Today we introduced the partial fraction decomposition. This is an algebra technique that helps integrate rational functions. Here is a video example. We applied the technique to solve the following integrals:
\(\displaystyle \int \frac{x+5}{x^2+x-2} \, dx\).
\(\displaystyle \int \frac{1}{P(1-P)} \, dP\).
Partial fractions works for any rational function where the degree of the numerator (top) is less than the degree of the denominator (bottom) and the denominator factors into linear (i.e., degree 1) factors that all have different roots. If the degree of the numerator is bigger than or equal to the degree of the denominator, then use polynomial long division first. Here is a video review of how polynomial long division works.
We did these examples in class:
\(\displaystyle \int \frac{x^3 + x}{x-1} \, dx\).
\(\displaystyle \int \frac{x^4-5x^2+4x}{x^2-4} \, dx\).
We finished by talking about how to do partial fractions when the denominator has repeated roots. You can do this, but you need a separate partial fraction for each power of the repeated factor up to its actual power. We did this example:
Today we switched from integrals to limits. We will be using limits a lot in the next few weeks, so we need a fast way to compute them. One option that works frequently (but not always) is L’Hospital’s rule. Here is a quick video explanation. We did the following examples:
\(\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}\).
\(\displaystyle \lim_{x \rightarrow 5^+} \frac{x}{x-5}\). (You can’t use L’Hospital’s rule here! Why not?)
\(\displaystyle \lim_{x \rightarrow \infty} \frac{x^2}{e^x}\).
\(\displaystyle \lim_{x \rightarrow 3^-} \frac{x^2 - 2x - 3}{x^2 - 5x +6 }\).
\(\displaystyle \lim_{x \rightarrow \infty} \frac{\ln x}{1-x}\).
\(\displaystyle \lim_{x \rightarrow \pi^+} \frac{\sin x}{1-\cos x}\). (Watch out, this one isn’t L’Hospital’s rule either!)
\(\displaystyle \lim_{x \rightarrow \infty} \frac{3x^{8}-4x^{7}+3x^{4}-5x}{2x^{7}+5x^{5}-3x+12}\) (Hint: The key to this one is to factor out the highest power of \(x\) in the numerator & denominator).
In addition to the indeterminant forms \(\frac{\infty}{\infty}\) and \(\frac{0}{0}\), there are other indeterminant forms that come up when working with limits. These include: \[0 \cdot \infty, ~ \infty - \infty, ~ 0^0, ~ 1^\infty, ~ 0^\infty\] Here are some ideas on how to deal with these when they pop up in limit calculations:
\(\displaystyle \lim_{x \rightarrow 0^+} x \ln x\). Hint: re-write the problem as \(\displaystyle \lim_{x \rightarrow 0^+} \dfrac{\ln x}{x^{-1}}\) so you can apply L’Hospital’s rule.
\(\displaystyle \lim_{x \rightarrow \tfrac{\pi}{2}^-} (\sec x - \tan x)\). Hint: combine the secant and tangent into one fraction by switching to sines and cosines.
\(\displaystyle \lim_{n \rightarrow \infty} \left(1+ \frac{r}{n} \right)^n\) where \(r\) is any constant. Hint, let \(\displaystyle A = \lim_{n \rightarrow \infty} \left(1+ \frac{r}{n} \right)^n\) and take the natural log of both sides.
Today we talked about improper integrals which are integrals involving infinity (either in the bounds, or because of a vertical asymptote). You deal with these just like any other definite integral, except you might have to calculate a limit to find the answer. We did several examples:
\(\displaystyle \int_1^{\infty} \frac{1}{x^2} \, dx\). This one turns out to just be 1, which illustrates a very important idea: The area under an infinite curve can still be finite!
\(\displaystyle \int_2^3 \frac{1}{\sqrt{x-2}} \, dx\). You could integrate this one without even realizing that it is technically an improper integral. But it you look at the graph, you’ll see why it counts as an improper integral.
\(\displaystyle \int_2^\infty \frac{1}{x} \, dx\).
\(\displaystyle \int_{-\infty}^0 x e^{x} \, dx\).
If the value of the integral is a finite number, we say it converges. Otherwise, it diverges. An improper integral can diverge for two reasons. It might have infinite area, or there might not be one number that the area converges to (which is why \(\displaystyle \int_0^{\infty} \sin x \, dx\) diverges).
We also talked about another simple idea known as the comparison test. If \(f(x)\) and \(g(x)\) are nonnegative functions and \(f(x) \le g(x)\), then \(\int f \le \int g\). This has two consequences:
If the bigger function has a finite integral, then the smaller integral must be finite too.
If the smaller function has an infinite integral, then so does the bigger one.
Compare the integrals \(\displaystyle \int_1^e \frac{1}{x \ln x} \, dx\) and \(\displaystyle \int_1^e \frac{1}{\ln x}\) One of these integrals is easy if you make the u-subsitution \(u = \ln x\). Does that integral converge? What about the other integral? Can you use the comparison test here?
Use the comparison test to show that \(\displaystyle \int_{\pi}^\infty \left( \frac{\sin x}{x} \right)^2 \, dx\) converges.