Assignment 6

course Mth 174

Xϳ|ɮassignment #006

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Physics II

02-05-2007

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16:51:08

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

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17:02:23

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

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Least to greatest:

RIGHT(n): less than exact when decreasing on interval

TRAP(n): less than exact when concave down on interval

EXACT

MID(n): greater than exact when concave down on interval

LEFT(n): greater than exact when decreasing on interval

Based on shape of curve, RIGHT(n) is less than TRAP(n) and MID(n) is less than LEFT(n)

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17:02:38

between which approximations does the actual integral lie?

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See previous response

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17:06:50

Explain your reasoning

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Based on the shape of the curve (using n=1 as an example), RIGHT(n) produces a much smaller area than TRAP(n), with both under the exact, and MID(n) produces a smaller area than LEFT(n), with both over the exact. Therefore, the exact must lie between TRAP(n) and MID(n).

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17:11:36

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

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The trapezoidal rule underestimates the integral because straight lines are drawn from point to point (the points where the value of the function intersects the x-values of the ends of the intervals), and these lines are always below the function (because it curves upward away from the lines), thus excluding part of the area under the function.

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17:18:01

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

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The midpoint rule overestimates the the integral because the function curves below the tangent line at the midpoint on both sides of the midpoint, and the area of the trapezoid created by the tangent line and the vertical lines marking the ends of the interval is equal to the area of the rectangle created by the horizontal line through the midpoint to the same vertical lines. Thus, since the trapezoid excludes some of the area under the function on the interval, so does the rectangle.

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17:20:53

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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17:29:31

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

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Guess from information given:

The area of a trapezoid is equal to the average of the two bases (which essentially converts the trapezoid into a rectangle of equal area with the base equal to the average) multiplied by the altitude. In thi case, (L1 + L2)/2 must be the average of the bases, and the delta x must be (h - 0).

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17:35:34

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

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Guess based on given information:

This appears to be a LEFT(1) situation, which results in a rectangle with vertices of (0,0), (h,0), (h,f(0)), and (0,f(0)).

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17:38:40

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

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This appears to be a RIGHT(1) situation creating a rectangle with vertices: ((0,0), (h,0), (h,f(h)), (0,f(h)).

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17:41:12

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

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Appears to be MID(1) situation: rectangle with vertices (0,0), (h,0), (h,f(h/2)), (0,f(h/2)).

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17:47:42

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

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Appears to be TRAP(1), creating a rectangle with vertices: (0,0), (h,0), (h,[f(0)+f(h)]/2), (0,[f(0)+f(h)]/2).

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17:58:49

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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This appears to be TRAP(2) situation: creates two rectangles:

First: (0,0), (h/2,0), (h/2,[f(0)+f(h/2)]/2), (0,[f(0)+f(h/2)]/2)

Second: (h/2,0), (h,0), (h,[f(h/2)+f(h)]/2), (h/2,[f(h/2)+f(h)]/2)

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18:14:37

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

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C is the TRAP(1) rule, E is the LEFT(1) rule, and F is the RIGHT(1) rule. By definition, the trapezoid rule results in the average of the left and right rules.

Algebraically:

(Left rule + right rule)/2 = trapezoid rule

Applying the distributive rule to distribute the ""h"" on the right side below would produce the left side.

(h*f(0) + h*(f(h)) /2 = h*[f(o)+f(h)]/2

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20:01:10

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

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This states that the result of the TRAP(2) rule is equal to the average of the midpoint and TRAP(1) rules.

Algebraically,

Does (h/2)*[f(0)+f(h/2)]/2 + (h/2)*[f(h/2)+f(h)]/2 = {[h*f(h/2) + h*(f(0)+f(h)]/2}/2

Cleaning this up does indeed produce an equality, so the result of the TRAP(2) rule is equal to the average of the midpoint and TRAP(1) rules.

Logically: The sums of the areas above and below the rectangles created by the TRAP(2) rule (i.e., divided at h/2) that are created by the MID(1) and TRAP(1) rules are equal and opposite (i.e., one side above TRAP(2) and one side below) on either side of h/2. Therefore, averaging those two rules is adding and subtracting equal areas from what is the area produced by the TRAP(2) rule.

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20:06:54

Is E or F the better approximation to the area?

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As long as the curve is concave up the right approximation (F) will be a better approximation. If the curve is a straight line, they should have equal and opposite errors, and when the curce is concave down, the left will become the better approximation.

You clearly understand. Another good explanation:

** F is closer. The area 'wrongly included' under the graph of E is greater than the area 'wrongly excluded' by the graph of F because the average thickness of the former region is greater than that of the latter. The upward concavity of f means that it's closer to the right-hand approximation than to the left-hand approximation for most of the interval. **

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20:12:42

Is R or C the better approximation to the area?

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The MID(1) rule (R) is a better approximation than the TRAP(1) rule (C).

Excellent answers on this series of questions, concisely stated and correct throughout.

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20:15:14

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

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21:14:44

Explain why the equation must hold.

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21:26:38

Explain why the equation must hold.

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To prove the equation for #25 (which is what you asked for, not #24 as listed in the question), first #24 is proved:

The difference between the left and right sums is the left sum includes f(a)delta x and the right sum includes f(b) delta x, all the other products in the sums are the same. So if the remaining products = Q, then:

RIGHT(n) = Q + f(b)*delta x

LEFT(n) = Q + f(a)*delta x

Subtracting:

RIGHT(n) - LEFT(n) = f(b)*delta x - f(a)* delta x, or:

RIGHT(n) = LEFT(n) + f(b)*delta x - f(a)* delta x

TRAP(n) = [LEFT(n) + RIGHT(n)]/2 (by definition)

Substituting from above for RIGHT(n):

TRAP(n) = [LEFT(n) + LEFT(n) + f(b)*delta x - f(a)* delta x]/2

Simplifying:

TRAP(n) = LEFT(n) + [f(b)*delta x - f(a)*delta x]/2

TRAP(n) = LEFT(n) + (1/2)[f(b) - f(a)]*delta x

Which is the equation in #25

Excellent solution. Again, just FYI:

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.

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21:43:12

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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For each interval of the function, the area of the trapezoid (TRAP(n) rule) is the average of the areas of the rectangles for the LEFT(n) and RIGHT(n) rules--or the area of the LEFT(n) rectangle plus half the difference between the two rectangles (LEFT(n) and RIGHT(n). Overall, TRAP(n) is the average of LEFT(n) and RIGHT(n), or LEFT(n) plus half the difference between LEFT(n) and RIGHT(n). f(b) - f(a) is the difference between LEFT(n) and RIGHT(n), so 1/2 of f(b) - f(a) added to LEFT(n) equals TRAP(n).

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Excellent work. Let me know if you have questions.