3. the Antiderivative date: 1/30 hw: p225 #1-41 EOO

6. lesson check exponent properties

7. which is the derivative?

8. antiderivative an antiderivative of a given function f is a function F whose derivative is equal to fderivative F′ = f.

9. anti means “opposite” the derivative: Now, given 3x 2, find an antiderivative graph 3 antiderivatives the antiderivative of 3x 2 is of the form:

10. 1

11. solving a differential equation find a function whose derivative is 5 the general solution of the differential equation:

12. notation for antiderivatives deriv: antideriv: the integral of f(x) with respect to x our 2 examples:

13. Power Rule for Integration add one to the exponent and divide by the new exponent! EXAMPLE

14. property 1 you may pull the constant out of an integral. deriv: integral: EXAMPLE:

15. property 2 when integral has terms added or subtracted, then those terms can be split apart into separate integrals

16. rewriting before integrating

19. the Antiderivative 2 date: 1/31 hw: p225 #43-63 EOO, 67,69

20. lesson check

21. finding a particular solution

22. visualizing differential equations: SLOPE FIELDS

23. vertical motion problem

24. writing about concepts

25. approximating area under a curve

26. check equation of a circle of radius, r

27. exact areas “under a curve” domain range

28. approximating area Riemann sums: Left, Right, Midpoint.

29. find the left Riemann sum would the right Riemann sum be more or less?

30. handout

31. the Trapezoid rule hw p314 #1,9,11,21,39,51

32. a trapezoid a trapezoid is a ________ with exactly one pair of ____ sides. the area of a trapezoid is _______

33. compare methods! upper sums (circumscribed rectangles) trapezoidal rule

34. Riemann sums?? we can do better load example e^(x^2) on [0,1] upper, lower, middle, trapezoid n=100 calc: 1.4626517

35. the bases have length __ and __, and the height, h, is d – c or ___

36. the trapezoid rule the area of the first trapezoid ________ the area of the 2 nd trapezoid ________ a wxyz b

37. the trapezoid rule find the width of one interval and divide it by 2. multiply by the sum of all the function values doubled, except for the first and last values f(a) and f(b) x1x1 x2x2 x3x3 x4x4 a b

38. example p234 Apex example

39. definite integrals hw p280 #13-49 odd 55,57

40. Definition of a Definite Integral

41. so what’s happening?!

42. Continuity Implies Integrability

43. The Definite Integral as the Area of a Region

44. example the region bounded by the graph of f and the x axis

45. areas of common geometric figures

46. However, remember that area is always considered to be positive and so if the function f is negative, the term represents the additive inverse of the area of a small rectangle.

50. worksheet

51. FUNFUNdamental theorem of calculus

53. Figure 4.27

54. Theorem 4.9 The Fundamental Theorem of Calculus

55. Guidelines for Using the Fundamental Theorem of Calculus

56. The first part of this theorem tells us how to evaluate a definite integral provided that f has an indefinite integral. Translation: to evaluate the definite integral, find the antiderivative of f(x),then plug the upper bound into the antiderivative. you will get a number. then from that number subtract the result of plugging the lower bound into the antiderivative.

60. wiki Suppose a particle travels in a straight line with its position given by x(t) where t is time. The derivative of this function is equal to the infinitesimal change in x per infinitesimal change in time (of course, the derivative itself is dependent on time). Let us define this change in distance per time as the speed v of the particle. In Leibniz's notation:Leibniz's notation – Rearranging that equation, it is clear that: – By the logic above, a change in x, call it Δx, is the sum of the infinitesimal changes dx. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. Clearly, this operation works in reverse as we can differentiate the result of our integral to recover the speed function.

61. Suppose you have an object, say a train, that is in motion. You will recall that if x(t) is its position as a function of time, t, and v(t) is its velocity as a function of time, then dx v(t) = x'(t) = eq. 10.2-1 dt If v were constant, you would be able to find the position of the train easily if you knew only where it started from, x0, and what the time, t, was. You have the familiar motion equation that you learned in algebra: x = x0 + vt eq. 10.2-2 But if the train's velocity is constant, say 30 miles per hour, how are the passengers ever going to get on or off the train? So to add some realism to the situation, let's suppose that starting at the moment the train begins pulling out of the station, its velocity is given by v(t) = at eq. 10.2-3 where a is a constant acceleration. Now how do we find the train's position as a function of time? Clearly from equation 10.2-1, x(t) must be an antiderivative of v(t). But look at what's going on in more detail. At time, t = 10, the train is moving at a velocity of v(t) = 10a. And at that moment the train is at some position, x10. Now suppose that for the second beginning at t = 10, the trains velocity is frozen at v = 10a. During the interval of time from 10 seconds to 11 seconds, we will evaluate the train's position using your old motion equation, 10.2-2. And in particular, we are interested in where it would end up at t = 11 seconds based on our assumption of constant velocity for that one second. x11 = x10 + (1 second × v) = x10 + (1 second × 10a) eq. 10.2-4 Next we allow that at the moment of t = 11 seconds, the velocity is frozen at v = 11a, and we can do the same sum for this next second x12 = x11 + (1 second × 11a) eq. 10.2-5 If you imagine holding the velocity constant throughout each second and doing the sum for each second the train travels, the equation for the kth second is xk = xk-1 + (1 second × (k-1)a) eq. 10.2-6 Can you see that what you really end up with is a Riemann sum? The equation is saying that this x is the sum of something plus the last x. Of course that last x was the sum of something plus the x before that, and so on. So really, this x is the sum of all those somethings, going back to the very first x. So this x is the sum: n xn = x0 + ∑ 1 second × (k-1)a eq. 10.2-7 k=1 where n is the number of seconds elapsed since the train left the station. Of course the train's velocity is not constant, even over the period of just a second. So our assumption that it is will lead to some inaccuracy in the approximation, xn, of the train's position after n seconds. But you will agree that trains' velocities don't change a whole lot in the span of a second. And if this method leads to too much inaccuracy, you could always do it by a shorter interval instead -- say by the millisecond. Since there are a thousand milliseconds in each second, you will have to sum a thousand times more terms, but that's the price you pay for accuracy, right? Of course you could take the limit as the short interval of time goes to zero and the number of intervals goes to infinity. That should give you a perfect approximation of position as a function of time. But isn't that exactly what we were doing in the last section? The point is, we know that position as a function of time is an antiderivative of velocity. And we know that we can also get position from velocity by taking the limit of a Rieman sum. This suggests that the limit of the Riemann sum of velocity is always an antiderivative of velocity. Which is the thrust of the Fundamental Theorem of Calculus.

62. fun2 applet applet2

63. fundamental