INTEGRALS 287 7 Chapter INTEGRALS Just as a mountaineer climbs a mountain – because it is there, so a good mathematics student studies new material because it is there. — JAMES B. BRISTOL  7.1 Introduction Differential Calculus is centred on the concept of the derivative. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions. If a function f is differentiable in an interval I, i.e., its derivative f ′exists at each point of I, then a natural question arises that given f ′at each point of I, can we determine the function? The functions that could possibly have given function as a derivative are called anti derivatives (or G .W. Leibnitz primitive) of the function. Further, the formula that gives (1646-1716) all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration. Such type of problems arise in many practical situations. For instance, if we know the instantaneous velocity of an object at any instant, then there arises a natural question, i.e., can we determine the position of the object at any instant? There are several such practical and theoretical situations where the process of integration is involved. The development of integral calculus arises out of the efforts of solving the problems of the following types: (a) the problem of finding a function whenever its derivative is given, (b) the problem of finding the area bounded by the graph of a function under certain conditions. These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus. 2019-20 288 MATHEMATICS There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability. In this Chapter, we shall confine ourselves to the study of indefinite and definite integrals and their elementary properties including some techniques of integration. 7.2 Integration as an Inverse Process of Differentiation Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti differentiation. Let us consider the following examples: d We know that (sin x) = cos x ... (1) dx d x3 ( ) = x2 ... (2) dx 3 d and (ex)=ex ... (3) dx We observe that in (1), the function cos x is the derived function of sin x. We say x3 that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3), and 3 ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note that for any real number C, treated as constant function, its derivative is zero and hence, we can write (1), (2) and (3) as follows : d d x3 d (sin x+C)=cosx, ( +C)=x2and (ex +C)=ex dx dx 3 dx Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by choosing C arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. d More generally, if there is a function F such that F(x)= f (x), ∀x ∈ I (interval), dx then for any arbitrary real number C, (also called constant of integration) d [F(x)+C] = f(x), x ∈ I dx 2019-20 INTEGRALS 289 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Remark Functions with same derivatives differ by a constant. To show this, let g and h be two functions having the same derivatives on an interval I. Consider the function f = g – h defined by f(x) = g(x) – h(x), ∀x ∈ I df Then =f′ = g′ – h′ giving f′ (x) = g′ (x) – h′ (x) ∀x ∈ I dx or f′ (x) = 0, ∀x ∈ I by hypothesis, i.e., the rate of change of f with respect to x is zero on I and hence f is constant. In view of the above remark, it is justified to infer that the family {F + C, C ∈ R} provides all possible anti derivatives of f. We introduce a new symbol, namely, ∫ f(x)dx which will represent the entire class of anti derivatives read as the indefinite integral of f with respect to x. Symbolically, we write ∫ f(x)dx=F(x)+C. dy Notation Given that = f (x), we write y = ∫ f (x)dx. dx For the sake of convenience, we mention below the following symbols/terms/phrases with their meanings as given in the Table (7.1). Table 7.1 Symbols/Terms/Phrases Meaning ∫ f(x)dx Integral of f with respect to x f(x) in ∫ f(x)dx Integrand x in ∫ f(x)dx Variable of integration Integrate Find the integral An integral of f A function F such that F′(x) = f (x) Integration The process of finding the integral Constant of Integration Any real number C, considered as constant function 2019-20 290 MATHEMATICS We already know the formulae for the derivatives of many important functions. From these formulae, we can write down immediately the corresponding formulae (referred to as standard formulae) for the integrals of these functions, as listed below which will be used to find integrals of other functions. Derivatives Integrals (Anti derivatives) (i) d xn+1=xn ; ∫xn dx= x n+1 +C, n ≠ –1 dx n+1 n+1 Particularly, we note that d (x)=1 ; ∫dx= x+C dx d (ii) (sin x)=cos x ; ∫cosxdx=sin x+C dx d (iii) (–cosx)=sin x ; ∫sin xdx=–cos x+C dx d (iv) (tan x)=sec2x ; ∫sec2 xdx=tan x+C dx d (v) (–cot x)=cosec2x ; ∫cosec2 xdx=–cot x+C dx d (vi) (secx)=secxtan x ; ∫sec xtanxdx=secx+C dx d (vii) (–cosecx)=cosecxcot x ; ∫cosecxcotxdx=–cosec x+C dx d 1 dx (sin–1 x)= ∫ = sin–1 x+C (viii) ; dx 1– x2 1– x2 d 1 dx (–cos–1 x)= ∫ =–cos–1 x+C (ix) ; dx 1– x2 1– x2 d 1 dx (x) (tan–1 x)= ; ∫ =tan–1 x+C dx 1+ x2 1+ x2 d 1 dx (xi) (–cot–1 x)= ; ∫ =–cot–1 x+C dx 1+ x2 1+ x2 2019-20 INTEGRALS 291 d 1 dx (sec–1 x)= ∫ =sec–1 x+C (xii) ; dx x x2 –1 x x2 –1 d 1 dx (–cosec–1 x)= ∫ =–cosec–1x+C (xiii) ; dx x x2 –1 x x2 –1 d (xiv) (ex)=ex ; ∫exdx=ex +C dx d 1 1 (xv) (log|x|)= ; ∫ dx=log|x|+C dx x x d  ax  ax (xvi)  =ax ; ∫axdx= +C dx log a log a   Note In practice, we normally do not mention the interval over which the various functions are defined. However, in any specific problem one has to keep it in mind. 7.2.1 Geometrical interpretation of indefinite integral Let f(x) = 2x. Then ∫ f(x)dx= x2 +C. For different values of C, we get different integrals. But these integrals are very similar geometrically. Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals. By assigning different values to C, we get different members of the family. These together constitute the indefinite integral. In this case, each integral represents a parabola with its axis along y-axis. Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin. The curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along y-axis in positive direction. For C = – 1, y = x2 – 1 is obtained by shifting the parabola y = x2 one unit along y-axis in the negative direction. Thus, for each positive value of C, each parabola of the family has its vertex on the positive side of the y-axis and for negative values of C, each has its vertex along the negative side of the y-axis. Some of these have been shown in the Fig 7.1. Let us consider the intersection of all these parabolas by a line x = a. In the Fig 7.1, we have taken a > 0. The same is true when a < 0. If the line x = a intersects the parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P , P , P , P , P etc., 0 1 2 –1 –2 dy respectively, then at these points equals 2a. This indicates that the tangents to the dx curves at these points are parallel. Thus, ∫2xdx=x2 +C=F (x)(say), implies that C 2019-20 292 MATHEMATICS Fig 7.1 the tangents to all the curves y = F (x), C ∈ R, at the points of intersection of the C curves by the line x = a, (a ∈ R), are parallel. Further, the following equation (statement)∫ f(x)dx=F(x)+C= y(say), represents a family of curves. The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. This is the geometrical interpretation of indefinite integral. 7.2.2 Some properties of indefinite integral In this sub section, we shall derive some properties of indefinite integrals. (I) The process of differentiation and integration are inverses of each other in the sense of the following results : d ∫ f(x)dx = f(x) dx and ∫ f′(x)dx = f(x) + C, where C is any arbitrary constant. 2019-20 INTEGRALS 293 Proof Let F be any anti derivative of f, i.e., d F(x) = f(x) dx Then ∫ f(x)dx = F(x) + C d d Therefore ∫ f(x)dx = (F(x)+C) dx dx d = F(x)= f(x) dx Similarly, we note that d f′(x) = f(x) dx and hence ∫ f′(x)dx = f(x) + C where C is arbitrary constant called constant of integration. (II) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. Proof Let f and g be two functions such that d d ∫ f(x)dx = ∫g(x)dx dx dx d or ∫ f(x)dx–∫g(x)dx = 0 dx  Hence ∫ f(x)dx –∫g(x)dx= C, where C is any real number (Why?) or ∫ f(x)dx = ∫g(x)dx+C { } So the families of curves ∫ f(x)dx+C ,C ∈R 1 1 { } and ∫g(x)dx+C ,C ∈R are identical. 2 2 Hence, in this sense, ∫ f(x)dxand∫g(x)dx are equivalent. 2019-20 294 MATHEMATICS  Note The equivalence of the families {∫ f(x)dx+C ,C ∈R} and 1 1 { } ∫g(x)dx+C ,C ∈R is customarily expressed by writing ∫ f(x)dx=∫g(x)dx, 2 2 without mentioning the parameter. (III) ∫[f(x)+g(x)]dx=∫ f(x)dx+∫g(x)dx Proof By Property (I), we have d ∫[ f(x)+g(x)]dx = f(x) + g(x) ... (1) dx  On the otherhand, we find that d d d ∫ f(x)dx+∫g(x)dx = ∫ f(x)dx+ ∫g(x)dx dx   dx dx =f(x) + g(x) ... (2) Thus, in view of Property (II), it follows by (1) and (2) that ∫(f(x)+ g(x))dx= ∫ f(x)dx+∫g(x)dx. (IV) For any real number k, ∫k f(x)dx=k ∫ f(x)dx d Proof By the Property (I), ∫k f(x)dx=k f(x). dx d d Also k ∫ f(x)dx = k ∫ f(x)dx=k f(x) dx   dx Therefore, using the Property (II), we have ∫k f(x)dx=k ∫ f(x)dx. (V) Properties (III) and (IV) can be generalised to a finite number of functions f , f , ..., f and the real numbers, k , k , ..., k giving 1 2 n 1 2 n ∫[k f (x)+k f (x)+...+k f (x)]dx 1 1 2 2 n n = k ∫ f (x)dx+k ∫ f (x)dx+...+k ∫ f (x)dx. 1 1 2 2 n n To find an anti derivative of a given function, we search intuitively for a function whose derivative is the given function. The search for the requisite function for finding an anti derivative is known as integration by the method of inspection. We illustrate it through some examples. 2019-20 INTEGRALS 295 Example 1 Write an anti derivative for each of the following functions using the method of inspection: 1 (i) cos 2x (ii) 3x2 + 4x3 (iii) , x ≠ 0 x Solution (i) We look for a function whose derivative is cos 2x. Recall that d sin 2x =2 cos 2x dx 1 d d 1  or cos 2x = (sin 2x) =  sin2x 2 dx dx 2  1 Therefore, an anti derivative of cos 2x is sin2x. 2 (ii) We look for a function whose derivative is 3x2 + 4x3. Note that d (x3 +x4)=3x2 + 4x3. dx Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4. (iii) We know that d 1 d 1 1 (log x)= ,x>0and [log(–x)]= (–1)= ,x<0 dx x dx – x x d 1 Combining above, we get (log x)= ,x≠0 dx x 1 1 Therefore, ∫ dx=log x is one of the anti derivatives of . x x Example 2 Find the following integrals: (i) ∫x3 –1dx (ii) ∫(x23 +1)dx (iii) ∫ (x32 +2ex – 1)dx x2 x Solution (i) We have x3 –1 ∫ dx= ∫xdx– ∫x–2 dx (by Property V) x2 2019-20 296 MATHEMATICS x1+1   x–2+1  = 1+1 +C1––2+1+C2; C1, C2 are constants of integration x2 x–1 x2 1 = +C – –C = + +C –C 2 1 –1 2 2 x 1 2 x2 1 = + +C, where C = C – C is another constant of integration. 2 x 1 2 Note From now onwards, we shall write only one constant of integration in the final answer. (ii) We have 2 2 ∫(x3 +1)dx= ∫x3 dx+ ∫dx 2 +1 x3 3 5 = + x+C = x3 + x+C 2 5 +1 3 3 3 1 1 (iii) We have ∫(x2 +2ex – )dx= ∫x2 dx+ ∫2ex dx– ∫ dx x x 3 +1 x2 = +2ex –log x +C 3 +1 2 5 2 = x 2 +2ex –log x +C 5 Example 3 Find the following integrals: (i) ∫(sin x+cos x)dx (ii)∫cosecx(cosec x+cot x)dx 1–sin x (iii) ∫ dx cos2 x Solution (i) We have ∫(sin x+cosx)dx=∫sin xdx+ ∫cosxdx = –cosx+sin x+C 2019-20