Day 37, Mat129: Calculus I

• We'll meet here on Friday.
• Your 4.3 homework is due today.
• I've assigned 4.4 homework for next Monday, and 4.5 for next Wednesday.
• Your exams are returned.
  • The exam

  As I said, I'm going to drop the lowest of your three midterms. That being said, in many cases, this was your lowest score. And the scores were generally low.

  For that reason I'll allow you to get some of your points back by correcting your exams. You can get up to 50% of your missed points back, if you'll fix anything that you did wrong. The best thing to do would be to write me a perfect exam. This will also help you to prepare for the final.

  You will need to:

  1. Print off a fresh copy of the exam
  2. Redo everything that you didn't get perfect
  3. Submit both copies of the exam (old and new) by Wednesday, 4/20.
  Everything you've corrected will earn back half the points.

  You have to earn back the points. If you don't redo a problem, you will get no points back.

  You may work with anyone: your classmate, the calc lab, me, etc.

• Some comments:
  1. In spite of my efforts, many of you mistook this function for either $f$ or something else. It's a graph of $f'$.
  2. Slant asymptotes are important.
  3. In the study of a function I should see $f'$ at least! Also $f''$, asymptotes, domain, symmetry, etc.
  4. Draw a picture!
  5. Don't forget: the anti-derivative of a function is actually a family of functions.
  6. Signed area. Triangles.
  7. Riemann sums. Sines and cosines without arguments are both "sins". Both require parenthesis, and something to eat. They're both hungry, hungry operators.

This is our last technique of integration. Then we end the semester with some applications of integration.

• The substitution rule is the chain rule backwards.

  Every differentiation rule is written backwards to create an integration rule. Substitution is the flip side of the chain rule.

  • Q: What is the derivative of $\sin(x^3)$?

  • A: $\int{\cos(x^3)3x^2dx}=\sin(x^3)+C$

  That's a funny answer, but it suggests the general idea (but for a specific example). If we let

  $h(x)=\sin(x)$

  and

  $g(x)=x^3$

  then we can generalize this:

  • Q: What is the derivative of $h(g(x))$?

  • A: $\int{h'(g(x))g'(x)dx}=h(g(x))+C$

  So, in terms of a definite integral, the rule is that

  $h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx$

  and, even better, let $u=g(x)$; then

  $\frac{du}{dx}=g'(x)$ or $du=g'(x)dx$,

  so

  $h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx=\int_{g(a)}^{g(b)}h'(u)du$

  We re-express an integral in $x$ in terms of a variable $u$. Sometimes this method is called "u-substitution", in honor of our favorite substitution variable.

  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). Notice especially the change in the limits on the integral.

  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 hence that piece of the integrand disappears.

  The trick generally is to recognize the presence of a "chain-rule derivative" in the integrand -- that is a product that one can think of as $h'(g(x))g'(x)$.

  Indefinite integrals are a little simpler to think about: we might think of them this way: \[ \int f(g(x))g'(x)dx = \int f(u)du = F(u) + C = F(g(x)) + C \]

• Examples:
  • #2, p. 335
  • #5
  • #8
  • #19
  • #38
  • Then we'll start on your homework, if there's time remaining.... pp. 335--, #1, 3, 4, 7, 9, 11, 16, 20, 21, 22, 25, 32, 38, 39, 43, and 49 (due Wednesday, 4/20)

    Practice!