calculus - midpoint rule

[tmh88]

I'm kinda confused on one part of using the midpoint rule when using approximate integration.


Could someone explain to me how to solve for xi? Here is a scan from my book. I just dont understand what to plug in for xi if it doesn't even tell me what it is.

Say delta x is 1/4, n = 8, and the lower and upper limits are 0 and 2 respectively. It says xi is the chosen midpoint of the subinterval, but what are you supposed to use to get it?

24870

btw this book is essential calculus by james stewart if anyone is wondering...do i really need to cite everything?

[JB]

The midpoint rule lets you approximate the area under the curve (aka integral), by dividing the area into pieces and summing the individual areas. The division along the x-axis in your example is given by delta x. This means you will have sections [a, a+delta x], [a+delta x, a + 2(delta x)]...[a+(n-1)(delta x), b]. Therefore, you will need to calculate your xi in terms of a and delta x.

[tmh88]

I know how to use the midpoint rule(that is when I'm given the value of the increments that are being substituted).

I know how to use the trapezoid and simpsons rules without a problem, it's just finding the sections that I dont know how to do for the midpoint rule. I have no problem finding them with the trapezoid and simpsons rules, because they don't explain it in retarded terminology that relates to ZERO of their examples.

[JB]

This means you will have sections [a, a+delta x], [a+delta x, a + 2(delta x)]...[a+(n-1)(delta x), b]. Therefore, you will need to calculate your xi in terms of a and delta x.

So basically slice up your area into n pieces and find the midpoint of each piece. Your first post contains all the information you need...

[tmh88]

so say I'm using the lower and upper limits of 0 and 2, using n=8. So I take 2 and divide it into 8 pieces, giving me each one is 1/4. Then I find the middle of that which is 1/8?

[JB]

That is correct for the first slice.

[tmh88]

alright so....

for x2 would it be 1/8 + 1/4 = 3/8

x3 -->3/8 + 1/4 = 5/8

x4 -->5/8 + 1/4 = 7/8

x5 -->7/8 + 1/4 = 9/8

x6 -->9/8 + 1/4 = 11/8

x7 -->11/8 + 1/4 = 13/8

x8 --> 13/8 + 1/4 = 15/8

edit - I know the rules for applying it so I'm good in that department. Finding the x's was the problem I had... delta x(f(x1) + f(x2) ....+f(xn))

[JB]

yep

[tmh88]

I love the midpoint rule almost as much as I love partial fractions!

[JB]

if you are studying engineering you better love it all;)

[tmh88]

ISE...not as hard as say mechanical engineering, but definitely more so than your regular business majors....not trying to put anyone down on here, but I think its assumed engineering is one of the harder fields.

[JB]

what is that?

[tmh88]

industrial and systems engineering

just wondering, but what did you major in?

[JB]

Computer Engineering for undergraduate, working on the MS in Computer Science