1. Calculus July 19-21

2. Review of Week 1 Your thoughts, questions, musings, etc...

3. A problem that was around long before the invention of calculus is to find the area of a general plane region (with curved sides). And a method of solution that goes all the way back to Archimedes is to divide the region up into lots of little regions, so that you can find the area of almost all of the little regions, and so that the total area of the ones you can't measure is very small.

4. Ameba By Newton's time, people realized that it would be sufficient to handle regions that had three straight sides and one curved side (or two or one straight side - - the important thing is that all but one side is straight). Essentially all regions can be divided up into such regions.

5. These all-but-one-side-straight regions look like areas under the graphs of functions. And there is a standard strategy for calculating (at least approximately) such areas. For instance, to calculate the area between the graph of y = 4x - x 2 and the x axis, we draw it and subdivide it as follows:

6. Since the green pieces are all rectangles, their areas are easy to calculate. The blue parts under the curve are relatively small, so if we add up the areas of the rectangles, we won't be far from the area under the curve. For the record, the total area of all the green rectangles is: 24625 whereas the actual area under the curve is: Also for the record, 246/25 = 9.84 while 32/3 is about 10.6667.

7. We can improve the approximation by dividing into more rectangles: Now there are 60 boxes instead of 20, and their total area is: which is about 10.397. Getting better. We can in fact take the limit as the number of rectangles goes to infinity, which will give the same value as the integral. This was Newton's and Leibniz's great discovery -- derivatives and integrals are related and they are related to the area problem. Area 60 boxes 7018675

8. Limits of Riemann sums A kind of limit that comes up occasionally is an integral described as the limit of a Riemann sum. One way to recognize these is that they are generally expressed as, where the “something” depends on n as well as on i.

9. Green graph Again, recall that one way to look at integrals is as areas under graphs, and we approximate these areas as sums of areas of rectangles. This is a picture of the“right endpoint” approximation to the integral of a function.

10. approximating

11. Example...

12. solution

13. Area between two curves: A standard kind of problem is to find the area above one curve and below another (or to the left of one curve and to the right of another). This is easy using integrals. Note that the "area between a curve and the axis" is a special case of this problem where one of the curves simply has the equation y = 0 (or perhaps x=0 )

14. 1. Graph the equations if possible 2. Find points of intersection of the curves to determine limits of integration, if none are given 3. Integrate the top curve's function minus the bottom curve's (or right curve minus left curve).

15. Example: Find the area between the graphs of y=sin(x) and y=x(  -x)

16. It’s easy to see that the curves intersect on the x-axis, and the values of x are 0 and . The parabola is on top, so we integrate: And this is the area between the two curves.

17. An Area Question: Find the area of the region bounded by the curves y=4x 2 and y=x 2 +3. A. 1/2 B. 1 C. 3/2 D. 2 E. 5/2 F. 3 G.7/2 H. 4

18. Position, velocity, and acceleration: Since velocity is the derivative of position and acceleration is the derivative of velocity, Velocity is the integral of acceleration, and position is the integral of velocity. (Of course, you must know starting values of position and/or velocity to determine the constant of integration.)

19. Example... An object moves in a force field so that its acceleration at time t is a(t) = t -t+12 (meters per second squared). Assuming the object is moving at a speed of 5 meters per second at time t=0, determine how far it travels in the first 10 seconds. 2

20. First we determine the velocity, by integrating the acceleration. Because v(0) = 5, we can write the velocity v(t) as 5 + a definite integral, as follows: The distance the object moves in the first 10 seconds is the total change in position. In other words, it is the integral of dx as t goes from 0 to 10. But dx = v(t) dt. So we can write: (distance traveled between t=0 and t=10) = = = 3950/3 = 1316.666... meters. Solution...

21. Surfaces of revolution: Volume A "surface of revolution" is formed when a curve is revolved around a line (usually the x or y axis). The curve sweeps out a surface Interesting problems that can be solved by integration are to find the volume enclosed inside such a surface or to find its surface area.

22. Volumes You might already be familiar with finding volumes of revolution. Once a surface is formed by rotating around the x-axis, you can sweep out the volume it encloses with disks perpendicular to the x axis.

23. Here is the surface formed... Here is the surface formed by revolving around the x axis for x between 0 and 2, showing one of the disks that sweep out the volume:

24. enclosed inside the surface, we need to enclosed inside the surface, we need to add up the volumes of all the disks. add up the volumes of all the disks. The disks are (approximately) cylinders turned sideways, and the disk centered at (x,0) has radius and width (or height) dx. The volume of the disk is thus To find the total volume of the solid we have to integrate this quantity for x from 0 to 2. We get To calculate the volume, or cubic units cubic units

25. In general, if the piece of the graph of the function of y = f (x) between x = a and x = b is revolved around the x axis, the volume inside the resulting solid of revolution is calculated as: The same sort of formula applies if we rotate the region between the y-axis and a curve around the y-axis (just change all the x's to y's). A formula for volume:

26. The same region rotated around the y axis A different kind of problem is to rotate the region between a curve and the x axis around the y axis (or between a curve and the x axis around the y axis (or vice versa). For instance, let's look at the same region (between y=0 and y= for x between 0 and 2), but rotated around the y axis instead:

27. The generating curve Here is the surface being swept out by the generating curve:

28. Washers We could sweep out this volume with “washers” with inner radius y 2 and outer radius 2 as y goes from 0 to Each washer is (approximately) a cylinder with a hole in the middle. The volume of such a washer is then the volume of the big cylinder minus the volume of the hole.

29. For the washer centered at the point (0, y), the radius of the outside cylinder is always equal to 2, and the radius of the hole is equal to the corresponding x (which, since, is equal to y 2 ). And the height of the washer is equal to dy. So the volume of the washer is The volume of the washers...

30. Therefore the volume of the entire solid is cubic units

31. Cylindrical shells Another way to sweep out this volume is with "cylindrical shells".

32. The volume of a cylindrical shell Each cylindrical shell, if you cut it along a vertical line, can be laid out as a rectangular box, with length, with width laid out as a rectangular box, with length, with width and with thickness dx. The volume of the cylindrical and with thickness dx. The volume of the cylindrical shell that goes through the point (x,0) is thus So, we can calculate the volume of the entire solid to be: cubic units, which agrees with the answer we got the other way.

33. Another family of volume problems involves volumes of three-dimensional objects whose cross-sections in some direction all have the same shape. For example: Calculate the volume of the solid S if the base of S is the triangular region with vertices (0,0), (2,0) and (0,1) and cross sections perpendicular to the x-axis are semicircles. First, we have to visualize the solid. Here is the base triangle, with a few vertical lines drawn Other volumes with known cross sections on it (perpendicular to the on it (perpendicular to the x-axis). These will be diameters of the semicircles in the solid.

34. 3-D Now, we'll make the three-dimensional plot that has this triangle as the base and the semi-circular cross sections.

35. 3-D BOX From that point of view you can see some of the base as well as the cross section. We'll sweep out the volume with slices perpendicular to the x-axis, each will look like half a disk:

36. Since the line connecting the two points (0,1) and (2,0) has equation y = 1 - x/2, the centers of the half-disks are at the points (x, 1/2 - x/4), and their radii are likewise 1/2 - x/4. Therefore the little bit of volume at x is half the volume of a cylinder of radius 1/2 - x/4 and height dx, namely Therefore, the volume of the solid S is: The volume of that object:

37. Note that we could also have calculated the volume by noticing that the solid S is half of a (skewed) cone of height 2 with base radius = 1/2. Using the formula for a cone, we arrive at the same answer, cubic units. Finally...

38. Methods of integration Before we get too involved with applications of the integral, we have to make sure we're good at calculating antiderivatives. There are four basic tricks that you have to learn (and hundreds of ad hoc ones that only work in special situations): 1. Integration by substitution (chain rule in reverse) 2. Trigonometric substitutions (using trig identities to your advantage) 3. Partial fractions (an algebraic trick that is good for more than doing integrals) 4. Integration by parts (the product rule in reverse) We'll did #1 last week, and we’ll do the others this week. LOTS of practice is needed to master these!

39. Integration by parts This is another way to try and integrate products. It is in fact the opposite of the product rule for derivatives: Product rule for derivatives: d(uv) = u dv + v du. Integration by parts: Use this when you have a product under the integral sign, and it appears that integrating one factor and differentiating the other will make the resulting integral easier.

40. If we differentiate, we'll get 2x which seems simpler. And integrating doesn't change anything. If we differentiate x 2, we'll get 2x which seems simpler. And integrating e x doesn't change anything. Let and. Then and. Let u = x and dv = e x dx. Then du = 2x dx and v = e x. The integration by parts formula then gives us: Example: -- There's no other rule for this, so we try parts.

41. Continuing: We can use parts again on this integral (with and ) to get: (with u = 2x and dv = e x dx) to get:

42. You try one... …contrast this with

43. XOXXOX There is a method called the "tic-tac-toe" method for integration by parts. Different strategies for choosing and are called for to integrate or orsomething like Different strategies for choosing u and dv are called for to integrate x ln x or x tan -1 x or something like e x sin x x

44. A problem for you... Find the value of  0 x cos(x) dx A)  B) 2  C) 2 D) 0 E) -2 F) 1 G) 1/2 H)  /2

45. And another: Evaluate 1 0 x ln x dx A) -1/4 B) -1/2 C) 0 D) 1/4 E) 1/2

46. Trigonomeric substitutions These are just substitutions like the ordinary substitution method, but some of them are a little surprising. There are two kinds: To integrate products of powers of sine and cosine To integrate products of powers of sine and cosine (like sin (x) cos (x) ). You need the identities: (like sin (x) cos (x) ). You need the identities: sin (x) + cos (x) = 1 (everybody knows that one!), sin (x) + cos (x) = 1 (everybody knows that one!), and the double angle formulas: cos (x) = and the double angle formulas: cos (x) = and sin (x) = and sin (x) = 36 22 2 1+cos(2x) 2 1-cos(2x) 2 2 The first, 11

47. The trick is as follows... (a) If both the power of sine and the power of cosine are even, then use the double angle formulas to divide both in half. Keep doing this until at least one of the powers is odd. (b) Once at least one of the powers is odd, say it's the power of cosine, then let u = sin( -- ) -- then use the Pythagorean identity to convert all but one power of cosine to sines, and the last power of cosine will be the du in a substitution.

48. Let’s do one... Both the powers are even, so use the double angle formula trick: In the last two terms, use the identities again to get what’s on the next slide.

49. Trig identity city! …and (finally!) this is something we can integrate!

50. …and at last:

51. A) 2/15 B) 4/15 C) 2/5 D) 8/15 E) 2/3 F) 4/5 G) 14/15 Here’s one with an odd power for you to do:

52. Try this one:  0 cos x dx = 4 A) 2 B)  C)  - 1/2 D) 2  E) 3 

53. The second, -- and more important, kind of trig substitution happens when there is a sum or difference of squares in the integrand. Usually one of the squares is a constant and the other involves the variable. If some other substitution doesn't suggest itself first, then try the Pythagorean trig identity that has the same pattern of signs as the one in the problem.,, or Either 22

54. Example: If you think about it, the substitution u = 5 - x 2 won’t work, because of the extra factor of x in the numerator. If you think about it, the substitution u = 5 - x 2 won’t work, because of the extra factor of x in the numerator. But 5 - x 2 is vaguely reminiscent of 1 - sin 2, so let Then and

55. We make all the substitutions and get: This is a trig integral of the other sort. And we can use the double-angle formula to get:

56. Now we have to get back to x’s So far, where therefore and so and so Also, And from the triangle,

57. The conclusion is: Quite a bit of work for one integral!

58. Examples for you… Here are two for you to work on. Notice the subtle difference in the integrand that changes entirely the method used:

59. Partial Fractions Last but not least is integration by partial fractions. This method is based on an algebraic trick. It works to integrate a rational function (quotient of polynomials) when the degree of the denominator is greater than the degree of the numerator (otherwise "when in doubt, divide it out") and you can factor the denominator completely. In full generality, partial fractions works when the denominator has quadratic and repeated factors, but we will consider only the case of distinct linear factors (i.e., the denominator factors into linear factors and they're all different). It also uses the (easy) fact that

60. Partial Fractions (continued) The idea of partial fractions is to take a rational function of the form (where there can be more or fewer factors in the denominator, but the degree of p(x) must be less than the number of factors) and rewrite it as a sum: for some constants A, B, C (etc..).

61. For instance, You can rewrite, first by factoring the denominator: and then in partial fractions as: and then in partial fractions as:

62. Two questions... ? 1. Why do partial fractions? 2. How to do partial fractions?

63. Two reasons for why: First reason: It helps us to do integrals. From the previous example, we see that we don't know how to integrate the fraction right away, but...

64. Two reasons for why: Second reason: Partial fractions helps us to understand the behavior of a rational function near its "most interesting” points. For the same example, we graph the function in blue, Let’s take a closer look and the partial fractions and in red: and the partial fractions and in red:

65. red and blue curves Note that one or the other of the red curves mimics the behavior of the blue one at each singularity.

66. OK, now for the “how” First, we give the official version, then a short-cut. Official version: It is a general fact that the original fraction will break up into a sum with one term for each factor in the denominator. So (in the example), write where A and B are to be determined. To determine A and B, pick two values of x other than x=-3 or x=-1, substitute them into the equation, and then solve the resulting two equations for the two unknowns A and B. For instance, we can put x=0, and get and for x=1, get. Solve to get A = 5/2 and B = -1/2 as we did before.

67. “Short cut” 1. Write the fraction with denominator in factored form, and leave blanks in the numerators of the partial fractions: 2. To get the numerator that goes over x + 3, put your hand over the x + 3 factor in the fraction on the left and set x = -3 in the rest. You should end up with 5/2. 3. To get the numerator that goes over x + 1, cover the x + 1 and set x = -1 (and you get -1/2). It's that simple.

68. Another example: First, the numerator and denominator have the same degree. So we have to divide it out before we can do partial fractions. and we can use partial fractions on the second term

69. Partial fractions gives: Use either the official or short-cut method to get A = -1 and B = 2. Therefore:

70. An example for you to try: Find 3 2 dx dx x (x-1) A) 3/2 B) 4/3 C) ln 2 D) ln 3 E) ln (3/2) F) ln (4/3) G) ln (2/3) H) 3 ln (2)/2

71. Another example: 4 3 4x - 6 4x - 6 x - 3x +2 2 = A) ln (4/3) B) 2 + arctan (3) C) ln (9) D) ln (12/5) E)  /3 - arctan (1/4)

72. Arc length The length of a curve in the plane is generally difficult to compute. To do it, you must add up the little “pieces of arc”, ds. A good approximation to ds is given by the Pythagorean theorem: We can use this to find the length of any graph – provided we can do the integral that results!

73. Find the arclength of the parabola y = x 2 for x between -1 and 1. Since dy / dx = 2x, the element of arclength is so the total length is:

74. So far, we have that the length is To do this integral, we will need a trig substitution. But, appealing to Maple, we get that

75. Can we do the integral? The arc length integral from before was: This is a trig substitution integral of the second kind: With the identity tan 2 + 1 = sec 2 in mind, let What about dt ? Since we have that These substitutions transform the integral into This is a tricky integral we need to do by parts!

76. To integrate Let Then But tan 2 = sec 2 - 1, so rewrite the last integral and get

77. Still going…. It’s remarkable that we’re almost done. The integral of secant is a known formula, and then you can add the integral of sec 3 to both sides and get So we’ve got this so far for where

78. We need a triangle! So far, with From the triangle, So

79. Definite integral: So far, we have Therefore To get the answer Maple got before, we’d have to rationalize the numerator inside the logarithm.

80. The area of a surface of revolution is calculated in a manner similar to the volume. The following illustration shows the paraboloid based on (for x=0..2) that we used before, together with one of the circular bands that sweep out its surface area. Surface Area

81. To calculate the surface area To calculate the surface area, we first need to determine the area of the bands. The one centered at the point (x,0) has radius and width equal to. has radius and width equal to. Since we will be integrating with respect to x (there is a band for each x), we'll factor the dx out of ds and write. So the area of the band. So the area of the band centered at (x,0) is equal to: Thus, the total surface area is equal to the integral

82. The surface area turns out to be: 133 

83. Good night!…. DON’T FORGET: 1. Send in your homework. 2. Meet Nakia and me in the chatroom. 3. Keep in touch by email! 4. Have a great week!