Day 37 in Mat129

Last time: Fundamental Theorem (Part II)

Next time: Area between curves

Announcements
- Next exam: tomorrow, 4/20. It will cover chapter 4, and 5.1, 5.2 (that's a lot!). Study hard....
- Mon 4/19 Section 5.5 Read 5.6 for Thursday
  Begin on-line homework "FTC and Area Between Two Curves" (due 4/25)
Last time: The Fundamental Theorem of Calculus, Part II
- The news isn't earth-shaking: but it's certainly an interesting idea!
- Now we arrive at the FTC, Part II:
- Examples:
  - #1, p. 282
  - #8, p. 282
  - #16, p. 282
  - #20, p. 282
  - #23 p. 282
  - #46 p. 284
Section 5.5: Some examples of Integrals
- #1
- #13
- #19
- #21
Section 5.6: The substitution rule
- The substitution rule is the chain rule backwards: I'd prefer that we think as follows: ${h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx}$ and, even better, ${h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx=\int_{g(a)}^{g(b)}h'(u)du}$
  Forgetting for a moment that we might know how to solve this!;), we can always do the change of variables
  ${\int_{a}^{b}f(g(x))g'(x)dx=\int_{g(a)}^{g(b)}f(u)du}$ and hope that the integral on the right is easier to solve (certainly less cluttered).
  Writing it in this last way may be mysterious, because of the change of variable to u (and the change in the limits); but it's the disappearance of g'(x) that's really curious. It falls right out of the change of variables, however:
  ${u=g(x)\Longrightarrow\frac{du}{dx}=g'(x)}$ Solving for du, we get du=g'(x)dx, and that piece disappears.
- Examples:
  - #2, p. 295
  - #7, p. 295
  - #13, p. 296
  - #31, p. 296
  - #41, p. 296
  - #57, p. 296
  - #63, p. 297
  - #64, p. 297

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Updated on 04/19/2010 11:53:14