course MTH 174 8:31 p.m. 6/21/10. This QA took a little work to find. The link you made for it on
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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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Self-critique (if necessary): This answer is wrong because the graph is said to be concave up in the question but the solution has the graph as being concave down. This effectively reverses the TRAP and MID positions in the answer.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This problem matches the solution given above which is: RIGHT < TRAP < exact value < MID < LEFT. 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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********************************************* Question: **** between which approximations does the actual integral lie? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TRAP and MID
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Trapeziod and midpoint
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**** Explain your reasoning: TRAP and MID are the two most exact methods here and we know and when f(x) is concave down, as we have here, then TRAP(n) ≤ Int(f(x), x, a,b) ≤ MID(n).
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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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********************************************* Question: **** if you have not done so explain why when a function is concave down the trapezoidal rule UNDERestimates the integral YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What happens is the function is arched opening down when you draw a line between any two points on the curve the line will lie below the curve, which is a chord, the area under the line is therefore less than the area under the curve.
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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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********************************************* Question:**** if you have not done so explain why when a function is concave down the midpoint rule OVERrestimates the integral YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here we can look at the midpoint of the rectangular interval as being tangent to the curve. If the curve is concave down then the endpoints of the rectangle lie above the curve giving an increase in the area of the rectangle over the area under the curve.
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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 Self-critique rating #$&* 3 ********************************************* Question:**** Problem 7.5.18 (actually Problem 4 from as 7.5.18 is problem 5 in the homework) graph positive, decreasing, concave upward over interval 0 < x < h. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L1 corresponds to f(0) and L2 to f(h) which are the intervals endpoints ordinates and are viewed as the bases of a trapezoid with h being the height and thus h(L1+L2)/2 follows from h(b1+b2)/2. This gives a TRAP(1) (over) approximation of the area under the curve. Confidence rating #$&*3
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Given Solution: 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. ********************************************* Question:**** why is the area of the trapezoid h (L1 + L2) / 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Explained above. (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 fuction. 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. ********************************************* Question:**** Describe how you sketched the area E = h * f(0) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This gives the are of a rectangle whose length is h and whose height is f(0). This corresponds to a LEFT(1) (over) approximation. 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. Question: **** Describe how you sketched the area F = h * f(h) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This area is a rectangle with length h and height of f(h). This corresponds to a RIGHT(1) (under) approximation. 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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********************************************* Question: **** Describe how you sketched the area R = h*f(h/2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is a rectangle whose length is h and height is f(h/2). Since h/2 is the midpoint of the interval this area corresponds to a MID(1) (under) approximation. 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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Question:**** Describe how you sketched the area C = h * [ f(0) + f(h) ] / 2 Your solution: This is just like the first part of the question with L1 and L2 replaced by f(0) and f(h), respectively, and like above gives a TRAP(1) (over) approximation. 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. Question:**** Describe how you sketched the area N = h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) } f(h) ] / 2 Your solution: This equation and the original problem both seem to have typos but seem to want to mean: N = h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2)+ f(h) ] / 2 (the homework lacked the final /2 and the above lacks the rightmost +). Regardless, this is the area of two trapezoids whose h= h/2 and bases are f(0) and f(h/2) then f(h/2) and f(h). This is equivalent to a TRAP(2) (still over) approximation. 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. ********************************************* Question: **** why is C = ( E + F ) / 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of a trapezoid, which the C formula is, is the average of the area two rectangles, which E and F are. Using a little factoring algebra you can do this: (h*f(0)+h*f(h))/2 = h*((f(0)+f(h))/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. ********************************************* Question: **** Why is N = ( R + C ) / 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My gut reaction from the graph is to say it is an average of the two areas but I can't get this to work out algebraically. N is an avg of the graphs of R and C. This can be observed by looking at the corresponding graphs ********************************************* Question: **** Is E or F the better approximation to the area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F includes almost totally 'actual area' where E contains a lot of 'extra' 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. ** ********************************************* Question:**** Is R or C the better approximation to the area? Your solution: Here again we can say that the trapezoid contains noticeably more 'extra' area than the midpoint formula does. So R is the better method. ** 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. ** .....!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK except for the N= (R+C)/2. Self-critique rating #$&* 3 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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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I couldn't get the problem in this section to work out. And this equation differs from the one in the homework which is: TRAP(n) = LEFT(n) + 1/2(f(h) + f(0))* `dx. The one here makes more sense than the other one and I know that somehow it can be derived from the TRAP= (LEFT + RIGHT)/2 formula but just can't get it to work out quite right. Confidence rating #$&* 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay this last part helped me see where I was going wrong with the algebra so I got it. Self-critique rating #$&* 3 **** Query Add comments on any surprises or insights you experienced as a result of this assignment. ……!!!!!!!!……………………………. This was a good refresher from Calculus I and the Riemann Sums that eluded me on the first test.
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