Day 12 in Mat229

• I created this quiz to reflect the problems of the text, and what we did in class (e.g. asymptotic quadratics).

  You should probably go back and review the agendas in general, if you're not already doing that. They tend to contain pointers to what I think is most important.

  Also make sure that you're attempting the assigned exercises. I'll make sure to include at least one on each quiz.

• quiz 3 and key
• Median: 7
• Individual problem thoughts:
  1. Two strategies -- properties of logs makes it a little easier! (What's the best way to express the answer?)
  2. What's the best tangent line equation? (Dust off your product rule, and chain rule.)
  3. Long division is a lost art?
  4. For the bonus, I expected you to use the inverse formula....

• M-R, 10-3
• F, 10-noon

Consider the method of u-substitution, for example. That's just the chain rule in reverse.

We were working on one of those last time, but several of them could be worked that way. Let's convert the following to integrals we know how to do:

• $\displaystyle \int^{\frac{\sqrt{2}}{2}}_0 \frac{x}{\sqrt{1-x^2}}dx$

• $\displaystyle \int^{0}_{-2} x e^{x^2} dx$

• $\displaystyle \int^{10}_{1} \frac{\ln(x)}{5x}dx$

1. $\displaystyle \int^{e}_{1} \frac{3}{x^2}dx$ (power law)

2. $\displaystyle \int^{e}_{1} \frac{3}{x}dx$ (derivative of \(\ln(x)\))

3. $\displaystyle \int^{\frac{\sqrt{2}}{2}}_0 \frac{1}{\sqrt{1-x^2}}dx$ (derivative of \(\arcsin(x)\) -- but note that Mathematica goes with a related antiderivative! In their step-by-step description they mention arcsine....)

4. $\displaystyle \int^{10}_{0} \frac{2}{1+x^2}dx$ (derivative of \(\arctan(x)\))

5. $\displaystyle \int^{10}_{-10} x^{3001} dx$ (zero, by virtue of odd symmetry)

• Inverse functions are reflections of one-to-one functions over the line $y=x$.
• Properties of logs are really reflections of properties of exponentials. E.g. \[ e^{a+b} = e^ae^b \] (The exponential of a sum is the product of the exponentials). Take logs of both sides: \[ \ln(e^{a+b}) = a+b = \ln(e^a) + \ln(e^b) = \ln(e^ae^b) \] The log of a product is the sum of the logs. \[ \ln(\alpha) + \ln(\beta) = \ln(\alpha \beta) \]

• Every time you learn a new derivative, you can solve a new integral.

Here's a video clip of Fred Astaire and Ginger Rogers. They say Fred Astaire was pretty good (well, "the best dancer ever"); but Ginger Rogers did everything Fred did, only backwards (and in heels...). I especially like this moment.

Now the integral of a sum is just the sum of the integrals, so we can split up the two parts, and treat them separately.

This is the key: you've got two integrals, and you can trade one of those integrals you don't like for one that is more palatable....

\[ \int_{}^{}f(x)g'(x)dx = f(x)g(x) - \int_{}^{}f'(x)g(x)dx \] or \[ \int_{}^{}udv = uv - \int_{}^{}vdu \]

1. Let's start off with a straightforward one: \[ \int x e^{2x} d x \] Which part of the product would you prefer to differentiate, and which integrate, in order to leave you with an integral you'd prefer to do?

2. How about a silly one? \[ \int x^2 d x \]

3. One new anti-derivative we can get using this technique is the anti-derivative of arctan. Let's see how that works (Exercise #9 of this section). This one involves something of a trick:

   \[ \int \tan^{-1}(x) d x \]

   The problem is that you might not see the other function in the product....