Midpoint rule

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janeann

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using midpoint rule approximate the area under x^6on [0,1] using 6 rectangles.
so far i have the delta x=1/N
but i have no idea how to get the rest.
for the right hand approximation i have (N(N+1)(2n+1)(3N^4+6n^4-3N+1))/42.
 
There's only six pieces. List them all separately and see how it works.
 


The concept is the same with the midpoint method as with the leftpoint or rightpoint methods.

The difference is that each rectangle gets its height f(x) from the value of x located in the MIDDLE of its base.

You found the width of the rectangles to be 1/n, where n = 6, yes?

Next, you need to find the six midpoint values of x (shown with question marks below).

Then, you get the six rectangle heights by evaluating f(x) = x^6 at each of these midpoints.

Code: 
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0 ___ ? ___ 1/6 ___ ? ___ 2/6 ___ ? ___ 3/6 ___ ? ___ 4/6 ___ ? ___ 5/6 ___ ? ___ 6/6

 
janeann said:
for the question makes am i supposed to find what x is?

You are supposed to find the HEIGHTS of the six rectangles and add them together.
Click to expand...

The area of each rectangle is BASE × HEIGHT.

The BASE of each rectangle is 1/6.

The HEIGHT of each rectangle is f(x), where x is the value in the middle of the BASE.

I posted a diagram of the six rectangle bases. You need to first determine the values of x at the midpoint of each base (I labeled these midpoints with "?").

Next, substitute each midpoint value of x into f(x) = x^6 to get the six HEIGHTs.

Multiply their sum by 1/6, and you're done.

Did you check out the example at the link that I posted for you?

 


It's been awhile, since we gave you some guidance. I hope that you were able to finish this exercise. For the benefit of future readers of this thread, here is the answer that I got.

To find the midpoint between two points, we take their average.

(0 + 1/6)/2 = 1/12

(1/6 + 2/6)/2 = 3/12

(2/6 + 3/6)/2 = 5/12

Et cetera. See a pattern?

The midpoints of the six approximating rectangles are 1/12, 3/12, 5/12, 7/12, 9/12, and 11/12 (some of these fractions reduce, of course).

The heights of these rectangles are each x^6 at x = midpoint.

In other words, (1/12)^6, (1/4)^6, (5/12)^6, (7/12)^6, (3/4)^6, and (11/12)^6.

The Midpoint Rule tells us that the area underneath the curve of y = x^6 from 0 through 1 is approximately:

(1/6)(1/2985984 + 1/4096 + 15625/2985984 + 117649/2985984 + 729/4096 + 1771561/2985984) = (1/6)(1771561/2985984)

The approximate area is 0.1360 square units.

Does your result match mine?

Cheers 8-)

 
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