course Mth 173 Tޗrwpassignment #015
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13:19:12 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 --> x^ 2 + 1 0^2 + 1 = 1 1^2 + 1 = 2 2^2 + 1 = 5 3^2 + 1 = 10 4^2 + 1 = 17 5^2 + 1 = 26 6^2 + 1 = 37 left-hand: 2 * 1 + 2 * 5 + 2 * 17 = 46 right-hand: 2 * 5 + 2 * 17 + 2 * 37 = 118 Integral: 6^3 / 3 + 6 - (0^3/ 3 + 0) = 216/3 + 6 = 234/3 = 78 Sums differ 37 -1 = 36 * 2 = 72
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13:19:14 ** 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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13:28:45 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 shape of the graph the estimate would expect to be upward and increasing and would be closer to the left hand estimates because 78-46 = 32 and 118-78=42
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13:28:47 ** 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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15:11:36 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 --> The line curves around .6 and the integral is close to -.6 2 + -2 - .6 = -.6
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15:11:38 ** 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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15:13:21 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 --> e^10 - e^0 22026.46579 - 1. 22025.46579 22025.46579/10 = 2202.546579
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15:13:23 ** 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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15:14:29 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 --> height * length must equal the integral which means that the height of the rectangle would need to be the average
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15:14:31 ** 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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15:25:46 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 --> It states that when the velocities are negative she is going toward the lake and positive is going from the lake. By that information, I would say that at .7,30 to 1,0
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15:27:09 ** 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 --> By my assumption compared to the information stated here, I would think that my answer is correct with enough said to conclude I have a good understanding. However, I did not go about this the math way but by the reasoning way.
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15:27:10
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