course Mth 272 dƙū۟assignment #011
......!!!!!!!!...................................
15:12:21 5.5.4 (previous probme was 5.6.2 midpt rule n=4 for `sqrt(x) + 1 on [0,2]) 5.5.4 asks for an n = 4 midpoint-rule approximation to the integral of 1 - x^2 on the interval [-1, 1].
......!!!!!!!!...................................
RESPONSE --> ( 1 - (-1) ) / 4 = 1/2 so [-1, -.5], [-.5, 0], [0, 5], [.5,1] 1-x^2 to get the midpoints at -.75, .25, .25, .75 and y= .4375, .9375, .9375, .4375 so multiply through by 1/2 to get .21875, .46875, .46875, .21875, respectivly totaling to 1.375
.................................................
......!!!!!!!!...................................
15:29:03 ** Dividing [-1, 1] into four intervals each will have length ( 1 - (-1) ) / 4 = 1/2. The four intervals are therefore [-1, -.5], [-.5, 0], [0, 5], [.5,1]. The midpoints are -.75, .25, .25, .75. You have to evaluate 1 - x^2 at each midpoint. You get y values .4375, .9375, .9375 and .4375. These values will give you the altitudes of the rectangles used in the midpoint approximation. The width of each rectangle is the length 1/2 of the interval, so the areas of the rectangles will be 1/2 * .4375,1/2 * .9375, 1/2 * .9375 and 1/2 * .4375, or .21875, .46875, .46875, .21875. Adding these areas we get total area 1.375. The curve is concave down so the midpoints will give you values which are a little high. We confirm this by calculating the integral: The exact integral is integral(1 - x^2, x from 0 to 2). An antiderivative is x - 1/3 x^2; evaluating from -1 to 1 we find that the antiderivative changes from -2/3 to 2/3, a change of 4/3 = 1.333. So the accurate integral is 4/3 = 1.333 and our estimate 1.375 is indeed a little high. ** DER
......!!!!!!!!...................................
RESPONSE --> antiderivative is x - 1/3 x^2 the antiderivative changes from -2/3 to 2/3, a change of 4/3 = 1.333. So the accurate integral is 4/3 = 1.333 and our estimate 1.375, little high.
.................................................
......!!!!!!!!...................................
15:31:56 5.6.12 (was 5.6.10 midpt rule n=4 for x^2-x^3 on [-1,0]
......!!!!!!!!...................................
RESPONSE --> intervals: (-1, -.75), (-.75, -.5), (-.5, -.25). (-.25, 0) with midpoints of -.875, -.625, -.375, -.125 use x^2 - x^3 to get 1.44, 0.63, 0.19, 0.02 total area is (735 + 325 + 99 + 9) / 2048 = /2048 = .5703
.................................................
......!!!!!!!!...................................
15:32:47 ** The four intervals are (-1, -3/4), (-3/4, -1/2), (-1/2, -1/4) and (-1/4, 0); in decimal form these are (-1, -.75), (-.75, -.5), (-.5, -.25) and (-.25, 0). The midpoints of these intervals are-7/8, -5/8, -3/8 and -1/8; in decimal form we get -.875, -.625, -.375, -.125. The values of the rectangle heights at the midpoints are found by evaluating x^2 - x^3 at the midpoints; we get respectively 735/512, 325/512, 99/512 and 9/512, or in decimal form 1.435546875; 0.634765625; 0.193359375; 0.017578125. The approximating rectangles each have width 1/4 or .25 so the areas arerespectively 735/2048 325/2048, 99/2048, 9/2048, or in decimal form 0.3588867187; 0.1586914062; 0.04833984375; 0.00439453125. The total area is (735 + 325 + 99 + 9) / 2048 = /2048 = 73/128, or in decimal form approximately .5703. An antiderivative of the function is x^3 / 3 - x^4 / 4; evaluating from -1 to 0 we obtain 1/3 + 1/4 = 7/12 = .5833... . So the midpoint approximation is low by about .013 units. ** DER
......!!!!!!!!...................................
RESPONSE --> antiderivative is x^3 / 3 - x^4 / 4 so 1/3 + 1/4 = 7/12 = .5833 the approximation was a little low
.................................................
......!!!!!!!!...................................
15:34:50 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> there is more then one way to solve a problem. you can check how close your approximation is by using the antiderivative and evaluating. or use the antiderivative to find the exact answer
.................................................
KSywpӉǰ assignment #012 {hİ}˭ǚ Applied Calculus I 06-26-2006
......!!!!!!!!...................................
15:50:16 5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5]
......!!!!!!!!...................................
RESPONSE --> ( 5 - 1 ) / 4 = 1 so intervals are [1, 2], [2, 3], [3, 4], [4,5] endpoint values are 0, .5, .471, .433 and . 4 (0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417 totaling too 1.6
.................................................
......!!!!!!!!...................................
15:51:28 ** Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore [1, 2], [2, 3], [3, 4], [4,5]. The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore (0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417. Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417. The sum of these areas is the trapezoidal approximation 1.604. ** DER
......!!!!!!!!...................................
RESPONSE --> trapeoid areas are equal to average height * width since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417
.................................................
......!!!!!!!!...................................
15:54:21 5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft). How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater?
......!!!!!!!!...................................
RESPONSE --> average alitudes would be (0 + 50 / 2) = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 ...
.................................................
......!!!!!!!!...................................
15:55:02 ** Since widths at 20-ft intervals are 50, 54, 82, 82, 73, 75, 80 ft the pond area can be approximated by a series of trapezoids with these altitudes. The average altitudes are therefore respectively (0 + 50 / 2) = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 etc., with corresponding areas 25 * 20 = 500 52 * 20 = 1040 etc., all areas in ft^2. The total area, according to the trapezoidal approximation, will therefore be 20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 square feet. The midpoint widths would be calculated based on widths at positions 10, 30, 50, 70, ..., 150 ft. Due to the convex shape of the pond these estimates and will lie between the estimates made at 0, 20, 40, ..., 160 feet. The convex shape of the pond also ensures that the midpoint of each rectangle will 'hump above' the trapezoidal approximation, so the midpoint estimate will exceed the trapezoidal estimate. **
......!!!!!!!!...................................
RESPONSE --> then find the areas and the total area which is 9920 sq. ft.
.................................................
......!!!!!!!!...................................
15:55:47 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> different but consecutive steps are used for these types of problems.
.................................................
"