Midterm 2

Posted: 03/05/2025

I have made this page in response to a first-year student unable to attend the Civ Club Review Session for the second MAT187 midterm. This page is intended to be an overview of all the concepts learned since the first midterm. I did not have much time to assemble this page, so it lacks practice problems in some areas. If you have any questions about prior content or you would like some practice problems, please contact me. I can not guarantee I will be able to give some to you prior to your midterm as my Wednesdays and Thursdays are rather busy.

Integration Techniques

In MAT187 last year, this was the first thing we touched on. I am not sure why it is being taught so late now, I think they are a natural continuation of where we left off at the end of MAT186 (integration by substitution).

Integration By Parts

Suppose \(u\) and \(v\) are differentiable functions. You might recall the dervative of \(uv\): \[\frac{\mathrm d}{\mathrm dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\] If we integrate, we get: \[u(x)v(x) = \int v(x) \, \mathrm du(x) + \int u\, \mathrm dv(x)\] Rearranging, we get: \[\int u\, \mathrm dv(x) = u(x)v(x) - \int v(x) \, \mathrm du(x)\] This is the basis of integration by parts. If you need to find the integral of two functions multiplied together, the goal is to make it solvable using other methods. Let us look at an example: \[\int \ln x \,\mathrm dx\] You might remember the derivative of \(\ln x\) is \(\frac1x\). Looking at the rearranged formula for integration by parts, let us make \(ln x \) our \(u\). This means our \(dv\) must be \(1\). Plugging this into our formula, we get: \[\int \ln x \,\mathrm dx = x\ln x - \int \frac{x}{x} \, \mathrm dx\] \[\int \ln x \,\mathrm dx = x\ln x - \int 1 \, \mathrm dx\] \[\int \ln x \,\mathrm dx = x\ln x - x + C\] As an additional exercise to the reader, try out the following problem: \[\int_1^2 x^2e^x\,\mathrm dx\] Sometimes, things cancel out easily or the derivatives of \(u\) run out. Sometimes, neither occurs. Let us look at an example of this: \[\int e^x\sin x \,\mathrm dx\] You are probably not sure which one is \(u\) and which is \(v\). As it turns out, that does not matter much. Let us make \(\sin x\) our \(u\). \[\int e^x\sin x \, \mathrm dx\ = e^x\sin x - \int e^x \cos x \,\mathrm dx\] That did not get us to a solution, let's try again: \[\int e^x\cos x \, \mathrm dx\ = e^x\cos x + \int e^x \sin x \,\mathrm dx\] \[\int e^x\sin x \, \mathrm dx\ = e^x\sin x - e^x\cos x - \int e^x \sin x \,\mathrm dx\] Now, we can move that integral to the left side to get: \[2\int e^x\sin x \, \mathrm dx\ = e^x\sin x - e^x\cos x\] Then we can simply divide both sides by 2.

Trigonometric Substitution

This involves some trigonometric identities. I list the most useful ones below, as noted in last year's lecture slides: \[1 - \sin^2\theta = \cos^2\theta\] \[1 + \tan^2\theta = \sec^2\theta\] \[\sec^2\theta - 1 = \tan^2\theta\] Let us look at an example integral to see how these work: \[\int \frac1{\sqrt{x^2-4}}\, \mathrm dx\] Since there is a square root here, the method of solving this integral does not seem clear with prior knowledge. However, we can substitute \(x = 2\sec\theta\). As with a typical substitution problem, we must find a relation between \(\mathrm dx\) and \(\mathrm d\theta\). Taking the derivative, we find \(\mathrm dx = 2\sec\theta\,\tan\theta\). Thus, our integral becomes: \[\int \frac1{\sqrt{4\sec^2\theta - 4}}\,2\sec\theta \, \tan\theta\, \mathrm d\theta\] We can then use the second trigonometric identity above to rewrite the integral as: \[\int \frac1{\sqrt{4\tan^2\theta}}\,2\sec\theta \, \tan\theta\, \mathrm d\theta\] We can do some further simplification to obtain: \[\int \sec\theta \, \mathrm d\theta\] I do not know if the 2T8s are getting a formula sheet with integrals of trigonometric functions. Either way, feel free to search up the integral of \(\sec\theta\). I will leave substituting the \(x\) back into this equation as an exercise to the reader.

You might have noticed that I used a trigonometric identity to allow me to find the square root. This kind of simplification is typical of trigonometric substitution problems. In your own time, try solving the integral below. Remember the identities above and how you might use them. Please contact me if you have any questions. \[\int \sqrt{9 - x^2} \mathrm dx\]

Partial Fractions

This is more a trick of algebra, but it simplifies the solutions of many integrals you may encounter. For all steps when solving an integral using this method, when you have \(\int \frac{P(x)}{Q(x)} \,\mathrm dx\), the degree of \(P(x\) must be less than that of \(Q(x)\). If it is not, you will need to first use polynomial long division and make the remainder your new \(P(x)\).

When using this method, you factorize the denominator, then split the fraction into ones with one factor in the denominator, added together. The following example is from last year's lecture slides: \[\int \frac{34x^4 + 6x^3 + 89x^2 + 26x - 16}{3x^5 + 5x^4 +10x^3 + 20x^2 - 8x} \,\mathrm dx\] Let us first factor the denominator: \[\int \frac{34x^4 + 6x^3 + 89x^2 + 26x - 16}{x(3x-1)(x+2)(x^2+4)}\,\mathrm dx\] You might recall from long ago (perhaps 8th or 9th Grade) that when adding fractions together, you will find the least common factor of all of the denominators and multiply the numerators accordingly. We will do that here. \[\int (\frac Ax + \frac B{3x-1} + \frac C{x+2} + \frac{Dx+E}{x^2 + 4}) \,\mathrm dx\] \[\int (\frac{A(3x-1)(x+2)(x^2+4)}{x(3x-1)(x+2)(x^2+4)} + \frac{Bx(x+2)(x^2+4)}{x(3x-1)(x+2)(x^2+4)} + \frac{Cx(3x-1)(x^2+4)}{x(3x-1)(x+2)(x^2+4)} + \frac{(Dx+E)x(3x-1)(x+2)}{x(3x-1)(x+2)(x^2+4)}) \,\mathrm dx\] I will leave the rest of the solution as an exercise to the reader, but the simplified version should be as shown below. Keep in mind that the sum of the numerators in the above integral must equal \(34x^4 + 6x^3 + 89x^2 + 26x - 16\). \[\int (\frac2x + \frac1{3x-1} + \frac7{x+2} + \frac{2x-3}{x^2 + 4}) \,\mathrm dx\] You can then use other methods to solve the integral. There is one other case I would like to highlight: Repeated factors. Let's say you have an integral: \[\int\frac{2x^2-3x+10}{(x-1)^2(x+2)}\,\mathrm dx\] The partial fraction decomposition would look like: \[\int(\frac A{(x-1)^2} + \frac B{x-1} + \frac C{x+2})\,\mathrm dx\] The rest of the solution is left as an exercise to the reader.

Improper Integrals

If you were present during the previous review session, you might remember me trying to talk about sequences and series using improper integrals. Back then, you probably had not learned about them, but now you have.

Improper integrals are integrals with some condition requiring a limit to be taken. This can be it running to \(\pm \infty\), a vertical asymptote, or maybe something else I forgot about. A limit must be taken because some number approaches infinity, preventing us from using direct substitution. Let's look at an example: \[\int_0^1\frac1{\sqrt x} \,\mathrm dx \] The function has a vertical asymptote at \(x = 0\), so it is improper. I will leave the solution as an exercise to the reader, but know that it is convergent. In fact, we have a general rule of convergence for these types of integrals (specifically \(\int_0^a \frac1{x^b}\,\mathrm dx\)), where \(a\) and \(b\) are real numbers. For the integral to be convergent, \(b < 1\). Otherwise, it is divergent, and the limit you get when you solve the integral goes to infinity. There is a similar rule for integrals that go to infinity (\(\int_a^\infty \frac1{x^b}\,\mathrm dx\), where \(a\) and \(b\) are real numberes). In that case, it is divergent if \(0 < b \leq 1\). Now for an example:

Let's say you are building a rocket to explore the universe. You remember from Grade 12 physics that \(F_g = \frac{Gm_1m_2}{r^2}\), where \(G = 6.6743\times10^{-11} \mathrm{\frac{N\,m^2} {kg^2}}\), \(m_1\) and \(m_2\) are the two gravitationally attracted masses, and \(r\) is the distance between them. You also remember that work \(W = F\Delta r\). How much energy would it take for the rocket to escape Earth's gravitational pull after launching from Earth's surface. In other words, how much energy would be required to move the rocket from Earth's surface to an infinite distance away? Some important parameters:

Mass of the Earth: \(5.972 \times 10^{24}\, \mathrm{kg}\)
Mass of your ship: \(14,135\, \mathrm{kg}\)
\(r\) for Earth's surface: \(6.36\times10^6\, \mathrm{m}\)

As a bonus, try figuring out how much energy it would take you to fly a rocket from Earth's core to its surface, assuming you could travel through the planet like you can travel through space. If you have any questions about this problem or would like a solution, please contact me.

A potential mistake

What if there is a vertical asymptote in the middle of your integral? For example, \(\int_{-1}^1 \frac{1}{x^2}\,\mathrm dx\). If you evaluate as normal, it may seem like this integral converges, i.e. it has a solution. However, it does not. To ensure you know whether your integral converges, you will need to split it at the asymptote, i.e. making it \(\int_{-1}^0 \frac{1}{x^2}\,\mathrm dx + \int_0^1 \frac{1}{x^2}\,\mathrm dx\).

Power Series

A power series is a series of the form \( c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + ... = \Sigma_{i=0}^{\infty} c_i(x-a)^i \), where \(a\) and \(c\) are constant real numbers. All Taylor series are power series, and any power series is a Taylor Series representing some function. All power series converge at \(x = a\). For other real numbers, you will need to determine the power series' radius of convergence \(R\), the distance from \(a\) up to which the power series is convergent. For values of \(x\) which are \(R\) away from \(a\), the value may or may not be convergent (you will need to determine this with a convergence test). The interval of convergence describes the interval where the power series is convergent.

You can find the interval of convergence using the ratio test:
If you have a series \(\Sigma_{i=0}^{\infty} a_i\) and we let \(r = \lim_{n\to\infty} | \frac{a_{n+1}}{a_n}|\), then the series converges when \(r < 1\) and diverges when \(r > 1\). If \(r = 1\), we must use another test. For power series, this usually occurs when \(x\) is \(R\) away from \(a\). Recall the convergence tests you learned before Midterm 1.

I unfortunately did not learn how to find the radius of convergence of the power series representing the function \(\frac{1}{p(x)}\) using the roots of polynomial \(p(x)\). Please review it, as it is on the list of concepts for your midterm.

Last updated: 03/06/2025