#$&* course mth 173 7/31 5 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** An antiderivative of x^2 + 1 is x^3 / 3 + x. The change in the antiderivative from x = 0 to x = 6 is equal to the integral: Integral = 6^3 / 3 + 6 - (0^3 / 3 + 0) = 216/3 + 6 = 234/3 = 78. The values of the function at 0, 2, 4, 6 are 1, 5, 17 and 37. The left-hand sum would therefore be 2 * 1 + 2 * 5 + 2 * 17 = 46. The right-hand sum would be 2 * 5 + 2 * 17 + 2 * 37 = 118. The sums differ by ( f(b) - f(a) ) * `dx = (37 - 1) * 2 = 72.** From the shape of the graph which estimate would you expect to be low and which high, and what property of the graph makes you think so? ** The graph is increasing so the left-hand sum should be the lesser. the graph is concave upward and increasing, and for each interval the average value of the function will therefore be closer to the left-hand estimate than the right. This is corroborated by the fact that 46 is closer to 78 than is 118. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The left hand sum will be closer because the graph is concave up. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q query problem 5.2.32 f(x) piecewise linear from (-4,1) to (-2,-1) to (1,1) to (3,0) then like parabola with vertex at (4,-1) to (5,0). What is your estimate of the integral of the function from -3 to 4 and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is not 5.2.32 in the 5th edition. confidence rating #$&*: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query problem 5.3.10 was 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = e^t F'(x) = f(x) F(x) = e^t a = 0, b = 0 (F(b) - F(A))/(b-a) = (e^10 -e^0)/(10-0) = 2202.55 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** An antiderivative of f(t) = e^t is F(t) = e^t. Using the Fundamental Theorem you get the definite integral by calculating F(10) - F(0). This gives you e^10 - e^0 = 22025, approx. The average value of the function is the integral divided by the length of the interval. The length of the interval is 10 and 22025 / 10 is 2202.5. ** What would be the height of a rectangle sitting on the t axis from t = 0 to t = 10 whose area is the same as that under the y = e^t graph from t=0 to t=10? ** The height of the rectangle would be the average value of the function. That would have to be the case so that height * length would equal the integral. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query problem 5.3.38, was 5.3.26 v vs. t polynomial (0,0) to (1/6,-10) thru (2/6,0) to (.7,30) to (1,0). When is the cyclist furthest from her starting point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the graph we see that at x = 1/6 the displacement is roughly (3/4)(1/6)(-10) = -1.25 From x = 1/6 to x = 2/6 the displacement is roughly .5(1/6)(-10) = -.83 So from x = 0 to x = 1/3 there is a total displacement of roughly -2 miles. 5-2 = 3 so The cyclist is now 3 miles from the lake From this point the velocity stays positive until x = 1 so the maximum displacement will occur at x = 1 V(1/3) = 0 V(2/3) = 30 average velocity = 30/2 = 15 15*1/3 = 5 miles of displacement v(2/3) = 30 v(5/6) = 25 v(1) = 0 18.33 * (1/3) = 6 miles of displacement approximately s(1) = 3 + 5 + 6 = 14 miles from the lake confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer is slightly different but I believe I used the sam principles just approximated differently. ------------------------------------------------ Self-critique Rating: 3