ASSN 6

course Mth 174

Test 1 should be in the mail no later than Tuesday

Physics II02-14-2008

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14:57:21

Query problem 7.5.13 (3d edition #10) graph concave DOWN and decreasing (note changes indicated by CAPS)

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14:58:31

list the approximations and their rules in order, from least to greatest

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For concave down: RIGHT < TRAP < EXACT < MID < LEFT

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14:58:48

between which approximations does the actual integral lie?

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Between TRAP and MID

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15:01:50

Explain your reasoning

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By using the rules for aproixmating intervals on pg 341 -2.

Concave down: TRAP< MID

RIGHT < LEFT

RIGHT < TRAP < LEFT

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15:02:36

if you have not done so explain why when a function is concave down the trapezoidal rule UNDERestimates the integral

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Because the curve in the graph is concave down so it rises above the trapazoid line on an interval.

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15:04:04

if you have not done so explain why when a function is concave down the midpoint rule OVERrestimates the integral

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Because the midpoint touches and makes a tangent to the highest part of the curve, on the interval while the graph continues down.

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15:04:19

Query NOTE: this problem has been left out of the new edition of the text, which is a real shame; you can skip on to the next problem (was problem 7.5.18) graph positive, decreasing, concave upward over interval 0 < x < h

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15:04:26

why is the area of the trapezoid h (L1 + L2) / 2?

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15:04:35

Describe how you sketched the area E = h * f(0)

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15:04:39

Describe how you sketched the area F = h * f(h)

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15:04:43

Describe how you sketched the area R = h*f(h/2)

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15:04:46

Describe how you sketched the area C = h * [ f(0) + f(h) ] / 2

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15:04:49

Describe how you sketched the area N = h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) } f(h) ] / 2

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15:04:52

why is C = ( E + F ) / 2?

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15:04:55

Why is N = ( R + C ) / 2?

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15:04:57

Is E or F the better approximation to the area?

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15:05:00

Is R or C the better approximation to the area?

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15:05:11

query problem 7.5.24 show trap(n) = left(n) + 1/2 ( f(b) - f(a) ) `dx

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15:05:21

Explain why the equation must hold.

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Because the (f(b) - f(a)) 'dx cancells out the difference between RIGHT and LEFT

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15:05:31

In terms of a graph describe how trap(n) differs from left(n) and what this difference has to do with f(b) - f(a).

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TRAP is the mid point between LEFT and RIGHT, which I'm going to is (f(b) -f(a)) 'dx, because it makes the difference between RIGHT and LEFT.

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GEOMETRIC SOLUTION:

First suppose n = 1. You have only one trapezoid. In this case left(n) represents the altitude of a rectangle constructed on the left-hand side of the trapezoid, so left(n) * `dx represents the area of this rectangle.

| f(b) - f(a) | is the altitude of the triangle formed by the top part of the trapezoid. Assuming b > a then if the slope of the top of the trapezoid is positive the area of the triangle is added to that of the rectangle to get the area of the trapezoid. If the slope is negative the area is subtracted.

The area of the triangle is 1/2 * | f(b) - f(a) | * `dx.

1/2 ( f(b) - f(a) ) * `dx is equal to the area if the slope is positive so that f(b) > f(a), and to the negative of the area if the slope is negative so that f(b) < f(a).

It follows that left(n) * `dx + 1/2 ( f(b) - f(a) ) * `dx = area of the trapezoid.

If you have n trapezoids then if a = x0, x1, x2, ..., xn = b represents the partition of the interval [ a, b ] then the contributions of the small triangles at the tops of the trapezoids are 1/2 ( f(x1) - f(x0) ), 1/2 ( f(x2) - f(x1) ), ..., 1/2 ( f(xn) - f(x(n-1)) ). These contributions add up to

1/2 ( (f (x1) - f(x0)) + (f (x2) - f(x1)) + ... + (f (xn) - f(x(n-1))) ). Rearranging we find that all terms except f(x0) and f(xn) cancel out:
1/2 ( -f(x0) + f(x1) - f(x1) + f(x2) - f(x2) + ... + f(x(n-1)) - f(x(n-1) + f(xn) ) =
1/2 (f(xn) - f(x0) ) = 1/2 ( f(b) - f(a) ).

Left(n) is the sum of all the areas of the left-hand rectangles, so left(n) + 1/2 ( f(b) - f(a) ) * `dx represents the areas of all the rectangles plus the contributions of all the small triangles, which gives us the trapezoidal approximation.**

SYMBOLIC SOLUTION: Assuming a < b and partition a = x0, x1, x2, ..., xn = b we have

left(n) = [ f(x0) + f(x1) + ... + f(x(n-1) ] * `dx

and

trap(n) = 1/2 [ f(x0) + 2 f(x1) + 2 f(x2) + ... + 2 f(x(n-1)) + f(xn) ] * `dx.

So

trap(n) - left(n) = [-1/2 f(x0) + 1/2 f(xn) ] * `dx. Since x0 = a and xn = b this gives us

trap(n) = left(n) + 1/2 ( f(b) - f(a) ) * `dx.