1. ## Midpoint Rule Help

Using the midpoint rule, find the area with two rectangles and 4 rectangles:
Problem: f(x) = x^3 btwn x = 0 and x = 1
My Process:
i) Two rectangles:
Δ x = (b-a)/n = (1-0)/2 = 1/2
A ˜ 1/2 [ f (1/4) + f (3/4) ]
A ˜ 1/2 [ (1/4)^3 + (3/4)^3 ] = . 21875
ii) Four rectangles:
Δ x = (b-a)/n = (1-0)/4 = 1/4
A ˜ 1/4 [ f (1/8) + f (3/8) + f (5/8) + f (7/8) ]
A ˜ 1/4 [ (1/8)^3 + (3/8)^3 + (5/8)^3 + (7/8)^3 ] = .2422
Is this right because when I tried doing the same process for the next problem, we got a different answer.

2. That is the correct sum you should get by breaking the area into a set of rectangles with the same base, and whose height is the height of the function at the midpoint of the base.

3. Originally Posted by slider142
That is the correct sum you should get by breaking the area into a set of rectangles with the same base, and whose height is the height of the function at the midpoint of the base.
So how come on question # 7,

which asks f(x) = 1/x between x = 1 and x = 5

i) two rectangles:

Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

A approx. = 2 [f(1) + f(3)] = 2 [ 1 + 1/3] = 2.66666....

The way the solution manual did it was...

Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

Delta x = 1/2 x 2 + 1/4 x 2....So confused at this step???
= 2 [1/2 + 1/4]
= 2 [ 3/4]
= 3/2
What's going on?

4. Originally Posted by Nimmy
So how come on question # 7,

which asks f(x) = 1/x between x = 1 and x = 5

i) two rectangles:

Delta X = (b-a)/n = (5-1)/2 = 4/2 = 2

A approx. = 2 [f(1) + f(3)] = 2 [ 1 + 1/3] = 2.66666....
This is not correct, as 1 is not the midpoint of the base of the first rectangle. The base of the first rectangle is the interval [1, 1 + Deltax] = [1, 3]. The midpoint is thus (1 + 3)/2 = 2.
Look again at the first problem. The midpoints are correct there.

5. midpoint Riemann sum, two subintervals.

$\int_1^5 \frac{1}{x} \, dx \approx \frac{5-1}{2}\left[f(2) + f(4)\right] = 2\left(\frac{1}{2} + \frac{1}{4}\right)$