#$&* course MTH 173 11/6 00:10
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Given that the graph is piece wise linear from x = -3 to x = 3, thus the integral which is the area under the curve can be directly calculated by calculating the area under the curve. From (-3,0) to (-2,-1) the area of the triangle will be 0.5 square unit, and since the triangle is below the x axis the value of the integral will be negative. The straight line from (-2,-1) to (-.5,0) will form a triangle who area will be the value of the integral and since the triangle is below the x axis the integral will be negative. Thus the area of the triangle will be 0.75 square unit and thus the value of the integral will be -0.75 for x values from x = -2 to x = -0.5. The triangle formed by joining the points (-.5,0) to (1,1) the value of the area will be the value of the integral from x = -.5 to x = 1. The area is 0.75 square unit and the value of the integral will be 0.75 for x = -.5 to x = 1. The area of the triangle formed by joining (1,1) to (3,0) will be 1 square unit and thus the integral of the function from x = 1 to x = 3 will be 1. For the x value from 3 to 4 the function is a curve, if a straight line is considered the area will be 0.5 square unit and thus the area of the curve since it is below the straight line will be slightly more than 0.5 which is around 2/3 square unit, and thus the value of the integral will be approximately -2/3. Thus the net value of the integral will be -0.5 -0.75 +0.75 +1 -2/3 = -1/6 approximately. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** **** revised solution ... **** It appears that the solution given above doesn't match the problem. The graph as described would pass through (-3, 0), (-2, -1), (-1/2, 0), (1, 1), (3, 0) and (4, -1). You should sketch the graph and check to see that these points are consistent with the given description. The regions between the graph and the x axis will all be triangles, except for the last, which will have a curved boundary. Except for the last interval the graph would consist of broken straight lines. I believe the areas of the resulting triangular regions would be, respectively, 1/2, 3/4, 3/4 and 1. The last region would encompass more area than just the triangle defined by (3, 0) and (4, -1), so would be greater than 1/2. Let's call it 2/3. Taking account of whether each region is above or below the x axis, we have the following contributions to the integral: -1/2, -3/4, 3/4, 1 and -2/3. This would result in a total integral of -1/6. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: The given function is e^t over the interval of [0,10]. The average value of the function will be the value of the integral divided by the length of the interval. The value of the integral over this interval will be e^10 - 1 and the length of the interval will be 10 units. Thus the average value of the function over the given interval will be (e^10 - 1)/10 (e^10 - 1)/10 = 2202.546579 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: The given polynomial passes through (0,0), (1/6,-10), (2/6,0), (.7,30) and (1,0). The value of the average velocity will be the integral divided over the time duration. For the time duration from t = 0 to t = 2/6 the average velocity is about 6 mph and the time duration is 1/3 hours, thus the value of the integral is 2 mph and since the curve is below the x axis the integral is negative and thus -2 mph. Thus the displacement of the cyclist will be -2 mph and thus she will be 3 miles away. For the next curve from t = 1/3 hours to t = 1 hour, the average velocity over this interval would be around 18 mph. Thus the integral over this will be equal to 18 mph * 2/3 hour = 12 mph. The displacement of the cyclist is 12 mph and thus the final displacement of the cyclist will be 15 mph. Thus the cyclist will be furthest from the starting point at the end, which is when time is 1 hour. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** STUDENT QUESTIONS I know that you can find the change in position will be the integral multiplied by the change in time How did you find the average velocity over the intervals? From the points listed above? Integrals are really difficult for me to understand INSTRUCTOR RESPONSES
.............................................
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. ** STUDENT QUESTIONS I know that you can find the change in position will be the integral multiplied by the change in time How did you find the average velocity over the intervals? From the points listed above? Integrals are really difficult for me to understand INSTRUCTOR RESPONSES