query 6

#$&*

course mth 174

5/8 8:30 pm

Calculus IIAsst # 6

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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:

Since the graph is decreasing, we know that the right approximation is low, and the left approximation is high. We also know that since the trapezoidal approximation is the mean of the left and right we know that it lies in between the two

So far we have from least to greatest

Right, trapezoidal, left.

We also have to decide about the actual integral. We know that the actual value is greater than the trapezoidal value since when we draw a trapezoid in the graph it is below the slope of the line.

We also know that since the graph is concave down, the midpoint value is greater than the trap value

We also know that concave down in the midrange being more than the exact value.

so we can now say from least to greatest

Right, Trap, Actual, Mid, Left

The actual value lies between the trap value and midrange level.

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:

RIGHT < TRAP < exact < MID < LEFT.

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23:04:39

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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:

Now we have a concave down decreasing graph.

We know that for concave down graphs the trap approx. is high

For decreasing graphs we know that the left is high and the right is low. We also know that trap is the mean of the left and right so it is in the middle. So far from least to greatest we have

Right, trap, left

We also know that for concave down, the trap. Value is less than the midrange so now we have

Right, trap, midrange, left

We also know that in concave down the midrange is more than the exact so we finally have

Right, trap, exact, midrange left

The actual value lies between the trap and midrange

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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11:20:12

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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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11:26:14

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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 - I used the rules in the book to make my determinations as well as a sketch of a concave down graph. For the trapezoid estimate by drawing a trapezoid I see that the trapezoid remains under the line so I know that the trapezoid value is underestimated and that the actual value would be greater than the trapezoid value. I also can see that the midpoint is an overestimate by drawing the midpoint rectangle, because it extends beyond the line so the midpoint is more than the actual.

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Self-critique Rating:3

**** 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?

###Because the trapezoidal approximation is the mean of L1 and L2 with the understanding that sometimes the trapezoidal estimate is overestimated or underestimated. This trapezoidal value is also equivalent to the midpoint value.

(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)

### when I sketched E=h*f(0) I know that the h is the width and that f(0) is the height and F(0) is the left bound and is equal to L1 which is 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)

###this is the right height of the rectangle with width h and height L(2) which is 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)

### h/2 would be the midpoint between zero and h. so h*f(h/2) would be the midpoint multiplied by the height so it would be the height of the midpoint.

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=h*[f(0) + f(h)] /2 Is the mean or trapezoidal approximate value of the area from f(0) to f(h) on the x axis with a height of h. so C is the trapezoidal approx. value.

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

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

####When I drew this it looked the same as the original graph but I see that the values are halved. Which seems to checking to see if the trapezoidal estimate was over or underestimated.

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?

###We determined above that C is the mean so E and F must be the left and right rectangles of the graph thus the right plus left divided by 2 would give you c or the trapezoidal estimate.

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?

I see above that N is used to give a better approximation of the trapezoidal value by dividing the trapezoid in 2, I don’t understand how N is representative of the average???

@&

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

Expanding this expression and simplifying we get

N = (h/2 f(0) + h/2 f(h) + h f(h/2)) / 2

= ( h * (f(0) + f(h)) / 2 + h f(h/2) ) / 2

which is the average of

h * (f(0) + f(h)) / 2 and h f(h/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?

###Since the Graph is decreasing and E represent the left and F represents the right, then we know that left is greater than right so we would say that there is more area wrongly included from the left side than there is area wrongly left out of the right side. So that would mean that F is a better approximation of the area and closer to the actual value.

** 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?

### C is the trapezoidal value, R is the average value or midpoint. In drawing the graph of concave up and decreasing graphs I can see that the trap value is overestimated so it wrongly includes area. For the midpoint, the value is underestimated which means that area was wrongly left out. It appears to me that there is more area left out of the midpoint underestimate than area wrongly included in the trapezoid overestimate so I believe the midpoint is closer to the actual 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, I think I’m ok but let me know if I am misunderstanding anything????

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Self-critique Rating:

@&

Good, but check my one note in answer to your question.

*@

**** 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:

I know that ‘dx is equal to the value under the graph for a given integral. The equation reads trap(n) where n is the number of integrals or trapezoids I will let n = 1. F(b) would be the value of the area of the right hand integral (which is represented by drawing a rectangle in this area) and f(a) would be the value of the area of the left hand integral (which is represented by drawing a rectangle in this interval)

The normal equation for the trap is trap(n) = [(f(a) + f(b)))/2 ] *1

Left n is equal to f(a)*1 + f(b)*1

So if we have trap(1) = left(1) + 1/2 ( f(b) - f(a) ) ] `dx

We can substitute in that

[(f(a) + f(b)))/2 ] *1= f(a)*1 + f(b)*1+1/2 ( f(b) - f(a) ) ] `dx

So simplifying I get

.5f(a) + .5f(b) = f(a) + f(b) + .5f(b) - .5f(a)

.5f(a) + .5f(b) = .5f(a) + 1.5f(b)

.5f(b) = 1.5f(b)

###I’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

@&

`dx is the change in the x value. This would correspond to the width of the trapezoid.

In other words `dx is equivalent to the h used previously.

*@

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

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Self-critique Rating:3

**** 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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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I know that ‘dx is equal to the value under the graph for a given integral. The equation reads trap(n) where n is the number of integrals or trapezoids I will let n = 1. F(b) would be the value of the area of the right hand integral (which is represented by drawing a rectangle in this area) and f(a) would be the value of the area of the left hand integral (which is represented by drawing a rectangle in this interval)

The normal equation for the trap is trap(n) = [(f(a) + f(b)))/2 ] *1

Left n is equal to f(a)*1 + f(b)*1

So if we have trap(1) = left(1) + 1/2 ( f(b) - f(a) ) ] `dx

We can substitute in that

[(f(a) + f(b)))/2 ] *1= f(a)*1 + f(b)*1+1/2 ( f(b) - f(a) ) ] `dx

So simplifying I get

.5f(a) + .5f(b) = f(a) + f(b) + .5f(b) - .5f(a)

.5f(a) + .5f(b) = .5f(a) + 1.5f(b)

.5f(b) = 1.5f(b)

###I’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

@&

`dx is the change in the x value. This would correspond to the width of the trapezoid.

In other words `dx is equivalent to the h used previously.

*@

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

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Self-critique Rating:3

#*&!

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

......!!!!!!!!...................................

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I know that ‘dx is equal to the value under the graph for a given integral. The equation reads trap(n) where n is the number of integrals or trapezoids I will let n = 1. F(b) would be the value of the area of the right hand integral (which is represented by drawing a rectangle in this area) and f(a) would be the value of the area of the left hand integral (which is represented by drawing a rectangle in this interval)

The normal equation for the trap is trap(n) = [(f(a) + f(b)))/2 ] *1

Left n is equal to f(a)*1 + f(b)*1

So if we have trap(1) = left(1) + 1/2 ( f(b) - f(a) ) ] `dx

We can substitute in that

[(f(a) + f(b)))/2 ] *1= f(a)*1 + f(b)*1+1/2 ( f(b) - f(a) ) ] `dx

So simplifying I get

.5f(a) + .5f(b) = f(a) + f(b) + .5f(b) - .5f(a)

.5f(a) + .5f(b) = .5f(a) + 1.5f(b)

.5f(b) = 1.5f(b)

###I’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

@&

`dx is the change in the x value. This would correspond to the width of the trapezoid.

In other words `dx is equivalent to the h used previously.

*@

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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’m not sure what I am doing wrong here and I’m not sure what to do about the ‘dx. I think ‘dx represents the area under the graph but I’m not sure what that value would be with 1 interval????

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Self-critique Rating:3

#*&!#*&!

&#Good responses. See my notes and let me know if you have questions. &#