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Math Help - Riemann Sums

  1. #1
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    Riemann Sums

    Let f(x)= (x-(1/2))^2 + 3, 0<= x <= 1. If the interval [0,1] is partitioned into 4 subintervals of equal length, then what is the value of the smallest Riemann sum for f(x) and this partition?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chaddy View Post
    Let f(x)= (x-(1/2))^2 + 3, 0<= x <= 1. If the interval [0,1] is partitioned into 4 subintervals of equal length, then what is the value of the smallest Riemann sum for f(x) and this partition?
    they want you to do the lower Riemann sum

    so you would partition the interval into [0,1/4], [1/4,1/2], [1/2, 3/4], and [3/4,1]

    in each of these intervals, find the point that makes f(x) the smallest, and then use those points as your x_i's in the formula

    \int_0^1 f(x) \approx \sum_{i = 1}^{4} f(x_i) \Delta x

    where, of course, \Delta x = \frac {1 - 0}4 = \frac 14
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  3. #3
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    I wound up getting <br />
\sum_{i = 1}^{4} \frac{11}{16}<br />
for the first interval,
    <br />
\sum_{i = 1}^{4} \frac{47}{64}<br />
for the second interval,
    <br />
\sum_{i = 1}^{4} \frac{3}{4}<br />
for the third interval, and
    <br />
\sum_{i = 1}^{4} \frac{49}{64}<br />
for the fourth interval,

    I'm not sure what to do from there?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chaddy View Post
    I wound up getting <br />
\sum_{i = 1}^{4} \frac{11}{16}<br />
for the first interval,
    <br />
\sum_{i = 1}^{4} \frac{47}{64}<br />
for the second interval,
    <br />
\sum_{i = 1}^{4} \frac{3}{4}<br />
for the third interval, and
    <br />
\sum_{i = 1}^{4} \frac{49}{64}<br />
for the fourth interval,

    I'm not sure what to do from there?
    there is one summation, how exactly did you get 4?

    \int_0^1 f(x) \approx \sum_{i = 1}^{4} f(x_i) \Delta x = \Delta x \sum_{i = 1}^{4} f(x_i) = \Delta x [f(x_1) + f(x_2) + f(x_3) + f(x_4)]
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