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Week 2: Calculus II Notes

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Week 2: Calculus II Notes Christian Parkinson GRE Prep: Calculus II Notes 1 Week 2: Calculus II Notes We concluded the Calculus I notes with Riemann integration, Fundamental Theorem of Calculus and some helpful integration techniques. However, ofter times, you will be asked to identify whether an integral converges or diverges even when you cannot find the value. This is especially true for improper integrals: those in which the integrand had a singularity or the range goes to infinity. We begin the Calculus II notes by discussing some basic convergence results. Proposition 1 (p-test). Let p 2 R. The integral Z 1 dx p 1 x converges if p > 1 and diverges otherwise. The integral Z 1 dx p 0 x converges if p < 1 and diverges otherwise. Note that as a consequence, the integral Z 1 dx p 0 x never converges. These can all be verified by a straight-forward calculation and hint at some good intuition. The borderline case is 1=x where neither integrals converge, so as a general rule, for Z 1 f(x)dx 1 to converge, we will need f(x) ! 0 faster than 1=x as x ! 1 and if g(x) has a singularity at x = 0, then for Z 1 g(x)dx 0 to converge, we need g(x) to blow up slower than 1=x as x ! 0+. Note, these \general rules" should not be seen has hard-and-fast truths, but rather heuristics which can often make it easy to verify if an integral converges or diverges. The following two theorems give more rigorous forms of these general rules. Theorem 2 (Comparison Test). Suppose that f; g : [0; 1) ! [0; 1) are continuous and that f(x) ≤ g(x) for all x 2 [0; 1). Then Z 1 Z 1 (a) if g(x)dx converges, then so does f(x)dx, 0 0 Christian Parkinson GRE Prep: Calculus II Notes 2 Z 1 Z 1 (b) if f(x)dx diverges, then so does g(x)dx. 0 0 Theorem 3 (Limit Comparison Test). Suppose that f; g : [0; 1) ! [0; 1) are contin- uous and that f(x) lim x!1 g(x) exists as a number in (0; 1)[excluding the endpoints]. Then Z 1 Z 1 f(x)dx and g(x)dx both converge or both diverge. 0 0 Note, the previous two theorems also hold for improper integrals where the functions have singularities at the same point but the limit in the second must be taken at the singularity. This last theorem tells us that the only important thing in determining convergence of an integral with 1 as a limit is the asymptotic behavior of the integrand. We have a special notation for this. If f; g are such that f=g ! L as x ! 1 where L is a finite non-zero number, then we write f ∼ g [note: in most contexts, L is required to be 1 in order to say f ∼ g; for our purposes, it is fine to allow L to be finite and non-zero]. We typically read this as \f is asymptotic to g." Note that if f ∼ g, then there are constants c; C > 0 such that for sufficiently large x, we have cg(x) ≤ f(x) ≤ Cg(x). This shows that we can easily derive the Limit Comparison Test from the ordinary Comparison Test. A relaxed version of this requirement leads to a new definition. If there is C > 0 such that f(x) ≤ Cg(x) for all x sufficiently large, then we write f = O(g) [and say \f is big-oh of g"]. Thus f ∼ g iff f = O(g) and g = O(f). Specifically, f = O(g) iff f=g remains bounded as x ! 1. Finally, note that we can define the same notion for x approaching some other point. For example, sin(x) ∼ x as x ! 0. In order to apply these rules, it is nice to know something about the growth rates of different common functions at 1. We discuss a few here. Proposition 4 (Growth Rates of x; ex; log(x) at 1). Let α; β; γ > 0. Then 1 (a) lim = 0; x!1 log(x)α log(x)α (b) lim = 0; x!1 xβ xβ (c) lim = 0: x!1 eγx What these tells us is that asympotically as x ! 1, any power of x is smaller than any exponential and any power of a logarithm is smaller than any power of x. In asymptotic notation, we write 1 log(x)α xβ eγx where f g means that f=g ! 0 at 1. We can compound these into new relationships xβ xβ log(x)α. Many more such relationships can be proved by simply by taking the required limit. Note that if f g, then f = O(g); Christian Parkinson GRE Prep: Calculus II Notes 3 the converse need not be true. Example 5. Determine the convergence/divergence of the following integrals Z 1 1 (a) p dx. 2 x + x + log(x) Z 1 x6 + 4x3 + 3x2 + 2 (b) 9 dx 1 7x + x + 10 Z 1 (c) csc(x)1=2dx: 0 Solution. p p (a) Since x; log(x) x, there are constants c1; c2 such that for x large enough, x ≤ c1x, log(x) ≤ c2x. Thus Z 1 dx Z 1 dx p ≥ C + 2 x + log(x) + x M (1 + c1 + c2)x where C; M are some constants. The former diverges since the latter does. (b) Notice that if we divide the integrand by 1=x3 (hence multiply by x3) and take the limit, we see x9 + 4x6 + 3x5 + 2x3 1 + 4=x3 + 3=x4 + 2=x6 lim = lim = 1=7: x!1 7x9 + x + 10 x!1 7 + 1=x8 + 10=x9 R 1 dx Thus by the Limit Comparison Test, our integral converges iff 1 x3 converges. The latter converges, so the integral in (b) converges as well. p (c) Since sin(x) ∼ x as x ! 0, we see that csc(x)1=2 ∼ 1= x. Thus the given integral converges iff Z 1 dx p 0 x converges which it does. Next, we discuss applications of the integral. We have already established that it can be used to calculate area under curves but there are a few other applications that show up often on the math subject GRE; most prominently arc-length calculations and surfaces/volumes of revolution. We discuss the formulas for these briefly. Proposition 6 (Arc Length). Let f :[a; b] ! R be continuously differentiable. Then the arc-length of the graph (x; f(x)) between a and b is given by Z b L = p1 + f 0(x)2dx: a Christian Parkinson GRE Prep: Calculus II Notes 4 More generally, if x; y :[a; b] ! R2 are continuously differentiable, then the length of the curve parameterized by (x(t); y(t)) for t 2 [a; b] is given by Z b L = px0(t)2 + y0(t)2dt: a Proposition 7 (Surface / Volume of Revolution). Suppose that f :[a; b] ! R is continuous. If we were tor rotate the graph (x; f(x)) about the x-axis this will create a surface. The surface area of this surface is Z b A = 2π f(x)p1 + f 0(x)2dx a and the volume encapsulated by this surface is Z b V = π f(x)2dx: a These formulas are easy to derive when you have a good picture in your mind; otherwise you can use them as a black box. Note that they agree with our intuition. p Example 8. What is the length of the graph of f(x) = 1 − x2 for x 2 [−1; 1]? Solution. Intuitively, this graph is half of a unit circle so the length should be π. Using the formula, we see Z 1 r x2 Z 1 dx L = 1 + dx = p = arcsin(1) − arcsin(−1) = π=2 − (−π=2) = π; 2 2 −1 1 − x −1 1 − x where the antiderivative can easily be found using the trig. substitution x = sin(t). Example 9. Let f(x) = x for x 2 [0; 1] What surface area and volume result from rotating the graph of f about the x-axis? Solution. Intuitively, this will create a right, circular cone of \height" 1 and radius 1. You π 2 π may recall from geometry, the volume is 3 r h = 3 . According to our formula, Z 1 π V = π x2dx = : 0 3 Likewise, the surface areap (not includingp the \bottom" of the cone; i.e. the unit circle which forms the base) will be πr r2 + h2 = π 2 and our formula gives Z 1 p p A = 2π x 1 + 12 dx = π 2: 0 Note: much like the arc-length formula, the surface area and volume formulas can be adjusted to account for parameterized coordinates or to account for rotation around the Christian Parkinson GRE Prep: Calculus II Notes 5 y-axis or any other line ax + by = c. Also, you may be asked for the volume created when the area between two graphs (x; f(x)) and (x; g(x)) is rotated around the x-axis. This will be given by Z b 2 2 π f(x) − g(x) dx: a The reasoning should be somewhat clear from the geometry. From here we move onto sequences and series. Definition 10 (Sequence). A sequence is a map a : N ! R which assigns a real number to each natural number. The value that a assigns to n 2 N is typically denoted by an rather than a(n) though both may be used.
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