Integral Ch 7 - NCERT

INTEGRALSChapter2877INTEGRALSv Just as a mountaineer climbs a mountain – because it is there, soa good mathematics student studies new material becauseit is there. — JAMES B. BRISTOL v7.1 IntroductionDifferential Calculus is centred on the concept of thederivative. The original motivation for the derivative wasthe problem of defining tangent lines to the graphs offunctions and calculating the slope of such lines. IntegralCalculus is motivated by the problem of defining andcalculating the area of the region bounded by the graph ofthe functions.If a function f is differentiable in an interval I, i.e., itsderivative f ′ exists at each point of I, then a natural questionarises that given f ′ at each point of I, can we determinethe function? The functions that could possibly have givenG .W. Leibnitzfunction as a derivative are called anti derivatives (or(1646 -1716)primitive) of the function. Further, the formula that givesall these anti derivatives is called the indefinite integral of the function and suchprocess of finding anti derivatives is called integration. Such type of problems arise inmany practical situations. For instance, if we know the instantaneous velocity of anobject at any instant, then there arises a natural question, i.e., can we determine theposition of the object at any instant? There are several such practical and theoreticalsituations where the process of integration is involved. The development of integralcalculus 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 certainconditions.These two problems lead to the two forms of the integrals, e.g., indefinite anddefinite integrals, which together constitute the Integral Calculus.

288MATHEMATICSThere is a connection, known as the Fundamental Theorem of Calculus, betweenindefinite integral and definite integral which makes the definite integral as a practicaltool for science and engineering. The definite integral is also used to solve many interestingproblems from various disciplines like economics, finance and probability.In this Chapter, we shall confine ourselves to the study of indefinite and definiteintegrals and their elementary properties including some techniques of integration.7.2 Integration as an Inverse Process of DifferentiationIntegration 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 originalfunction. Such a process is called integration or anti differentiation.Let us consider the following examples:We know thatd(sin x) cos xdxd x3( ) x2dx 3and. (1). (2)d x( e ) ex. (3)dxWe observe that in (1), the function cos x is the derived function of sin x. We sayx3and3ex are the anti derivatives (or integrals) of x2 and ex, respectively. Again, we note thatfor any real number C, treated as constant function, its derivative is zero and hence, wecan write (1), (2) and (3) as follows :that sin x is an anti derivative (or an integral) of cos x. Similarly, in (2) and (3),dd xd x3(sin x C) cos x ,( C) x2 and(e C) exdxdx 3dxThus, anti derivatives (or integrals) of the above cited functions are not unique.Actually, there exist infinitely many anti derivatives of each of these functions whichcan be obtained by choosing C arbitrarily from the set of real numbers. For this reasonC is customarily referred to as arbitrary constant. In fact, C is the parameter byvarying which one gets different anti derivatives (or integrals) of the given function.dF (x ) f ( x) , x I (interval),More generally, if there is a function F such thatdxthen for any arbitrary real number C, (also called constant of integration)d[F (x) C] f (x), x Idx

INTEGRALS289{F C, C R} denotes a family of anti derivatives of f.Thus,Remark Functions with same derivatives differ by a constant. To show this, let g and hbe 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 Idf f′ g′ – h′ giving f′ (x) g′ (x) – h′ (x) x IdxThenorf ′ (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 ) dxwhich will represent the entireclass of anti derivatives read as the indefinite integral of f with respect to x.Symbolically, we writeNotation Given that f (x ) dx F (x) C .dy f (x ) , we write y dx f (x) dx .For the sake of convenience, we mention below the following symbols/terms/phraseswith their meanings as given in the Table (7.1).Table 7.1Symbols/Terms/PhrasesMeaning f (x ) dxIntegral of f with respect to xf (x) inx in f (x) dx f (x ) dxIntegrandVariable of integrationIntegrateFind the integralAn integral of fA function F such thatF′(x) f (x)The process of finding the integralIntegrationConstant of IntegrationAny real number C, considered asconstant function

290MATHEMATICSWe 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 belowwhich will be used to find integrals of other functions.DerivativesIntegrals (Anti derivatives)d xn 1 n x ;(i)dx n 1 n x dx x n 1 C , n –1n 1Particularly, we note thatd( x) 1 ;dxd(sin x) cos x ;(ii)dx(iii)d( – cos x ) sin x ;dxd( tan x) sec2 x ;dxd( – cot x ) cosec 2 x ;(v)dx(iv) dx x C cos x dx sin x C sin x dx – cos x C sec2x dx tan x C cosec2x dx – cot x C(vi)d(sec x ) sec x tan x ;dx sec x tan x dx sec x C(vii)d( – cosec x) cosec x cot x ;dx cosec x cot x dx – cosec x Cd1–1(viii) dx sin x ;1 – x2()d1–1(ix) dx – cos x ;1 – x2() dx1–x2dx1–x2 sin – 1 x C – cos(x)d1tan – 1 x ;dx1 x2 1 x2 tan(xi)d1– cot – 1 x ;dx1 x2 1 x2 – cot(())dxdx–1–1x Cx C–1x C

INTEGRALSd1–1(xii) dx sec x ;x x2 – 1 xdxd1–1(xiii) dx – cosec x ;x x2 – 1) xdxd x( e ) ex ;dxd1( log x ) ;(xv)dxx e(()(xiv)(xvi)d ax x a ;dx log a x291 sec– 1 x C2x –1 – cosec– 1 x C2x –1dx ex C1 x dx log x Cx a dx ax Clog aNote In practice, we normally do not mention the interval over which the variousAfunctions are defined. However, in any specific problem one has to keep it in mind.7.2.1 Geometrical interpretation of indefinite integral2Let f (x) 2x. Then f (x ) dx x C . For different values of C, we get differentintegrals. But these integrals are very similar geometrically.Thus, y x2 C, where C is arbitrary constant, represents a family of integrals. Byassigning different values to C, we get different members of the family. These togetherconstitute the indefinite integral. In this case, each integral represents a parabola withits axis along y-axis.Clearly, for C 0, we obtain y x2 , a parabola with its vertex on the origin. Thecurve y x2 1 for C 1 is obtained by shifting the parabola y x2 one unit alongy-axis in positive direction. For C – 1, y x2 – 1 is obtained by shifting the parabolay 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 fornegative values of C, each has its vertex along the negative side of the y-axis. Some ofthese 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 theparabolas y x2, y x2 1, y x2 2, y x2 – 1, y x2 – 2 at P0 , P1, P2, P–1 , P–2 etc.,dyrespectively, thenat these points equals 2a. This indicates that the tangents to thedx2curves at these points are parallel. Thus, 2 x dx x C FC ( x) (say), implies that

292MATHEMATICSFig 7.1the tangents to all the curves y F C (x), C R, at the points of intersection of thecurves 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 differentmembers of this family and these members can be obtained by shifting any one of thecurves parallel to itself. This is the geometrical interpretation of indefinite integral.7.2.2 Some properties of indefinite integralIn this sub section, we shall derive some properties of indefinite integrals.(I) The process of differentiation and integration are inverses of each other in thesense of the following results :df ( x) dx f (x)dx and f ′(x) dx f (x) C, where C is any arbitrary constant.

INTEGRALS293Proof Let F be any anti derivative of f, i.e.,dF(x) f (x)dx f (x ) dx F(x) CThenThereforeddx f (x) dx d( F ( x) C )dxdF (x ) f ( x)dxSimilarly, we note thatf ′(x) df ( x)dx f ′(x) dx f (x) Cand hencewhere C is arbitrary constant called constant of integration.(II) Two indefinite integrals with the same derivative lead to the same family ofcurves and so they are equivalent.Proof Let f and g be two functions such thatddf ( x) dx g (x ) dx dxdx ord f ( x) dx – g (x) dx 0 dx Hence f (x ) dx – g (x) dx or f (x ) dx g (x) dx C{ f (x) dx C , C R}{ g(x) dx C , C R} are identical.So the families of curvesandC, where C is any real numberHence, in this sense,2112 f (x ) dx and g (x) dx are equivalent.(Why?)

294MATHEMATICSA Note The equivalence of the families { f (x) dx C ,C R} and{ g(x) dx C ,C R} is customarily expressed by writing f (x ) dx g (x) dx ,1212without mentioning the parameter.(III) [ f (x) g (x )] dx f (x) dx g (x) dxProofBy Property (I), we haved [ f ( x) g (x )] dx f (x) g (x) dx On the otherhand, we find thatd df ( x) dx g ( x) dx dxdx. (1)d f (x) dx dx g (x) dx f (x) g (x)Thus, in view of Property (II), it follows by (1) and (2) that. (2) ( f (x) g (x) ) dx f (x ) dx g (x) dx .(IV) For any real number k, k f ( x) dx k f ( x) dxProof By the Property (I),Alsod kdx dk f ( x) dx k f (x) .dx f (x) dx kddx f (x) dx kTherefore, using the Property (II), we have kf ( x)f ( x) dx k f (x) dx .(V) Properties (III) and (IV) can be generalised to a finite number of functionsf1, f2 , ., fn and the real numbers, k1, k2, ., kn giving [ k1 f1 (x ) k2 f2 (x) . kn f n (x)] dx k1 f1 ( x) dx k2 f 2 ( x) dx . kn fn (x) dx .To find an anti derivative of a given function, we search intuitively for a functionwhose derivative is the given function. The search for the requisite function for findingan anti derivative is known as integration by the method of inspection. We illustrate itthrough some examples.

INTEGRALS295Example 1 Write an anti derivative for each of the following functions using themethod of inspection:(i) cos 2x(ii) 3x2 4x3(iii)1,x 0xSolution(i) We look for a function whose derivative is cos 2x. Recall thatdsin 2x 2 cos 2xdxorcos 2x d 11 d (sin 2x) sin 2 x dx 22 dx 1sin 2 x .2(ii) We look for a function whose derivative is 3x2 4x3 . Note thatTherefore, an anti derivative of cos 2x isd 3x x 4 3x2 4x3.dxTherefore, an anti derivative of 3x2 4x3 is x3 x4 .(iii) We know that()d1d11(log x) , x 0 and [log ( – x)] ( – 1) , x 0dxxdx–xxCombining above, we getTherefore,1 x dx log xd1log x ) , x 0(dxxis one of the anti derivatives of1.xExample 2 Find the following integrals:(i) x3 – 1dxx2(ii) 2(x 33 1) dx(iii)1x (x 2 2 e – x ) dxSolution(i) We have x3 – 1dx x dx – x– 2 dx2x(by Property V)