course Mth 173
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21:26:26 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 --> Since integral (a to b) f(t) dt = F(b) - F(a), F(b) = integral (a to b) f(t) dt + F(a) Since F(a) = 0, F(b) is just the integral F(1) = 1 (square of sides = 1) F(2) = 1.5 (adds a triangle base and height 1) F(3) = 1 (subtracts a triangle which is symmetric to the previous one about the x-axis, thus cancels out) F(4) = 0 (symmetric about the x-axis) F(5) = -1 F(6) = -1.5
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21:27:56 ** 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 --> It appears the sixth trapezoid should have altitudes -1 and 0, not -1 and 1, and thus area .5; so F(6) = -1.5
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21:28:55 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 --> Areas would be as mentioned in the previous answer, and the accumulated area would just be each trapezoid added to the next until b. Trapezoids can be used to approximate the integral of a function.
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21:33:16 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 --> The integral of e^(x^2) is positive since the function is always positive, and is equal to 2 times that same integral from 0 to 1 since the function is even. when x = 0, e^(x^2) = 1 when x = 1, e^(x^2) = e (2.71...) We can picture a trapezoid of width 1 (from 0 to 1) approximating the area (and thus the integral) which would have average altitude (2.71 - 1), giving approximate area of 1.71, which is between 0 and 3.
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21:34:12 ** 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 --> A more elegant explanation. The integral of 3 is 3, and since the entirety of the fuction lies below this, its integral must be < 3.
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21:36:38 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Good for now.