course Mth 173 I plan on taking Test 1 tomorrow. Thanks! Zµ姖w˗PӴ\assignment #016
......!!!!!!!!...................................
22:32:32 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.
......!!!!!!!!...................................
RESPONSE --> Values in (x, y) format: (1, 1) Over this period the rate stayed at 1, so 1*1=1. (2, 1.5) Over this period the rate dropped to 0, so an average of .5 added to the prior position of 1. (3, 1) This period saw a further rate drop, this time into the negative, rendering a -.5 added to the existing 1.5. (4, 0) This period had a full -1*1 block, making the result 0. (5, -1) The rate stayed at -1, rendering the result -1 overall. (6, -1.5) The rate increased to -.5, adding -.5 to -1.
.................................................
......!!!!!!!!...................................
22:33:41 ** 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 --> In this graph, the sixth trapeziod had altitudes -1 and 0, rendering an average of -.5.
.................................................
......!!!!!!!!...................................
22:40:45 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?
......!!!!!!!!...................................
RESPONSE --> The area of each trapeziod is the average altitude multiplied by the run. Parts of the graph below the x-axis subtract from the total area, so the total area is: 1*1 + .5*1 + -.5*1 + -1*2 + -.5*1 = -1.5 total area
.................................................
......!!!!!!!!...................................
22:42:07 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?
......!!!!!!!!...................................
RESPONSE --> Probably the simplest way is to plug highest value of the domain into the function to see the highest possible value in the range. In this case. plugging 1 into e^(x^2) simplifies to e^1, or roughly 2.71.
.................................................
......!!!!!!!!...................................
22:42:24 ** 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.
.................................................
......!!!!!!!!...................................
22:42:45 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Interesting uses of the Riemann function.
.................................................