Riemann sum Question

**VkL** · November 11th 2009, 10:57 AM

Consider the given function.

f(x) = 4 - (1/4)x

Evaluate the Riemann sum for 2 ≤ x ≤ 4 , with six subintervals, taking the sample points to be left endpoints. (Give an exact answer.)
L6 =

How do I start this? and how do i know if its upper or lower bound?

**tttcomrader** · November 11th 2009, 11:19 AM

So you are calculating $L_6$ , meaning you have to divide the interval into six equal portions. And to find how long each of the sub-interval is going to be, you will have to use this formula: $\frac {b-a}{n}$

For $a \leq x \leq b$ dividing into n sub-intervals.

So, you should have $\frac {4-2}{6} = \frac {1}{3}$ as the length of your subs.

Now, you need to use the formula: $\frac {b-a}{n} [ f(x_0)+f(x_1)+...+f(x_{n-1}) ]$

Note that $x_0 = a =2$ and $x_n = b =4$ , so $x_{n-1}$ in this case is equal to $b - \frac {1}{3} = 4 - \frac {1}{3} = \frac {11}{3}$ , because we are using the left end points

So once you have plugged in all the value, you should have the following expression:

$\frac {1}{3} [ f(2)+f(2+ \frac {1}{3})+f(2+ \frac {2}{3}) + . . . + f( \frac {11}{3} ) ]$

Using $f(x)= 4 - \frac {1}{4}x$ , just plug in the values! Can you take it from here?

**VkL** · November 11th 2009, 08:18 PM

L6 = 17

L6 = 16.6

L6 = 127

L6 = 15

Umm, am i doing the algebra wrong? What am I doing wrong? if you can show me what to "plug" I would really appreaciate it. thank you.

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