# Midpoint Rieman problem

• Jul 10th 2010, 08:45 AM
JohnJames
Midpoint Rieman problem
Hello all,

My summer school classes just started this week, and unfortunately, math tutoring just so happens to be at the same time as my second class, so I can't get any help from the tutors.

Anyway, I am having problems grasping the Midpoint Riemann problem we have for homework:

Estimate the area under the graph of https://webwork2.uncc.edu/webwork2_f...e647256c81.png from https://webwork2.uncc.edu/webwork2_f...55eafd5481.png to https://webwork2.uncc.edu/webwork2_f...b9ec2b31a1.png using https://webwork2.uncc.edu/webwork2_f...6d7c90b481.png subintervals of equal length and taking the sample points to be midpoints.
The midpoint Riemann sum https://webwork2.uncc.edu/webwork2_f...8d232a3a71.png ?

$\displaystyle (f(1/2)+f(3/2)+f(5/2)+f(7/2))1= 24$

This is wrong, and I kinda understand why it's wrong, if x started at 0 this would be right, but because it starts at 5, the f(x) should be different numbers (if that makes sense), my problem is not knowing the proper calculations to figure that out.

In my book the formula is this:

$\displaystyle xi= 1/2(x(i-0)+xi) = midpoint of [x(1-0, xi]$

Which is extremely confusing to me.

• Jul 10th 2010, 08:58 AM
skeeter
let $\displaystyle f(x) = |8-x|$

break the interval from x = 5 to x = 9 into 4 pieces ... each piece has delta x = 1

the midpoint of each piece is x = 5.5, x = 6.5, x = 7.5, and x = 8.5

using a midpoint Riemann sum ...

$\displaystyle \displaystyle \int_5^9 f(x) \, dx \approx f\left(5.5) + f(6.5) + f(7.5) + f(8.5) = 5$
• Jul 10th 2010, 09:05 AM
JohnJames
Yes that is the correct answer, Thank you Skeeter!

but I still don't understand how you came up with the numbers for the functions. Can you break it down to the formula for me? because I have to really understand how to do this before I move on.

Thanks again Skeeter.
• Jul 10th 2010, 09:51 AM
skeeter
look at the attached sketch and note that the height of each rectangle is the function value at the midpoint of each sub-interval ...

area of each rectangle = base * height = (1) * f(midpoint of the sub-interval)
• Jul 10th 2010, 10:38 AM
JohnJames
Ohhh OK, I think I understand the midpoint a little more now.

Thanks again.