course Mth 173 Calculus I10-29-2007
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21:39:49 query problem 5.2.23 integral of x^2+1 from 0 to 6. What is the value of your integral, and what were your left-hand and right-hand estimates?
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RESPONSE --> for the formula x^2+1 I got x y 0 1 1 2 2 5 3 10 4 17 5 26 6 37 My left sum was 61 and my right was 97 with a difference of 36 between them.
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21:59:02 ** An antiderivative of x^2 + 1 is x^3 / 3 + x. The change in the antiderivative from x = 0 to x = 6 is equal to the integral: Integral = 6^3 / 3 + 6 - (0^3 / 3 + 0) = 216/3 + 6 = 234/3 = 78. The values of the function at 0, 2, 4, 6 are 1, 5, 17 and 37. The left-hand sum would therefore be 2 * 1 + 2 * 5 + 2 * 17 = 46. The right-hand sum would be 2 * 5 + 2 * 17 + 2 * 37 = 118. The sums differ by ( f(b) - f(a) ) * `dx = (37 - 1) * 2 = 72.**
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RESPONSE --> I forgot the most important part the intrgral sorry! x^2+1 Antiderivative= x^3 /3+x The intergral from 0->6 is (6^3 / 3+6) - (0^3 / 3 +0) = 78
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22:03:40 From the shape of the graph which estimate would you expect to be low and which high, and what property of the graph makes you think so?
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RESPONSE --> The graph would be a parabola with concave up, so the graph would get closer to the left hand sum during each intreval due to the concavity.
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22:04:10 ** The graph is increasing so the left-hand sum should be the lesser. the graph is concave upward and increasing, and for each interval the average value of the function will therefore be closer to the left-hand estimate than the right. This is corroborated by the fact that 46 is closer to 78 than is 118. **
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RESPONSE --> okay
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22:18:29 query problem 5.2.32 f(x) piecewise linear from (-4,1) to (-2,-1) to (1,1) to (3,0) then like parabola with vertex at (4,-1) to (5,0). What is your estimate of the integral of the function from -3 to 4 and how did you get this result?
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RESPONSE --> x y -4 1 -2 -1 1 1 3 0 4 -1 Left hand sum= (1*2)+(-1*2)+(1*2)+(0*2)= 2 Right hand sum= (-1*2)+(1*2)+(0*2)+(-1*2)= -2 So the integral lies between the two sums (if I did everything correctly) Since the the graph is liner untill (3,0) the integral should lie pretty close to the bi-sector of the rieman sum box, so it would be zero but the graph is parabolic on the last point and would lie closer to the lower sum, so for the intrgral I would estimate about -.5
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22:24:53 ** We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly. From x=-3 to x=0 the area between the graph and the x axis is 2 (divide the region into two triangles and a rectangle and find total area, or treat it as a trapezoid). The area is below the x axis so the integral from x = -3 to x = 0 is -2. From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2. If the graph from x=3 to x=4 was a straight line from (3,0) to (4,-1) the area over this interval would be .5 and the integral would be -.5. Since the graph curves down below this straight line the area will be more like .6 and the integral closer to -.6. The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6. **
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RESPONSE --> So I used the rieman sums and still ended up with the same answer because we were so close to the x-axis and everything essentially cancelled out, so when should I find the area of the trapazoid and when should I use rieman sums?
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22:32:28 query problem 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval?
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RESPONSE --> I am assuming that you mean t= 0 -> 10 x y 0 1 1 2.7 2 7.4 3 20.1 4 54.6 5 148.4 6 403.4 7 1096.6 8 2981.0 9 8103.1 10 22026.5 By averaging the terms we get 17422.4 for the average y value.
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22:38:17 ** An antiderivative of f(t) = e^t is F(t) = e^t. Using the Fundamental Theorem you get the definite integral by calculating F(10) - F(0). This gives you e^10 - e^0 = 22025, approx. The average value of the function is the integral divided by the length of the interval. The length of the interval is 10 and 22025 / 10 is 2202.5. **
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RESPONSE --> again I've done things wrong: using the idea that the 'd integral / 'dx is the average of the function. The antiderivative of the e^t function is itself so we have the same numbers as before but to make things easier we can just use the first and last terms e^(10) - e^(0) = 22025 since this is the integral we divide by 'dx which is 10 so we get 2202.5
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22:54:09 10-29-2007 22:54:09 What would be the height of a rectangle sitting on the t axis from t = 0 to t = 10 whose area is the same as that under the y = e^t graph from t=0 to t=10?
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NOTES -------> y= e^t x y avg H Area 0 1 2 7.4 4.2 8.4 4 54.6 31 62 6 403.4 229 124 8 2981.0 1692.2 3384.4 10 22026.5 12503.8 25007.6 Total Area under e^t= 28586.4 A= h * 'dx 28586.4 = h *10 2858.64 = h
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22:54:14 What would be the height of a rectangle sitting on the t axis from t = 0 to t = 10 whose area is the same as that under the y = e^t graph from t=0 to t=10?
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RESPONSE --> y= e^t x y avg H Area 0 1 2 7.4 4.2 8.4 4 54.6 31 62 6 403.4 229 124 8 2981.0 1692.2 3384.4 10 22026.5 12503.8 25007.6 Total Area under e^t= 28586.4 A= h * 'dx 28586.4 = h *10 2858.64 = h
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22:55:22 ** The height of the rectangle would be the average value of the function. That would have to be the case so that height * length would equal the integral. **
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RESPONSE --> The average value of the function is the integral / 'dx which is about what I did when I tried to find the height of the box
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23:00:36 query problem 5.3.26 v vs. t polynomial (0,0) to (1/6,-10) thru (2/6,0) to (.7,30) to (1,0). When is the cyclist furthest from her starting point?
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RESPONSE --> The cyclest would be farthest from home at the maxium value of Y which is 1.
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23:17:37 ** The change in position of the cyclist from the start to any instant will be the integral of the v vs. t graph from the t = 0 to that instant. From t = 0 to t = 1/3 we see that v is negative, so the cyclist is getting further from her starting point. The average velocity over this interval appears to be about -6 mph, so the integral over this segment of the graph is about -6 * 1/3 = -2, indicating a displacement of -2 miles. At t = 1/3 she is therefore 3 miles from the lake. For the remaining 2/3 of an hour the cyclist is moving in the positive direction, away from the lake. Her average velocity appears to be a bit greater than 15 mph--say about 18 mph--so the integral over this segment is about 18 * 2/3 = 12. She gets about 12 miles further from the lake, ending up about 15 miles away. The key to interpretation of the integral of a graphed function is the meaning of a rectangle of the Riemann Sum. In this case any rectangle in the Riemann sum has dimensions of velocity (altitude) vs. time interval (width), so each tiny rectangle represents a product of velocity and time interval, giving you a displacement. The integral, represented by the total area under the curve, therefore represents the displacement. **
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RESPONSE --> I was having some truble interperting the graph upon the reading I understand the whole concept of having a negative distance was confusing. The velocity to t=1/3 is negative and takes her toward the lake so by calculating the area between this and the x axis we get: -2 so it will take +2 units of the area under the positive velocity to take the cylist back to the original point. Since the total area under the positive velocity is 12, we have to take away the original two units spent going towards the lake so she is 10 miles from her starting point at five miles past the lake and therefore she is 15 miles from the lake.
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23:18:06
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RESPONSE --> okay
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23:21:37 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> When you are finding the area under a graph you are finding the displacement. When you use the riemann sums you are finding the possible values for the integral. Are there times when it is necessary to find the displacement when trying to find the integral? Meaning are the two seperate completely or do they merge together to help to achieve the integral?
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