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course Mth 173 Question: `q query problem 5.2.24 was 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?.............................................
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): ------------------------------------------------ Self-critique Rating:
********************************************* 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: The integral from x= -3 to x=0 is -2, because the area between the graph and the x-axis is 2, and the area is below the x axis, resulting in a negative integral. The integral from x=0 to x=3 is 2, since the area is now above the x axis. The integral from x=3 to x=4 is -.6. This is because the graph is not a straight line anymore. And it is negative, because the line is below the x axis. To get the total integral, we just add the 3 partial integrals to get -.6, since the 2 and -2 cancel each other out. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: ** We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly. From x=-3 to x=0 the area between the graph and the x axis is 2 (divide the region into two triangles and a rectangle and find total area, or treat it as a trapezoid). The area is below the x axis so the integral from x = -3 to x = 0 is -2. From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2. If the graph from x=3 to x=4 was a straight line from (3,0) to (4,-1) the area over this interval would be .5 and the integral would be -.5. Since the graph curves down below this straight line the area will be more like .6 and the integral closer to -.6. The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q query problem 5.3.10 was 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can find the average value by using the equation (f(10) - f(0)) / 10. This gives us 22,025.5, roughly. Divided by 10 that gives us an average value of roughly 2,202.55. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating:
********************************************* 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: During the first leg of the ride, t=0 to t=2/6 or 1/3, v is negative. The average v over this time is roughly -6mph. The integral for that time would be -6 * 1/3 which is -2. So, the cyclist has moved 3 miles from the lake. During the second leg, 1/3 to 1, or 2/3, v is positive. The cycolist has an average v of somewhere near 18 mph. The integral for this time would be 18 * 2/3 = 12. In this time she gets about 12 miles further, which adds up to be 15 miles away. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! `gr31#$&* course Mth 173 Question: `q query problem 5.2.24 was 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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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: The integral from x= -3 to x=0 is -2, because the area between the graph and the x-axis is 2, and the area is below the x axis, resulting in a negative integral. The integral from x=0 to x=3 is 2, since the area is now above the x axis. The integral from x=3 to x=4 is -.6. This is because the graph is not a straight line anymore. And it is negative, because the line is below the x axis. To get the total integral, we just add the 3 partial integrals to get -.6, since the 2 and -2 cancel each other out. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly. From x=-3 to x=0 the area between the graph and the x axis is 2 (divide the region into two triangles and a rectangle and find total area, or treat it as a trapezoid). The area is below the x axis so the integral from x = -3 to x = 0 is -2. From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2. If the graph from x=3 to x=4 was a straight line from (3,0) to (4,-1) the area over this interval would be .5 and the integral would be -.5. Since the graph curves down below this straight line the area will be more like .6 and the integral closer to -.6. The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query problem 5.3.10 was 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can find the average value by using the equation (f(10) - f(0)) / 10. This gives us 22,025.5, roughly. Divided by 10 that gives us an average value of roughly 2,202.55. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: During the first leg of the ride, t=0 to t=2/6 or 1/3, v is negative. The average v over this time is roughly -6mph. The integral for that time would be -6 * 1/3 which is -2. So, the cyclist has moved 3 miles from the lake. During the second leg, 1/3 to 1, or 2/3, v is positive. The cycolist has an average v of somewhere near 18 mph. The integral for this time would be 18 * 2/3 = 12. In this time she gets about 12 miles further, which adds up to be 15 miles away. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!