple from my book:

    Use the limit process to find the area of the region between the graph of the function and the x-axis over the given interval.

    1. y=-2x+3, [0,1]

    2. 64-x[sup:1skslo47]3[/sup:1skslo47], [1,4]

    Also how do you determine if the left-endpoint or right-endpoint is better to use? Any help is greatly appreciated!

  2. #2
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    Re: Limit process to find area of regions

    1. Check whether or not the graph crosses the x axis in that interval (it doesn't).
    You do this to ensure you calculate area using integration.
    2. Is f(x) + or - from 0 to 1? (it's +). If it was -, you'd reverse the sign.

    Then your limits of integration are 0 and 1,
    so now you integrate the function from x=0 to x=1.
    First, write the integral of 3-2x.
    The constant of integration is not required when evaluating area.

    You will be subtracting the integral evaluated with x=0 from the integral evaluated at x=1.
    Basically, you will be subtracting zero.

    Try that and see if you can progress to the second problem.

  3. #3
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    Re: Limit process to find area of regions

    [tex]!) \ y \ = \ -2x+3, \ [0,1], \ always \ use \ the \ right\ \ endpoint \ unless \ instructed \ otherwise \ as \ it \ is \ the[/tex]

    [tex]\ easiest \ (less \ grunt \ work).[/tex]

    [tex]\Delta x \ = \ \frac{b-a}{n} \ = \ \frac{1-0}{n} \ = \ \frac{1}{n}.[/tex]

    [tex]Right \ endpoint \ = \ M_i \ = \ 0+i(1/n) \ = \ \frac{i}{n}[/tex]

    [tex]Hence, \ \lim_{n\to\infty}S(n) \ = \ \ \lim_{n\to\infty} \sum_{i=1}^{n}f(M_i)\Delta x[/tex]

    [tex]= \ \lim_{n\to\infty}\sum_{i=1}^{n}\bigg[-2\bigg(\frac{i}{n}\bigg)+3\bigg]\frac{1}{n}[/tex]

    [tex]= \ \lim_{n\to\infty} \frac{1}{n^{2}} \sum_{i=1}^{n}[-2i+3n}][/tex]

    [tex]= \ \lim_{n\to\infty}\frac{-2}{n^{2}}\sum_{i=1}^{n}i+\lim_{n\to\infty}\frac{3n }{n^{2}}\sum_{i=1}^{n}1[/tex]

    [tex]= \ \lim_{n\to\infty}\frac{-2}{n^{2}}\bigg[\frac{n(n+1)}{2}\bigg]+3[/tex]

    [tex]= \ \lim_{n\to\infty}-\frac{n}{n}-\frac{1}{n}+3 \ = \ 2 \ QED[/tex]

    [tex]2) \ \Delta x \ = \ \frac{3}{n} \ and \ M_i \ = \ 1+\frac{3i}{n}, \ I'll \ let \ you \ finish \ it.[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

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