course mth 173 assignment #016016. `query 16
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22:22:50 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 --> b f 1 1 (given) 2 0 3 -1(given) 4 -1 5 -1(given) 6 1(given) I found the two values of f that where not given by finding the point in between the two points by averaging them.
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22:26:51 ** 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. **
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RESPONSE --> So in this istance since the given value is a rate function then the area under the graph represents the actual displacement so we would use the area instead of rieman sums?
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22:31:29 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 --> The accumulated area would be -1 and the area under each of the trapazoids is: f(1)=1 f(2)=.5 f(3)= -.5 f(4)= -1 f(5)= -1 f(6)= 0 (there is an area but it cancels itself out by having the same number of positive and negative area units.
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22:44:47 Query 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?
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RESPONSE --> y= e^(x^2) x y 0 1 .25 1.06 .5 1.28 .75 1.76 1 2.72 By using rieman sums we get 1.28 for the lower sum and 1.71 for the upper sum, both of which are between 0 and 3 and are probobly fairly close due to the small 'dx
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22:45:58 ** 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. **
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RESPONSE --> okay
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22:50:41 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I asked the queston in the last assignment about when to use riemann sums and when to use the area under of the trapazoids. After thinking about it some I think I may have figured it out. We use the riemann sums to estimate where the integral will be, and by using a smaller 'dx we get closer and close to the actual integral. And the area in the trapazoids is when we are trying to find out the displacement of the integral rather than the integral itself. Correct, or close to it? It takes a while for everything to sink in with me, usually while I'm working on it it doesn't make much sence then it will ""hit"" me after I think about a little while.
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