COALITION MATH 152, SPRING 1998 (S. A. Fulling, assisted by Vera Rice) Day 20.T 1. The usual paper shuffling: Old homework, CAPA, revised new homework, lab sheet if ready. Announce: A. CAPA deadline is Tuesday 5:00; BUT bug reports will not be received after Tuesday class. There will be bonus points for bug reports received early (so that the problem can be corrected). B. Major change in syllabus and homework this week: i. Omit Sec. 7.2 (hard trig integrals); it will be in 251 anyway. If such an integral is encountered in later homework, look it up or use Maple. ii. Some volume problems moved to this week from next week. I will try to "push" the lecture schedule slightly between now and next test. iii. I added some stars as well as removing many. 2. Integration by parts A. Not every integral can be found by a substitution. Example: \int x^2 sin(3x) dx (we'll do it later). B. The formulas (u dv = d(uv) - v du and its integral form) and their origin in the product rule. C. The strategic principle for integration by parts: Choose u to be something whose DERIVATIVE is simple; choose dv/dx to be something whose ANTIDERIVATIVE is simple. (Then \int vu' dx will be easier than \int uv' dx.) There are two main classes of such integrals: D. Case 1: A polynomial times an exponential or trig function. Choose u = poly, v' = rest. INDIVIDUAL EXERCISE: Implement this for \int x^2 sin(3x) dx. (For this demonstration, please don't use "tabular method" you may have learned in high school. We want all students to understand the steps!) E. Case 2: A transcendental function whose derivative is algebraic (such as a log or inverse trig), times something else that you can integrate. [Attach definitions of "algebraic" and "transcendental".] Choose u = transcendental, v' = rest (opposite of the previous prescription!). TEAM EXERCISE: \int sin^{-1}(x) dx F. In the long run, probably the main importance of integration by parts is not in integrating particular functions, but in more general arguments. i. Example: Prove that if f(0) = 0 and f(1) = 0 [and f" exists] then \int_0^1 f(x) f"(x) dx is a negative number or zero. ii. This sort of thing comes up in physics and engineering in dealing with the energy density in a fluid or an electric field, etc. 3. Test results A. Median = 59. Top score = 94 (tie). Papers back in recitation tomorrow. B. EARN YOUR CURVE i. Everybody will get 5 more points outright. ii. On Thursday, Feb. 26 (21.R), turn in the following on separate (but attached) sheets of paper: a. (signed) Write out a solution of the problem on which you got the lowest score, or the problem you omitted. b. (unsigned, but indicate which problem you're talking about) Write a few thoughtful sentences: 1) How could the course presentation (Web pages, lecture, book, homework, lab, ...) have been improved, so that you would have done better on this problem? 2) What could YOU have done that would have improved your score on this problem (or overall)? BEFORE ANSWERING, LOOK AGAIN AT THE WEB PAGES, BOOK, .... iii. Andrew and/or Vera will verify the existence of a suitable submission and record more points: 2 pts if your score was in the 90s, 3 pts if in the 80s, etc. I will read the anonymous essays and hope to learn something from them. 4. Introduction to volumes (following prepared slides with suitable graphics) A. Recall that to find an AREA we typically have a choice of methods: i. Slice vertically (integrate over dx). ii. Slice horizontally (integrate over dy). B. For volumes of solids of revolution we have a choice of methods: IT'S NECESSARY HERE TO PEEK AHEAD AT THE SLIDES FOR POINTS D AND E, TO MAKE THE TERMS CLEAR. i. Disks and washers (integrate over the location of the disk along the axis). ii. Cylindrical shells (integrate over the radius -- the distance perpendicular to the axis). These are TWO WAYS TO DO THE SAME PROBLEM. C. But also, for a given curve, we can study i. The solid formed by rotation about the horizontal axis. ii. The solid formed by rotation about the vertical axis. These are TWO DIFFERENT PROBLEMS. EACH could be done EITHER by disks OR by cylinders. GOT THIS FAR. D. Disk method: V = \int (area of disk at u)(thickness) = \int_{umin}^{umax} (\pi r^2) (du) where r = f(u) = radius of disk at u. (For washers, area = \pi (r_{outer}^2 - r_{inner}^2.) E. Cylinder method: V = \int (circumference)(height) (thickness) = \int_{rmin}^{rmax} (2\pi r)(umax - umin)(dr) where umax = g(r) = right or top boundary of shell of radius r, umin = g(r) = left or bottom boundary. (Often rmin = 0.) F. Example: Implement these (set up all four integrals) for the curve x = 1 - y^2.