Question

Let f(x) = 4 - 12/x^2. Calculate the lower and upper sum estimates of the area under the curve over the interval [2,5] using subintervals of width ½.
I tried working it out but I screw up because I get a negative number.

Answer 1

what do you mean by lower and upper sums? im not familiar with this fancy terminology..

Answer 2

Maybe my textbook is weird

Answer 3

right and left hand sums

Answer 4

So wait are these Riemann sums o.o

Answer 5

yeah it seems so
look under methods to see what I think they mean by under and over or whatever...
http://en.wikipedia.org/wiki/Riemann_sum

Answer 6

these problems take a good long time to do...

Answer 7

@TuringTest is it just \[\huge \int_2^5 (4 - \frac{12}{x^2})dx\]

Answer 8

or no?

Answer 9

that is exact, we want approximations, from which we can derive that formula

Answer 10

that is the limit as the number of rectangles goes to infinity @lgbasallote

Answer 11

I have a feeling this is Riemann but I wasn't integrating just simplifying the summations and plugging in the value for n within the interval

Answer 12

my brain hurts

Answer 13

yes, it will just be a summation, which is usually more of a pain than integration

Answer 14

first let's look at the definition:
the area under a curve f(x) is approximated by\[A\approx\sum_{i=1}^nf(x_i^*)\Delta x\]

Answer 15

n is the number of rectangles we are going to use and \(\Delta x\) is the width of those rectangles

Answer 16

you should be able to figure out \(\Delta x\) at this point

Answer 17

oh it's given here
then from that you should be able to figure out n, the number of rectangles we will use

Answer 18

6, because the interval is [2,5] and the rectangles are 1/2 width.. yeah I have work for all of this already but I went wrong somewhere with the number crunching... the concept is just generally muddled confusion for me right now

Answer 19

so the left hand endpoints are going to be from 2 to 4.5
and the rh ones are from 2.5 to 5

Answer 20

left sum\[A\approx\sum_{i=0}^nf(x_i^*)\Delta x\]right sum\[A\approx\sum_{i=1}^nf(x_i^*)\Delta x\]and since \(\Delta x=\frac12\) and the starting point is \(a=2\) we have \[x_i^*=a+i\Delta x=2+\frac i2\]so we need to do two sums

Answer 21

...Crap, hang on, I'm checking my textbook and Riemann is in the next section

Answer 22

typo*
the left hand sum is\[A\approx\sum_{i=0}^{n-1}f(x_i^*)\Delta x\]the right hand sum is
\[A\approx\sum_{i=1}^nf(x_i^*)\Delta x\]

Answer 23

@TuringTest Here is my textbook example
I don't think it's Riemann exactly, but can you work off of that?

Answer 24

and we're solving for n =6

Answer 25

What is the solution?

Answer 26

@satellite73 I'm curious, what is the solution to this question?