Day 22 in MAT227

Last time: Section 7.8: Finished Inverse Trig Functions

Next time: Section 10.1: Solving Differential Equations

Today:

Announcements:
- Homework assignment 7.4 returned -- graded 22 and 24
  - #22:
    - Wise to start with initial value equal to 100, so that the value of P(t) is the percentage remaining!
  - #24:
    - Don't approximate too soon. You should have gotten ${\approx{9.509775004326938}}$ hours
    - How do you get the initial values?
    - What equation do you solve?
- Assignment 7.7 returned:
  - 1 -- don't be a limit dropper!
  - 7 -- do it twice (or use a trick!)
  - 45 -- some of you have really mastered your trig identities!;)
- Assignment 7.8 due
- FYI: second exam is next Monday, 11/15. I've decided to limit it to chapters 6 and 7.
- This Wednesday I'll stay after class until 10:00, if you want to do any extra problems in preparation for the upcoming exam.
Today: we're looking at two techniques. First off is Section 8.2: Integration by Parts

${\int_{}^{}u(x)v'(x)dx=u(x)v(x)-\int_{}^{}v(x)u'(x)dx}$

${\int_{}^{}udv=uv-\int_{}^{}vdu}$
- First of all, the big picture: integration by parts is just the product rule backwards: ${\left(u(x){\cdot}v(x)\right)'=u'(x)v(x)+u(x)v'(x)}$ This integration technique, like all integration techniques, is really just a differentiation technique in reverse.
  Example: Exercise #2, p. 428
- Integration by parts may need to be carried out multiple times: sometimes the idea is to simplify the integral each time, until a really simple one arises allowing us to calculate the final solution (e.g. Example 4, p. 425). Sometimes it's something of a trick: we compute the integral multiple times in order to return to the original integral, allowing us to solve an equation for the original integral (e.g. Example 5, p. 426).
  Example: Exercise #18, p. 428
- Suggested strategy:
  - Choose u so that u' is simpler than u itself.
  - Choose v' so that its antiderivative $v=\int_{}^{}v'dx$ is known, or can be evaluated.
  - Sometimes v'=1 is a good choice.
    Example: Exercise #25, p. 428
- Of course, all this works with definite integrals:
  
  Example: Exercise #49, p. 428
Section 9.1: Arc Length and Surface Area
- What kinds of problems are we looking at in this section? Arc length and surface area calculations for areas of revolution:
- Let's start with arc length:
  
  ${I=\int{dI}\approx\sum_i\Delta{I_i}}$
  The usual trick is to discretize something continuous, and then pass to a limit:
  
  Now hopefully we can figure out how the formula will look:
  
  So what's the general formula?
- For surface area (for areas of revolution)
  
  We discretize bits of the surface, and then make a series of little bands; take the limit, and get the general formula:
- Examples:
  - #5, p. 475
  - #12, p. 475
  - #39, p. 476
  - #51, p. 477

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