If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

012. `query 12

Question: `q5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]

Your solution:

Confidence Assessment:

Given Solution:

`a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore

[1, 2], [2, 3], [3, 4], [4,5].

The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore

(0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417.

Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417.

The sum of these areas is the trapezoidal approximation 1.604. **

DER

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Question: `q5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft).

How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater?

Your solution:

Confidence Assessment:

Given Solution:

`a Since widths at 20-ft intervals are 50, 54, 82, 82, 73, 75, 80 ft the pond area can be approximated by a series of trapezoids with these altitudes. The average altitudes are therefore respectively

(0 + 50 / 2) = 25

(50 + 54) / 2 = 52

(54 + 82) / 2 = 68

(82 + 82) / 2 = 82

etc., with corresponding areas

25 * 20 = 500

52 * 20 = 1040

etc., all areas in ft^2.

The total area, according to the trapezoidal approximation, will therefore be

20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 square feet.

Midpoint widths will be subject to some estimation errors.  Each midpoint width will be multiplied by the 20 ft width of the interval. 

The total area obtained from midpoint estimates will exceed that of the trapezoidal approximation.  The midpoint and trapezoidal rules won't differ much for the second through the seventh intervals, while on the first and the last the trapezoidal rule will seriously underestimate the area while the midpoint rule will seriously overestimate it.

The pond itself isn't convex, but the first and last region are, and this is the source of the main difference between the two estimates.

Put a little differently, the convex shape of these two regions ensures that the midpoint of the first and last regions will 'hump above' the trapezoidal approximation, so that for these two regions the midpoint estimate will very significantly exceed the trapezoidal estimate. **

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Question: `q Add comments on any surprises or insights you experienced as a result of this assignment.

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