6 Fundamental Theorems, Substitution, In-tegration by Parts, and Polar CoordinatesSo far we have separately learnt the basics of integration and differentiation.But they are not unrelated. In fact, they are inverse operations. Thisis what we will try to explore in the first section, via the two fundamentaltheorems of Calculus. After that we will discuss the two main methods oneuses for integrating somewhat complicated functions, namely integrationby substitution and integration by parts. The final section will discussintegration in polar coordinates, which comes up when there is radialsymmetry.6.1 The fundamental theoremsSuppose f is an integrable function on a closed interval [a, b]. Then we canconsider the signed area function A on [a, b] (relative to f) defined by thedefinite integral of f from a to x, i.e.,(6.1.1) A(x) =x∫af(t)dt.The reason for the signed area terminology is that f is not assumed to be≥ 0, so a priori A(x) could be negative.It is extremely interesting to know how A(x) varies with x. What condi-tions does one need to put on f to make sure that A is continuous, or evendifferentiable? The continuity part of the question is easy to answer.Lemma 6.1 Let f, A be as above. Then A is a continuous function on [a, b].Proof. Let c be any point in [a, b]. Then f is continuous at c iff wehavelimh→0A(c + h) = A(c).Of course, in taking the limit, we consider all small enough h for which c + hlies in [a, b], and then let h go to zero. By the additivity of the integral, we1have (using (6.1.1)),A(c + h) − A(c) =∫I(c,h)f(t)dt,where I(c, h) denotes the closed interval between c and c + h. Clearly, I(c, h)is [c, c + h], resp. [c + h, c], if h is positive, resp. negative. When h goes tozero, I(c, h) shrinks to the point {c}, and solimh→0A(c + h) − A(c) = 0,which is what we needed to show.2Remark 6.1.2: The general moral to remember is that, just as in real life,Integration is good and Differentiation is bad !Indeed, as seen in this Lemma, integration makes functions better; here ittakes an integrable, but not necessarily a continuous, function f, and fromit obtains a continuous function. The Theorem below says that if f is inaddition continuous, then its integral is even differentiable. Differentiation,on the other hand, makes functions worse. The derivative of a differentiablefunction f is often not differentiable (think of f(x) = sign(x)x2at x = 0); infact, f′need not even be continuous (think of f(x) = x2sin(1/x) when x = 0and = 0 when x = 0).A satisfactory answer to the question of differentiability of the integral isgiven by the following important result, which comes with an appropriatelyhonorific title:Theorem 6.2 (The first fundamental theorem of Calculus) Let f bean integrable function on [a, b], and let A be the function defined by (6.1.1).Pick any point c in (a, b), and suppose that f is continuous at c. Then A isdifferentiable there and moreover,A′(c) = f(c).Some would write this symbolically as(6.1.3)ddxx∫af(t)dt = f(x).2In plain words, this says that differentiating the integral gives back theoriginal as long as the original function is continuous at the point in question.Proof. To know if A(x) is differentiable at c, we need to evaluate thelimit(6.1.4) L = limh→0A(c + h) − A(c)h.By the additivity of the definite integral, we have(6.1.5) A(c + h) − A(c) =∫I(c,h)f(x)dx,where I(c, h) is as in the proof of Lemma 6.1.Denote by M(c, h), resp. m(c, h), the supremum, resp. infimum, of thevalues of f over I(c, h ). Then the following bounds evidently hold:(6.1.6) hm(c, h) ≤∫I(c,h)f(x)dx ≤ hM(c, h).Combining (6.1.4), (6.1.5) and (6.1.6), we get for all small h,(6.1.7) limh→0m(c, h) ≤ L ≤ limh→0M(c, h).But by hypothesis, f is continuous at c. Then both m(c, h) and M(c, h)will tend to f(c) as h goes to 0, which proves the Theorem in view of (6.1.7)and the squeeze theorem.(Draw pictures to convince yourselves that if f is not continuous, thenthese limits, even if they exist, need not equal f(c).)2Let f be any function on an open interval I. Suppose there is a differ-entiable function ϕ on I such that ϕ′(x) = f(x) for all x in I. Then we willcall ϕ a primitive of f on I. Note that the primitive is not unique. Indeed,for any constant α, the function ϕ + α will have the same derivative as ϕ.Intuitively, one feels immediately that the notion of a primitive should betied up with the notion of an integral. The following very important andoft-used result makes this expected relationship precise.3Theorem 6.3 (The second fundamental theorem of Calculus) Sup-pose f, ϕ are functions on [a, b], with f integrable on [a, b] and ϕ a primitiveof f on (a, b), with ϕ defined and continuous at the endpoints a, b. Thenϕ(b) − ϕ(a) =b∫af(x)dx.One can rewrite this, perhaps more expressively, asϕ(b) − ϕ(a) =b∫addxϕdx.Proof. Choose any partitionP : a = t0< t1< . . . < tn= b,and set, for each j ∈ {1, 2, . . . , n},(6.1.8) Mj= sup(f([tj−1, tj])) and mj= inf(f([tj−1, tj])).By definition,(6.1.9)n∑j=1(tj− tj−1)mj≤b∫af(x)dx ≤n∑j=1(tj− tj−1)Mj.On the other hand, the Mean Value Theorem gives us, for each j, anumber cjin [tj−1, tj] such that(6.1.10) ϕ′(cj) =ϕ(tj) − ϕ(tj−1)tj− tj−1.Since ϕ is by hypothesis the primitive of f on (a, b), f(cj) = ϕ′(cj) for eachj. Moreover,(6.1.11) mj≤ f(cj) ≤ Mj,and(6.1.12)n∑j=1ϕ(tj) − ϕ(tj−1) = ϕ(b) − ϕ(a).4Combining (6.1.10), (6.1.11) and (6.1.12), we obtain(6.1.13)n∑j=1(tj− tj−1)mj≤ ϕ(b) −ϕ(a) ≤n∑j=1(tj− tj−1)Mj.Since (6.1.9) and (6.1.13) hold for every partition P , and since f is inte-grable on [a, b], the assertion of the Theorem follows.26.2 The indefinite integralSuppose ϕ is a primitive of a function f on an open interval I, i.e., whichyields f back upon differentiation. It is not unusual to set, following Leibnitz,(6.2.1)∫f(x)dx = ϕ(x).This is called an indefinite integral because there are no limits and ϕ isnon-unique. So one can think of such an indefinite integral as a function ofx which is unique only up to addition of an arbitrary constant. One has, inother words, an equality for all scalars C∫f(x)dx =∫f(x)dx + C.It could be a bit unsettling to work with such an indefinite, nebulous functionat first, but one learns soon enough that it is a useful concept to be awareof.In many Calculus texts one finds formulas like∫cos xdx = sin x + Cand∫1xdx = log x + C .All they mean is that