Day 38, Mat129: Calculus I

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Today:

Announcements
- Your homework 4.2 is returned: graded 6, 36, 37
  - #6: $\Delta x=1$, so you're just adding up the correct representative function values for each subinterval.
  - #36 and 37: in both of these you are to interpret the integral as shapes for which you know how to find the areas.
    In #37, one part is a circle, radius 3.
  - #37: the exact answer is $3 + \frac{9\pi}{4}$, which is approximately equal to 10.07. When you circle an answer, circle the exact answer -- then approximate, for those who'd like some tangible idea of its size.
- Your homework 4.3 is due.
- There's a quiz at the end of the hour....
Today we finish off Section 4.4: the indefinite integral -- with one more problem. Then on to 4.5.
- 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.
- I want to bring the table of integrals to your attention at the back of the book. This is basically a table of functions and their antiderivatives.
- Notes:
  - if you can solve the indefinite integral problem, then you can immediately solve the definite integral problem. The anti-derivative solves the area problem -- that's key.
  - You'll notice that the "Net Change Theorem" (p. 324) was my starting point in the introduction to integration: \[ \int_{a}^{b}F'(x)dx = F(b)-F(a) \] I just wrote it with $p$ for $F$, and with $t$ as the dummy variable of integration.
  - Observe the displacement formula (total distance travelled, p. 325): \[ \int_{a}^{b}|p'(t)|dt \] The absolute value assures that you credit for moving with negative speed, as well -- obviously you cover just as much distance at negative speed, just in the opposite direction. So slapping the absolute value on the speed produces a positive answer in the end -- the distance travelled (rather than the net change in position).
- Examples:
  - #1, p. 326
  - #5
  - #45
  - #50; furthermore, what does
    \[ 100+\int_{0}^Tn'(t)dt \]
    represent?
  - #47
  - #48
  - #53
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: $\cos(x^3)3x^2$
  - Hence,
    \[ \int{\cos(x^3)3x^2dx}=\sin(x^3)+C \]
  That's the general idea. 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, \[ h(g(b))-h(g(a))=\int_{a}^{b}h'(g(x))g'(x)dx=\int_{g(a)}^{g(b)}h'(u)du \]
  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.
- Our author makes an important point on page 331: It is permissible to operate with $dx$ and $dy$ after integral signs as if they were differentials.
  One thing you mustn't do is drop them!
- Examples:
  - #2, p. 335
  - #5
  - #8
  - #19
  - #38
  - #52

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