Here is assignment 16. Let me know what i need to do next or when i should get started on it/have it completed by. I will be out of town for up and coming weekend beginning on friday and i'll be gone till Sunday (when both VHCC and Emory and Henry's spring break begin). Also, i am pretty sure i am submitting the assignment's the correct way but let me know if I am not. Thanks for your help! ˮʩ assignment #016 gmk۟ͯEwұҚR Calculus I 03-05-2006
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22:24:21 Problem 5.4.3 (was 5.4.2) fn piecewise linear from (0,1)-(1,1)-(3,-1)-(5,-1)-(6,1). F ' = f, F(0) = 0. Find F(b) for b = 1, 2, 3, 4, 5, 6.What are your values of F(b) for b = 1, 2, 3, 4, 5, 6 and how did you obtain them.
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RESPONSE --> f(1)= 1 f(2)=1.5 f(3)= 1 f(4)= 0 f(5) = -1 f(6) = -1.5
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22:24:43 ** f(x) is the 'rate function'. The change in the 'amount function' F(x) is represented as the accumulated area under the graph of f(x). Since F(0) = 0, F(x) will just the equal to the accumulated area.
We can represent the graph of f(x) as a trapezoidal approximation graph, with uniform width 1. The first trapezoid extends from x = 0 to x = 1 and has altitude y = 1 at both ends. Therefore its average altitude is 1 and its what is 1, giving area 1. So the accumulated area through the first trapezoid is 1. Thus F(1) = 1. The second trapezoid has area .5, since its altitudes 1 and 0 average to .5 and imply area .5. The cumulative area through the second trapezoid is therefore still 1 + .5 = 1.5. Thus F(2) = 1.5. The third trapezoid has altitudes 0 and -1 so its average altitude is -.5 and width 1, giving it 'area' -.5. The accumulated area through the third trapezoid is therefore 1 + .5 - .5 = 1. Thus F(3) = 1. The fourth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fourth trapezoid is therefore 1 + .5 - .5 - 1 = 0. Thus F(4) = 0. The fifth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fifth trapezoid is therefore 1 + .5 - .5 - 1 - 1 = -1. Thus F(5) = -1. The sixth trapezoid has altitudes 1 and -1 so its average altitude, and therefore it's area, is 0. Accumulated area through the sixth trapezoid is therefore 1 + .5 - .5 - 1 - 1 + 0 = -1. Thus F(6) = -1. **......!!!!!!!!...................................
RESPONSE --> i simply analyzed the areas under the curve
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22:26:31 If you made a trapezoidal graph of this function what would be the area of teach trapezoid, and what would be the accumulated area from x = 0 to x = b for b = 1, 2, 3, 4, 5, 6? What does your answer have to do with this question?
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RESPONSE --> it would be the sum of all of the rectangles.
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22:30:35 Problem 5.4.27 (was 5.4.12). integral of e^(x^2) from -1 to 1.
How do you know that the integral of this function from 0 to 1 lies between 0 and 3?......!!!!!!!!...................................
RESPONSE --> this integral is positive (2.93 to be exact). It is actually a parabola. It is symmetric around the Y axis. Therefore 2 times the integral value of one side (i.e. from 0 to 1) will give you the correct answer for the entire integral problem
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22:31:02 ** The max value of the function is e, about 2.71828, which occurs at x = 1. Thus e^(x^2) is always less than 3.
On the interval from 0 to 1 the curve therefore lies above the x axis and below the line y = 3. The area of this interval below the y = 3 line is 3. So the integral is less than 3. Alternatively e^(x^2) is never as great as 3, so its average value is less than 3. Therefore its integral, which is average value * length of interval, is less that 3 * 1 = 3. **......!!!!!!!!...................................
RESPONSE --> ok
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22:33:18 5.4.42. If the average value of v(x) on 1 <= x <= 6 is 4, and the average value on 6 <= x <= 8 is 5, then what is the average value of v(x) on the interval 1 <= x <= 8?
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RESPONSE --> 4.8? I am not quite sure as to the answer of this problem
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22:35:35 The average value of a function over an interval is equal to its integral over that interval, divided by the length of the interval.
To find the average value of v(x) on the interval 1 <= x <= 8 we first find its integral over that interval. The integral of v(x) from x = 1 to x = 8 is the sum of its integral from x = 1 to x = 6, and its integral from x = 6 to x = 8. Since the average value of v(x) on the first of these intervals is 4, its integral over that interval is equal to 4 times the length of the interval, or 4 * 5 = 20. Since the average value of v(x) on the second interval is 5, its integral over that interval is equal to 5 times the length of the interval, or 5 * 2 = 10. Thus the integral of v(x) from x = 1 to x = 8 is 20 * 10 = 30. The length of the interval being 8 - 1 = 7, the average value is 30 / 7 = 4 2/7, approximately 4.3.......!!!!!!!!...................................
RESPONSE --> i understand, i overcomplicated the problem by far
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22:36:06 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> the problems for this section were tough in the book however i very clearly understand the theorem's and concepts of this section.
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