Riemann Sums

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March 2nd 2008, 01:02 AM
chaddy

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?
March 2nd 2008, 10:44 AM
Jhevon

Quote:

Originally Posted by chaddy

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, $\Delta x = \frac {1 - 0}4 = \frac 14$
March 11th 2008, 01:23 AM
chaddy

I wound up getting $ \sum_{i = 1}^{4} \frac{11}{16} $ for the first interval,
$ \sum_{i = 1}^{4} \frac{47}{64} $ for the second interval,
$ \sum_{i = 1}^{4} \frac{3}{4} $ for the third interval, and
$ \sum_{i = 1}^{4} \frac{49}{64} $ for the fourth interval,

I'm not sure what to do from there?
March 11th 2008, 11:32 AM
Jhevon

Quote:

Originally Posted by chaddy

I wound up getting $ \sum_{i = 1}^{4} \frac{11}{16} $ for the first interval,
$ \sum_{i = 1}^{4} \frac{47}{64} $ for the second interval,
$ \sum_{i = 1}^{4} \frac{3}{4} $ for the third interval, and
$ \sum_{i = 1}^{4} \frac{49}{64} $ for the fourth interval,

I'm not sure what to do from there?

there is one summation, how exactly did you get 4?

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