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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- Mar 2nd 2008, 01:02 AMchaddyRiemann 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?

- Mar 2nd 2008, 10:44 AMJhevon
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 $\displaystyle f(x)$ the smallest, and then use those points as your $\displaystyle x_i$'s in the formula

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

where, of course, $\displaystyle \Delta x = \frac {1 - 0}4 = \frac 14$ - Mar 11th 2008, 01:23 AMchaddy
I wound up getting $\displaystyle

\sum_{i = 1}^{4} \frac{11}{16}

$ for the first interval,

$\displaystyle

\sum_{i = 1}^{4} \frac{47}{64}

$ for the second interval,

$\displaystyle

\sum_{i = 1}^{4} \frac{3}{4}

$ for the third interval, and

$\displaystyle

\sum_{i = 1}^{4} \frac{49}{64}

$ for the fourth interval,

I'm not sure what to do from there? - Mar 11th 2008, 11:32 AMJhevon