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Math 222 Second Semester Calculus MATH 222 SECOND SEMESTER CALCULUS Spring 2011 1 2 Math 222 – 2nd Semester Calculus Lecture notes version 1.7(Spring 2011) This is a self contained set of lecture notes for Math 222. The notes were written by Sigurd Angenent, starting from an extensive collection of notes and problems compiled by Joel Robbin. Some problems were contributed by A.Miller. The LATEX files, as well as the Xfig and Octave files which were used to produce these notes are available at the following web site www.math.wisc.edu/~angenent/Free-Lecture-Notes They are meant to be freely available for non-commercial use, in the sense that “free software” is free. More precisely: Copyright (c) 2006 Sigurd B. Angenent. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ”GNU Free Documentation License”. Contents 3 Chapter 1: Methods of Integration 1. The indefinite integral We recall some facts about integration from first semester calculus. 1.1. Definition. A function y = F (x) is called an antiderivative of another function y = f(x) if F ′(x)= f(x) for all x. 2 1.2. Example. F1(x)= x is an antiderivative of f(x)=2x. 2 F2(x)= x + 2004 is also an antiderivative of f(x)=2x. 1 G(t)= 2 sin(2t + 1) is an antiderivative of g(t) = cos(2t + 1). The Fundamental Theorem of Calculus states that if a function y = f(x) is continuous on an interval a x b, then there always exists an antiderivative F (x) of f, and one has ≤ ≤ b (1) f(x) dx = F (b) F (a). Za − The best way of computing an integral is often to find an antiderivative F of the given function f, and then to use the Fundamental Theorem (1). How you go about finding an antiderivative F for some given function f is the subject of this chapter. The following notation is commonly used for antiderivates: (2) F (x)= f(x)dx. Z The integral which appears here does not have the integration bounds a and b. It is called an indefinite integral, as opposed to the integral in (1) which is called a definite integral. It’s important to distinguish between the two kinds of integrals. Here is a list of differences: Indefinite integral Definite integral b f(x)dx is a function of x. a f(x)dx is a number. R b R By definition f(x)dx is any func- a f(x)dx was defined in terms of tion of x whoseR derivative is f(x). RRiemann sums and can be inter- preted as “area under the graph of y = f(x)”, at least when f(x) > 0. x is not a dummy variable, for exam- x is a dummy variable, for example, ple, 2xdx = x2 + C and 2tdt = 1 1 0 2xdx = 1, and 0 2tdt = 1, so t2 +CRare functions of diffferentR vari- R 1 1 R 0 2xdx = 0 2tdt. ables, so they are not equal. R R 4 2. You can always check the answer Suppose you want to find an antiderivative of a given function f(x) and after a long and messy computation which you don’t really trust you get an “answer”, F (x). You can then throw away the dubious computation and differentiate the F (x) you had found. If F ′(x) turns out to be equal to f(x), then your F (x) is indeed an antiderivative and your computation isn’t important anymore. 2.1. Example. Suppose we want to find ln x dx. My cousin Bruce says it might be F (x)= x ln x x. Let’s see if he’s right:R − d 1 (x ln x x)= x +1 ln x 1 = ln x. dx − · x · − Who knows how Bruce thought of this1, but he’s right! We now know that ln xdx = x ln x x + C. − R 3. About “+C” Let f(x) be a function defined on some interval a x b. If F (x) is an ≤ ≤ antiderivative of f(x) on this interval, then for any constant C the function F˜(x)= F (x)+ C will also be an antiderivative of f(x). So one given function f(x) has many different antiderivatives, obtained by adding different constants to one given antiderivative. 3.1. Theorem. If F1(x) and F2(x) are antiderivatives of the same function f(x) on some interval a x b, then there is a constant C such that F (x) = ≤ ≤ 1 F2(x)+ C. Proof. ′ ′ Consider the difference G(x)= F1(x) F2(x). Then G (x)= F1(x) ′ − − F2(x)= f(x) f(x) = 0, so that G(x) must be constant. Hence F1(x) F2(x)= C for some constant.− − It follows that there is some ambiguity in the notation f(x) dx. Two functions F1(x) and F2(x) can both equal f(x) dx without equalingR each other. When this happens, they (F1 and F2) differR by a constant. This can sometimes lead to confusing situations, e.g. you can check that 2 sin x cos x dx = sin2 x Z 2 sin x cos x dx = cos2 x Z − are both correct. (Just differentiate the two functions sin2 x and cos2 x!) These − two answers look different until you realize that because of the trig identity sin2 x+ cos2 x = 1 they really only differ by a constant: sin2 x = cos2 x + 1. − To avoid this kind of confusion we will from now on never forget to include the “arbitrary constant +C” in our answer when we compute an antideriv- ative. 1He integrated by parts. 5 4. Standard Integrals Here is a list of the standard derivatives and hence the standard integrals everyone should know. f(x) dx = F (x)+ C Z xn+1 xn dx = + C for all n = 1 Z n +1 6 − 1 dx = ln x + C Z x | | sin x dx = cos x + C Z − cos x dx = sin x + C Z tan x dx = lncos x + C Z − 1 dx = arctan x + C Z 1+ x2 1 π dx = arcsin x + C (= arccos x + C) Z √1 x2 2 − − dx 1 1 + sin x π π = ln + C for <x< . Z cos x 2 1 sin x − 2 2 − All of these integrals are familiar from first semester calculus (like Math 221), except a for the last one. You can check the last one by differentiation (using ln b = ln a ln b simplifies things a bit). − 5. Method of substitution The chain rule says that dF (G(x)) ′ ′ = F (G(x)) G (x), dx · so that ′ ′ F (G(x)) G (x) dx = F (G(x)) + C. Z · 5.1. Example. Consider the function f(x)= 2x sin(x2 +3). It does not appear in the list of standard integrals we know by heart. But we do notice2 that 2x = d (x2 + 3). So let’s call G(x)= x2 + 3, and F (u)= cos u, then dx − F (G(x)) = cos(x2 + 3) − and dF (G(x)) = sin(x2 + 3) 2x = f(x), dx · ′ F ′(G(x)) G (x) | {z } |{z} 2 You will start noticing things like this after doing several examples. 6 so that 2x sin(x2 +3)dx = cos(x2 +3)+ C. Z − The most transparent way of computing an integral by substitution is by in- troducing new variables. Thus to do the integral ′ f(G(x))G (x) dx Z where f(u) = F ′(u), we introduce the substitution u = G(x), and agree to write du = dG(x)= G′(x) dx. Then we get ′ f(G(x))G (x) dx = f(u) du = F (u)+ C. Z Z At the end of the integration we must remember that u really stands for G(x), so that ′ f(G(x))G (x) dx = F (u)+ C = F (G(x)) + C. Z For definite integrals this implies b ′ f(G(x))G (x) dx = F (G(b)) F (G(a)). Za − which you can also write as b G(b) ′ (3) f(G(x))G (x) dx = f(u) du. Za ZG(a) 5.2. Example. [Substitution in a definite integral. ] As an example we com- pute 1 x 2 dx, Z0 1+ x using the substitution u = G(x) =1+ x2. Since du = 2x dx, the associated indefinite integral is 1 1 x dx = 1 du. Z 1+ x2 2 Z u 1 du 1 2 u |{z} | {z } To find the definite integral you must compute the new integration bounds G(0) and G(1) (see equation (3).) If x runs between x = 0 and x = 1, then u = G(x)=1+x2 runs between u =1+02 = 1 and u =1+12 = 2, so the definite integral we must compute is 1 2 x 1 1 2 dx = 2 du, Z0 1+ x Z1 u which is in our list of memorable integrals. So we find 1 2 x 1 2 dx = 1 du = 1 ln u = 1 ln 2. Z 1+ x2 2 Z u 2 1 2 0 1 7 6. The double angle trick If an integral contains sin2 x or cos2 x, then you can remove the squares by using the double angle formulas from trigonometry. Recall that cos2 α sin2 α = cos2α and cos2 α + sin2 α =1, − Adding these two equations gives 1 cos2 α = (cos2α + 1) 2 while substracting them gives 1 sin2 α = (1 cos2α) .
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