Riemann Sum

[golflim]

suppose f(x) is increasing on (1,5)

when x=1,f(x)=0 x=1.8,F(x)=0.6 x=3.1, f(x)=1.2 x=4, F(x)=2.3 x=5, f(x)=6

find the riemann sum from the right on (1,5)

Is this an overestimate or underestimate for the integral of F(x) over 1 to 5?

Our book has nothing any help would be great. Thanks

 

[tkhunny]

1) You keep changing from F(x) to f(x) and back. Is this on purpose? It COULD make a difference. Sometimes "f(x)" MEANS the derivative of "F(x)".

2) You do not have to go to this trouble: "When x = 1, f(x) = 0". You can just write "f(1) = 0".

3) It's an INcreasing function. You are estimating from the Right (high) side. It's an OVERestimate.

 

[golflim]

Im sorry I meant f(x) to be F(x) but i still have no clue as to how to find the riemann sum.

 

[stapel]

i still have no clue as to how to find the riemann sum.

Didn't they cover that in your class and/or in your book?

Eliz.

http://www.stats.uwaterloo.ca/~cgsmall/riemann.html
http://en.wikipedia.org/wiki/Riemann_sum
http://archives.math.utk.edu/visual.cal ... nn_sums.4/
http://www.math.hmc.edu/calculus/tutori ... mann_sums/
http://mathworld.wolfram.com/UpperSum.html

 

[tkhunny]

1) You plaster your area with a bunch of rectangles, stuck up next to each other.
2) You calculate the area of each rectangle.
3) Adding up all the areas give the Riemann sum.

You must decide two things.
1) How wide are the rectangles.
2) How tall are the rectangles.

Width can be done many ways. It is common to pick the same width for each rectangle. This is probably simplest. So, each rectangle has width W, because I just decided that's how wide they all are. It's just an assumption for now.

This particular question is making you decide HOW to calculate the heights of the rectangles. Using the function F(x), we decide to make the rectangle get taller until the RiGHT hand side hits the function. We could have used the left hand side or the middle or anything else we wish.

The area of each rectangle is W*f(x_i), where 'i' is the number of the rectangle you are playing with.

You have:

F(x) is increasing on (1,5)
F(1)=0
F(1.8)=0.6
F(3.1)=1.2
F(4)=2.3
F(5)=6

Calculate the widths of the rectangles.

1.8 - 1 = 0.8
3.1 - 1.8 = 1.3
4 - 3.1 = 0.9
5 - 4 = 1

Note the given heights
F(1.8)=0.6
F(3.1)=1.2
F(4)=2.3
F(5)=6

We didn't use F(1). We would have used F(1) and NOT F(5) if we had been using left hand sides.

Put them all together:

0.8*0.6 + 1.3*1.2 + 0.9*2.3 + 1*6 = 0.48 + 1.56 + 2.07 + 6 = 10.11

There you go. Your first Riemann Sum.

Just for fun, look at what happens if we try a left-hand side sum

0.8*0 + 1.3*0.6 + 0.9*1.2 + 1*2.3 = 0 + 0.78 + 1.08 + 2.3 = 4.16

So, the REAL area under F(x), for 1 <= x <= 5 is somewhere between 4.16 and 10.11. Not a very good estimate for anything but teaching about Riemann sums. We couldn't quite asy this, if we didn't know F(x) was increasing on this interval.

What do you think? Are we getting anywhere?