Query 6

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course MTH 174

2/9 9:11 PM

Calculus IIAsst # 6

Optional preliminary questions:

Question: `q001. You will want to sketch the situations in these preliminary questions. Your sketches will probably form the basis for your answers.

Explain why, if f(x) is linear from x = a to x = b, the area of the trapezoid formed by f(x) on the interval is equal to the area of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Explain why, if the interval a <= x <= b is partitioned into n subintervals each of width `dx = (b - a) / n, the area of the trapezoid defined by the function on any given subinterval is equal to that of a rectangle whose width is `dx and whose altitude is equal to the value of f(x) at the midpoint of that subinterval.

Explain why the sum of the areas of all the subintervals is equal to the area of the single trapezoid of the first question.

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q002. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the trapezoid defined by the straight line from (a, f(a)) to (b, f(b)) is less than or greater than that of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q003. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the region below the graph is greater or less than that of the trapezoid defined by the straight line from (a, f(a)) to (b, f(b)).

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q004. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the region below the graph is greater or less than that of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Your solution:

Confidence Rating:

Self-critique Rating:

Questions from text assignment:

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Question: Section 7.5 Problem 3

4th edition 7.5.14 problem 7.5.7 graph concave up and decreasing

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

greatest

**** between which approximations does the actual integral lie?

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Your solution:

The left would be the biggest because of the decreasing nature of the graph. The midpoint would under estimate because the graph is concave up. The trapezoid would overestimate for that reason. The right would be the smallest because of the decreasing nature. So left > midpoint > trapezoid > right. That would put the exact measurement somewhere between the midpoint and trapezoid.

confidence rating #$&*:3

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Given Solution:

If the graph is decreasing then we know that the left end of an interval gives us a higher estimate than the right.

The trapezoidal estimate is the mean of the left and right estimates, so it will lie between the two. The trapezoidal estimate also corresponds to a straight-line estimate between the graph points at the left and right endpoing of each interval.

Since the graph is concave down, any straight line between two graph points will lie below the graph. In particular the value of the function at the midpoint will lie above the straight line. The trapezoidal estimate corresponds to the straight-line, estimate so in this case the midpoint estimate exceeds the trapezoidal estimate.

The straight-line trapezoidal area is also clearly less than the area beneath the curve, so the trapezoidal estimate is lower than the actual value of the integral.

The midpoint rule uses the function value at the midpoint for the altitude of the rectangle. Thus we imagine a rectangle whose top is a horizontal line segment through the midpoint value of the graph. You should sketch this.

If the function is concave downward, the part of the actual graph that lies between the endpoints and above the midpoint-value rectangle (this segment lies to the left of the midpoint) represents graph area under the actual graph which is left out of the midpoint approximation. The area under the horizontal segment at the midpoint which lies above the graph (this are lies to the right of the midpoint) represents area included in the midpoint approximation but which is not part of the area under the curve.

Since the function is concave down, the average altitude of the ‘left-out’ area to the right is greater than that of the ‘wrongly-excluded’ area to the left. Since the widths of the two regions are equal, the wrongly excluded area must be less than the wrongly-included area and the midpoint estimate must therefore be high. We conclude that the actual area is less than the midpoint area.

Since the trapezoidal approximation is less than the actual area, we have our final ordering:

LEFT < MID < exact < TRAP < RIGHT

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23:04:39

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Self-critique (if necessary):OK

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Self-critique Rating:OK

problem 4th edition 7.5.13 was 7.5.10 graph concave DOWN and decreasing (note changes indicated by CAPS) **** list the approximations and their rules in order, from least to greatest

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11:18:42

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Your solution:

Left > Midpoint > exact > Trapezoid > Right. Using similar reasoning as the last question. If it is decreasing than the left side is bigger than the right. If it is concave down than the midpoint overestimates and the trapezoid underestimates, leaving us with an exact value between them.

confidence rating #$&*:3

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Given Solution:

Right: if f is decreasing Right < `intf(x) < Left

Trapeziod: if f is concave down Trap < intf(x) < Mid

Exact: if f is concave down Trap < intf(x) < Mid and if f is decreasing Right < `intf(x) < Left, Trap and Mid are closer approximations than right and left

Mid: if f is concave down Trap < intf(x) < Mid

Left: if f is decreasing Right < `intf(x) < Left

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11:18:43

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**** between which approximations does the actual integral lie?

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11:18:54

Trapeziod and midpoint

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11:18:54

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**** Explain your reasoning

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11:20:11

Because those two approximations are closer approximations than right and left, because they have some areas on both sides of the function so the spots that are over and under balance out, whereas right and left approximations do not

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

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11:26:14

Because the corners touch the points a and b, and since the curve is pushed upward, there is a gap between the curve and the trapezoid

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

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11:30:21

The midpoint error is less clear, but to see that one, you have to draw a line tangent to the curve at the midpoint of the interval, then by shifting the area of the rectangle that is outside this tangent to make a rectangle to form a trapezoid along this tangent, you'll see that on a concave down curve, the tangent is on top of the curve therefore having area outside the curve, producing an overestimate.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

**** Query temporarily dumbed out problem 7.5.18 graph positive, decreasing, concave upward

over interval 0 < x < h.

The trapezoidal approximation to this graph consists of 'left altitude' L1, 'right altitude' L2 and 'width' h.

The 'left altitude' L1 corresponds to f(0), the value of the function at x = 0.

The 'right altitude' L2 corresponds to f(h), the value of the function at x = h.

The graph is decreasing so L1 > L2.

The graph is concave up so it 'dips below' the trapezoidal approximation.

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

(L1 + L2) / 2 is the 'average altitude', or 'midpoint altitude' of the trapezoidal approximation, and so is an approximation to the average value and midpoint value of the fucttion.

Note that this is only an approximation. Since the graph is curved the 'average altitude' or 'midpoint altitude' of the trapezoid does not correspond to either the average value or the midpoint value of the function on this interval.

The area of a trapezoid is the average altitude multiplied by the width of the corresponding interval.

We have used trapezoidal approximation graphs since the early part of MTH 173.

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

h * f(0) is the area of a rectangle of width h and altitude f(0), which is the 'left altitude' of the graph.

Since the graph is decreasing this is an upper bound for the area beneath the curve.

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

h * f(h) is the area of a rectangle of width h and altitude f(h), which is the 'right altitude' of the graph.

Since the graph is decreasing this is a lower bound for the area beneath the curve.

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23:09:09

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

x = 0 is the left end of the interval of the x axis, x = h the right end.

h / 2 is the midpoint of the interval.

f(h/2) is the 'midpoint altitude' of the actual graph.

Since the graph is concave up, this 'midpoint altitude' of the actual graph is less than the 'midpoint altitude' of the trapezoidal approximation.

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

C is merely the area of the trapezoidal approximation, [ f(0) + f(h) ] / 2 being the average of the altitudes over the interval h and h being the width of the interval.

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

STUDENT SOLUTION:

This is merely the sum of two areas, which makes it look like the previous graph, only it is cut in half at h/2, making it more accurate than the trap approx.

INSTRUCTOR COMMENT:

Right. If you take the original graph and approximate it by two trapezoids, one running from x = 0 to x = h/2 and the other from x = h/2 to x = h, the three 'graph altitudes' are f(0), f(h/2) and f(h). You get trapezoids with areas h/2 * [ f(0) + f(h/2) ] / 2 and h/2 * [ f(h/2) } f(h) ] / 2, with total area h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) + f(h) ] / 2.

The line segments from (0, f(0)) to (h/2, f(h/2)) to (h, f(h)) lie closer to the curve than the line segment from (0, f(0)) to (h, f(h)), so the area approximation h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) } f(h) ] / 2 is closer to the actual area beneath the curve than the original area approximation h * [ f(0) + f(h) ] / 2.

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

Geometrically, E is the area of the 'left rectangle' and F the area of the 'right rectangle'. The trapezoid 'splits the difference' between these areas, so its area lies halfway between those of the two rectangles. (E + F) / 2 is halfway between E and F.

Symbolically, E = h * f(0) and F = h * f(h) so (E + F) / 2 = (h * f(0) + h ( f(h) ) / 2 = h * ( f(0) + f(h) ) / 2, which is identical to C.

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

N is an avg of the graphs of R and C. This can be observed by looking at the

corresponding graphs

**** Is E or F the better approximation to the area?

** 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. **

**** Is R or C the better approximation to the area?

** This is the midpoint vs. trapezoidal approximation situation again. Sketch a picture similar to that described on the preceding problem and identify the regions corresponding to 'wrongly included' and 'wrongly excluded' areas under each of the approximations. Compare the two and see if you can figure out whether the underestimate of the midpoint graph is greater or less than the overestimate of the trapezoidal graph. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

**** query problem 7.5.22 show trap(n) = left(n) + 1/2 ( f(b) - f(a) ) `dx **** Explain why the equation must hold.

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Your solution: ?????

confidence rating #$&*:

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Given Solution:

I understand and see that the equation does indeed work, but I am having trouble explaining why. I can picture on a graph the fact that by taking f(b) - f(a) we are taking the difference in the value of the function at points a and b. We then multiply this value by 'dx, which is determined by the number of subdivisions chosen. After doing this, we multiply by 1/2 and arrive at the value representing the area of the trapezoid. This is the best I can come up with, but I'm almost sure there is a more rigorous explanation.

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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23:17:22

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Self-critique (if necessary): I see now.

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Self-critique Rating:3

**** Query Add comments on any surprises or insights you experienced

as a result of this assignment.

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23:19:34

This was a very informative exercise, and it shows how important learning

about that trapezoidal approximation graph was in early Calculus I. I am

interested in seeing how f'' affects the accuracy of these techniques, or

how it shows something more about them. I think I am supposed to find that

out in the next assignment, according to class notes. I find this part of

the course to be very interesting.

Show that for a continuous function f(x) on the interval a <= x <= b, trap(n) = right(n) - 1/2 ( f(b) - f(a) ) ] `dx

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Question: `q005. Show that for a continuous function f(x) on the interval a <= x <= b, trap(n) = right(n) - 1/2 ( f(b) - f(a) ) ] `dx

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Your solution:

confidence rating #$&*:

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Self-critique Rating:"

Self-critique (if necessary):

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Self-critique rating:

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#$&*

Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?

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Your solution:

since there are only 4 cards of each type (spades,clubs,diamonds,aces) and it doesn't specify of which type so it would be

c=(4,2)

then it asks for 3 of another so it would be

c=(4,3)

c=(4,2)*c(4,3)

but since there are two different types of card's probabilities being figured, we apply the 13 which is the complete number of cards in a suit, then there are 12 chances because the (4,2) took the 13 leaving just 12

for the (4,3)

meaning the problem is c(4,2)*c(4,3)*13*12 probabilities of a full house consisting of two of the same number and three of the same face card

Confidence Assessment- 2

Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses.

There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind.

Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.

Self-critique:

Self Critique Rating- ok

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Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?

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Your solution:

13 cards in each type of card (ace, clubs,spades,aces)

5 cards selected within each type

c(13,5)

since there are 4 types it would be

c(13,5)*4 possibility of getting a flush containing five cards of the same suit

Confidence Assessment- 1

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Given Solution: There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.

Self Critique:

Self Critique RAting : 2

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Question: `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?

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Your solution:

there are four chances for each 5,6,7,8,9

c(4,1)*c(4,1)*c(4,1)*c(4,1) *c(4,1) possible ways to get a straight consisting of 1 of 5,6,7,8,9

it's (4,1) because only one card is being chosen of the 4 of that number.

Confidence Assessment- 3

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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.

STUDENT QUESTION

not sure I understand why is it not C(20,5)

I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not…

INSTRUCTOR RESPONSE

There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations.

However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts.

That's not a straight, nor are most of the C(20, 5) combinations of these cards.

C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).<

Self-critique:

Self Critique Rating- ok

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Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?

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Your solution:

low card meaning anything between 1-10 no face cards are allowed. In some cases an ace can either mean a 1 or more than a king.

question 7 answered how many straights there were, 4^5

10*4^5 possible ways to get a straight that consists of 5 cards in consecutive numbering

Confidence Assessment- 3

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Given Solution: There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.

STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one?

INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card.

Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination.

Self Critique:

Self Critique Rating: ok

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Question: `q009. Using a standard deck, in how many ways is it possible to get a 5-card hand consisting of all face cards?

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Your Solution:

a complete deck consists of 52 cards in all

3 cards of each type are cae cards (jack,queen ,king)

5 cardss in each hand

c=n,r since the order doesn't matter

5 cards

3 types of face cards

5!/3!(5-3)!

5*4*3*2*1/3*2*1(2*1)

Simplify takiing out 3-1 on both sides

5*4/2

20/2

10 possible ways to get a 5 card hand that is all face cards

Confidence Assessment - 1

Self Critique

Self Critique Rating-1

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Question: `q010. Using a standard deck, in how many ways is it possible to get a 5-card hand which includes exactly two face cards?

(Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?)

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Your solution:

c=(12,2) 12 face cards with 2 face cards chosen, 4 sets * 3 faces=12

12!/2!(12-2)!

12!/2*1(10)!

simplify taking out of both sides 10-1 leaving

12*11/2*1

132/2

=66

52-12=40

c(40,3)

40!/3!(40-3)!

40!/3*2*1(37)!

simplify taking out 37-1 leaving

40*39*38/3*2*1

59280/6

c(40,3)*c(12,2)

confidence assessment-1

self critique

self critique rating -1

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Self-critique (if necessary):

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Self-critique rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#

Query 6

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course MTH 174

2/9 9:11 PM

Calculus IIAsst # 6

Optional preliminary questions:

Question: `q001. You will want to sketch the situations in these preliminary questions. Your sketches will probably form the basis for your answers.

Explain why, if f(x) is linear from x = a to x = b, the area of the trapezoid formed by f(x) on the interval is equal to the area of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Explain why, if the interval a <= x <= b is partitioned into n subintervals each of width `dx = (b - a) / n, the area of the trapezoid defined by the function on any given subinterval is equal to that of a rectangle whose width is `dx and whose altitude is equal to the value of f(x) at the midpoint of that subinterval.

Explain why the sum of the areas of all the subintervals is equal to the area of the single trapezoid of the first question.

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q002. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the trapezoid defined by the straight line from (a, f(a)) to (b, f(b)) is less than or greater than that of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q003. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the region below the graph is greater or less than that of the trapezoid defined by the straight line from (a, f(a)) to (b, f(b)).

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q004. If the graph of f(x) is positive and concave downward on the interval a <= x <= b, explain whether the area of the region below the graph is greater or less than that of a rectangle whose width is b - a and whose altitude is equal to the value of f(x) at the midpoint of the interval a <= x <= b.

Your solution:

Confidence Rating:

Self-critique Rating:

Questions from text assignment:

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Question: Section 7.5 Problem 3

4th edition 7.5.14 problem 7.5.7 graph concave up and decreasing

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

greatest

**** between which approximations does the actual integral lie?

.

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Your solution:

The left would be the biggest because of the decreasing nature of the graph. The midpoint would under estimate because the graph is concave up. The trapezoid would overestimate for that reason. The right would be the smallest because of the decreasing nature. So left > midpoint > trapezoid > right. That would put the exact measurement somewhere between the midpoint and trapezoid.

confidence rating #$&*:3

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Given Solution:

If the graph is decreasing then we know that the left end of an interval gives us a higher estimate than the right.

The trapezoidal estimate is the mean of the left and right estimates, so it will lie between the two. The trapezoidal estimate also corresponds to a straight-line estimate between the graph points at the left and right endpoing of each interval.

Since the graph is concave down, any straight line between two graph points will lie below the graph. In particular the value of the function at the midpoint will lie above the straight line. The trapezoidal estimate corresponds to the straight-line, estimate so in this case the midpoint estimate exceeds the trapezoidal estimate.

The straight-line trapezoidal area is also clearly less than the area beneath the curve, so the trapezoidal estimate is lower than the actual value of the integral.

The midpoint rule uses the function value at the midpoint for the altitude of the rectangle. Thus we imagine a rectangle whose top is a horizontal line segment through the midpoint value of the graph. You should sketch this.

If the function is concave downward, the part of the actual graph that lies between the endpoints and above the midpoint-value rectangle (this segment lies to the left of the midpoint) represents graph area under the actual graph which is left out of the midpoint approximation. The area under the horizontal segment at the midpoint which lies above the graph (this are lies to the right of the midpoint) represents area included in the midpoint approximation but which is not part of the area under the curve.

Since the function is concave down, the average altitude of the ‘left-out’ area to the right is greater than that of the ‘wrongly-excluded’ area to the left. Since the widths of the two regions are equal, the wrongly excluded area must be less than the wrongly-included area and the midpoint estimate must therefore be high. We conclude that the actual area is less than the midpoint area.

Since the trapezoidal approximation is less than the actual area, we have our final ordering:

LEFT < MID < exact < TRAP < RIGHT

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Self-critique (if necessary):OK

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Self-critique Rating:OK

problem 4th edition 7.5.13 was 7.5.10 graph concave DOWN and decreasing (note changes indicated by CAPS) **** list the approximations and their rules in order, from least to greatest

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Your solution:

Left > Midpoint > exact > Trapezoid > Right. Using similar reasoning as the last question. If it is decreasing than the left side is bigger than the right. If it is concave down than the midpoint overestimates and the trapezoid underestimates, leaving us with an exact value between them.

confidence rating #$&*:3

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Given Solution:

Right: if f is decreasing Right < `intf(x) < Left

Trapeziod: if f is concave down Trap < intf(x) < Mid

Exact: if f is concave down Trap < intf(x) < Mid and if f is decreasing Right < `intf(x) < Left, Trap and Mid are closer approximations than right and left

Mid: if f is concave down Trap < intf(x) < Mid

Left: if f is decreasing Right < `intf(x) < Left

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11:18:43

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**** between which approximations does the actual integral lie?

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11:18:54

Trapeziod and midpoint

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**** Explain your reasoning

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11:20:11

Because those two approximations are closer approximations than right and left, because they have some areas on both sides of the function so the spots that are over and under balance out, whereas right and left approximations do not

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

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11:26:14

Because the corners touch the points a and b, and since the curve is pushed upward, there is a gap between the curve and the trapezoid

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

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11:30:21

The midpoint error is less clear, but to see that one, you have to draw a line tangent to the curve at the midpoint of the interval, then by shifting the area of the rectangle that is outside this tangent to make a rectangle to form a trapezoid along this tangent, you'll see that on a concave down curve, the tangent is on top of the curve therefore having area outside the curve, producing an overestimate.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

**** Query temporarily dumbed out problem 7.5.18 graph positive, decreasing, concave upward

over interval 0 < x < h.

The trapezoidal approximation to this graph consists of 'left altitude' L1, 'right altitude' L2 and 'width' h.

The 'left altitude' L1 corresponds to f(0), the value of the function at x = 0.

The 'right altitude' L2 corresponds to f(h), the value of the function at x = h.

The graph is decreasing so L1 > L2.

The graph is concave up so it 'dips below' the trapezoidal approximation.

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

(L1 + L2) / 2 is the 'average altitude', or 'midpoint altitude' of the trapezoidal approximation, and so is an approximation to the average value and midpoint value of the fucttion.

Note that this is only an approximation. Since the graph is curved the 'average altitude' or 'midpoint altitude' of the trapezoid does not correspond to either the average value or the midpoint value of the function on this interval.

The area of a trapezoid is the average altitude multiplied by the width of the corresponding interval.

We have used trapezoidal approximation graphs since the early part of MTH 173.

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

h * f(0) is the area of a rectangle of width h and altitude f(0), which is the 'left altitude' of the graph.

Since the graph is decreasing this is an upper bound for the area beneath the curve.

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

h * f(h) is the area of a rectangle of width h and altitude f(h), which is the 'right altitude' of the graph.

Since the graph is decreasing this is a lower bound for the area beneath the curve.

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

x = 0 is the left end of the interval of the x axis, x = h the right end.

h / 2 is the midpoint of the interval.

f(h/2) is the 'midpoint altitude' of the actual graph.

Since the graph is concave up, this 'midpoint altitude' of the actual graph is less than the 'midpoint altitude' of the trapezoidal approximation.

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

C is merely the area of the trapezoidal approximation, [ f(0) + f(h) ] / 2 being the average of the altitudes over the interval h and h being the width of the interval.

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

STUDENT SOLUTION:

This is merely the sum of two areas, which makes it look like the previous graph, only it is cut in half at h/2, making it more accurate than the trap approx.

INSTRUCTOR COMMENT:

Right. If you take the original graph and approximate it by two trapezoids, one running from x = 0 to x = h/2 and the other from x = h/2 to x = h, the three 'graph altitudes' are f(0), f(h/2) and f(h). You get trapezoids with areas h/2 * [ f(0) + f(h/2) ] / 2 and h/2 * [ f(h/2) } f(h) ] / 2, with total area h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) + f(h) ] / 2.

The line segments from (0, f(0)) to (h/2, f(h/2)) to (h, f(h)) lie closer to the curve than the line segment from (0, f(0)) to (h, f(h)), so the area approximation h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) } f(h) ] / 2 is closer to the actual area beneath the curve than the original area approximation h * [ f(0) + f(h) ] / 2.

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

Geometrically, E is the area of the 'left rectangle' and F the area of the 'right rectangle'. The trapezoid 'splits the difference' between these areas, so its area lies halfway between those of the two rectangles. (E + F) / 2 is halfway between E and F.

Symbolically, E = h * f(0) and F = h * f(h) so (E + F) / 2 = (h * f(0) + h ( f(h) ) / 2 = h * ( f(0) + f(h) ) / 2, which is identical to C.

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

N is an avg of the graphs of R and C. This can be observed by looking at the

corresponding graphs

**** Is E or F the better approximation to the area?

** 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. **

**** Is R or C the better approximation to the area?

** This is the midpoint vs. trapezoidal approximation situation again. Sketch a picture similar to that described on the preceding problem and identify the regions corresponding to 'wrongly included' and 'wrongly excluded' areas under each of the approximations. Compare the two and see if you can figure out whether the underestimate of the midpoint graph is greater or less than the overestimate of the trapezoidal graph. **

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Self-critique (if necessary):ok

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**** query problem 7.5.22 show trap(n) = left(n) + 1/2 ( f(b) - f(a) ) `dx **** Explain why the equation must hold.

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Your solution: ?????

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Given Solution:

I understand and see that the equation does indeed work, but I am having trouble explaining why. I can picture on a graph the fact that by taking f(b) - f(a) we are taking the difference in the value of the function at points a and b. We then multiply this value by 'dx, which is determined by the number of subdivisions chosen. After doing this, we multiply by 1/2 and arrive at the value representing the area of the trapezoid. This is the best I can come up with, but I'm almost sure there is a more rigorous explanation.

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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23:17:22

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Self-critique (if necessary): I see now.

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Self-critique Rating:3

**** Query Add comments on any surprises or insights you experienced

as a result of this assignment.

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23:19:34

This was a very informative exercise, and it shows how important learning

about that trapezoidal approximation graph was in early Calculus I. I am

interested in seeing how f'' affects the accuracy of these techniques, or

how it shows something more about them. I think I am supposed to find that

out in the next assignment, according to class notes. I find this part of

the course to be very interesting.

Show that for a continuous function f(x) on the interval a <= x <= b, trap(n) = right(n) - 1/2 ( f(b) - f(a) ) ] `dx

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Question: `q005. Show that for a continuous function f(x) on the interval a <= x <= b, trap(n) = right(n) - 1/2 ( f(b) - f(a) ) ] `dx

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Your solution:

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