Day 35, Mat129: Calculus I

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Today:

Announcements
- Your 4.2 homework is returned: graded 8, 9, 37.
- Your quiz is returned.
- We have our third midterm exam coming up, on Monday.
  I'm going to drop the lowest of your three midterms. That's not an excuse to blow this one off, but rather an opportunity to improve on your test average.
  The test will cover from 3.3 through and including 4.2.
  - 3.3: How derivatives affect the shape of a graph
  - 3.4: Limits at infinity
  - 3.5: Curve Sketching
  - 3.7: Optimization
  - 3.9: Antiderivatives
  - 4.1: Areas and Distance
  - 4.2: The definite integral
- Today we'll begin with review. If you run out of questions, we'll move on.
Last time we wrapped up Section 4.4: the indefinite integral:
- Not too much to shout about here: the indefinite integral represents the family of anti-derivatives of a function:
  
  $\int{f(x)dx}=F(x)+C$
  where $F(x)$ is any particular anti-derivative.
- Let me bring to your attention the table of integrals on page 322, and at the back of the book (Reference Page 6 and following). This is a table of functions and their antiderivatives, and was the old way of evaluating integrals (before software got so good).
- Examples:
  - #1, p. 326
  - #5 (linearity of the integral)
  - #45. Think about it graphically.
  - #50; furthermore, what does
    
    $100+\int_{0}^Tn'(t)dt$
    represent?
Next up: Section 4.5: The substitution rule (for solving selected integrals analytically)
- 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 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
  
  $h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx=\int_{g(a)}^{g(b)}h'(u)du$
  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 $h'(g(x))g'(x)$.
  Let's do some examples!
- Examples:
  - #2, p. 335
  - #5
  - #8
  - #19
  - #38

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