DEFINITE INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS, ANTIDERIVATIVES

Transcription

1 DEFINITE INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS, ANTIDERIVATIVES MATH 152, SECTION 55 (VIPUL NAIK) Corresponding mteril in the book: Section 5.3, 5.4 Difficulty level: Hrd. Wht students should definitely get: Some results leding to nd including the fundmentl theorem of integrl clculus, the definition of ntiderivtive nd how to clculte ntiderivtives for polynomils nd the sine nd cosine functions. Wht students should hopefully get: The intuition behind the wy differentition nd integrtion relte; the concept of indetermincy up to constnts when we integrte. The reson for mking ssumptions such s continuity. Executive summry.1. Definite integrl, ntiderivtive, nd indefinite integrl. Words.. (1) We hve f(x) dx + c b f(x) dx = c f(x) dx. (2) We sy tht F is n ntiderivtive for f if F = f. (3) For continuous function f defined on closed intervl [, b], nd for point c [, b], the function F given by F (x) = x f(t) dt is n ntiderivtive for f. c (4) If f is continuous on [, b] nd F is function continuous on [, b] such tht F = f on (, b), then f(x) dx = F (b) F (). (5) The two results bove essentilly stte tht differentition nd integrtion re opposite opertions. (6) For function f on n intervl [, b], if F nd G re ntiderivtives, then F G is constnt on [, b]. Conversely, if F is n ntiderivtive of f, so is F plus ny constnt. (7) The indefinite integrl of function f is the collection of ll ntiderivtives for the function. This is typiclly written by writing one ntiderivtive plus C, where C is n rbitrry constnt. We write f(x) dx for the indefinite integrl. Note tht there re no upper nd lower limits. (8) Both the definite nd the indefinite integrl re dditive. In other words, f(x) dx + g(x) dx = f(x) + g(x) dx. The nlogue holds for definite integrls, with limits. (9) We cn lso pull constnts multiplictively out of integrls. Actions... (1) To do definite integrl, find ny one ntiderivtive nd evlute it between limits. (2) An importnt cvet: when using ntiderivtives to do definite integrl, it is importnt to mke sure tht the ntiderivtive is defined nd continuous everywhere on the intervl of integrtion. (Think of the 1/x 3 exmple). (3) To do n indefinite integrl, find ny ntiderivtive nd put +C. (4) To find n ntiderivtive, use the dditive splitting nd pulling constnts out, nd the fct tht x r dx = x r+1 /(r + 1)..2. Higher derivtives, multiple integrls, nd initil/boundry conditions. Actions... (1) The simplest kind of initil vlue problem ( notion we will encounter gin when we study differentil equtions) is s follows. The k th derivtive of function is given on the entire domin. Next, the vlues of the function nd the first k 1 derivtives re given t single point of the domin. We cn use this dt to find the function. Step by step, we find derivtives of lower orders. First, we integrte the k th derivtive to get tht the (k 1) th derivtive is of the form F (x) + C, where C is unknown. We now use the vlue of the (k 1) th derivtive t the given point to find C. Now, we hve the (k 1) th derivtive. We proceed now to find the (k 2) th derivtive, nd so on. 1

2 (2) Sometimes, we my be interested in finding ll functions with given second derivtive f. For this, we hve to perform n indefinite integrtion twice. The net result will be generl expression of the form F (x) + C 1 x + C 2, where F is function with F = f, nd C 1 nd C 2 re rbitrry constnts. In other words, we now hve up to constnts or liner functions insted of up to constnts s our degree of mbiguity. (3) More generlly, if the k th derivtive of function is given, the function is uniquely determined up to dditive differences of polynomils of degree strictly less thn k. The number of free constnts tht cn tke rbitrry rel vlues is k (nmely, the coefficients of the polynomil). (4) This generl expression is useful if, insted of n initil vlue problem, we hve boundry vlue problem. Suppose we re given G s function, nd we re given the vlue of G t two points. We cn then first find the generl expression for G s F + C 1 x + C 2. Next, we plug in the vlues to get system of two liner equtions, tht we solve in order to determine C 1 nd C 2, nd hence G. 1. Sttements of min results 1.1. The definite integrl: recll nd more detils. Recll from lst time tht for continuous (or piecewise continuous where ll discontinuities re jump discontinuities) function f on n open intervl [, b], the integrl of f over the intervl [, b], denoted: f(x) dx is kind of summtion for f for ll the rel numbers from to b. This integrl is lso clled definite integrl. The function is often termed the integrnd. The number b is termed the upper limit of the integrtion, nd the number is termed the lower limit of the integrtion. The vrible x is termed the vrible of integrtion. So fr, we hve mde sense of the expression s described bove with < b. We now dd in few detils on how to mke sense of two other possibilities: If = b, then, by definition, the integrl is defined to be zero. If > b, then, by definition, the integrl is defined s the negtive of the integrl f(x) dx. b Now, it mkes sense to consider the symbol f(x) dx without ny ordering conditions on nd b. With these definitions, we hve, for ny, b, c R: f(x) dx + c b f(x) dx = c f(x) dx 1.2. Definite integrls do exist for piecewise continuous functions. It is useful to know the following: (1) For continuous function f on closed nd bounded intervl [, b], the integrl exists nd is finite. In fct, the integrl over the intervl [, b] is bounded from bove by (b ) times the mximum vlue of the function nd from below by (b ) times the minimum vlue of the function. (Both of these exist by the extreme vlue theorem). (2) For piecewise continuous function f on closed nd bounded intervl [, b] such tht ll the onesided limits exist nd re finite t points of discontinuity, the integrl exists nd is finite. This follows from the previous prt, vi the intermedite step of breking [, b] into prts such tht the restriction of the function to ech prt is continuous nd extends continuously to the boundry of tht prt The definite integrl nd differentition. There is lso cler reltionship between the definite integrl nd differentition. In some sense, the integrl nd derivtive re inverses (opposites) of ech other. Let [, b] be n intervl. Suppose f is continuous function on [, b] nd c [, b] is ny number. Define the following function F on [, b]: F (x) := x c f(t) dt Note the wy the function is defined. t is the vrible of integrtion, nd F depends on x in the sense tht the upper limit of the intervl of integrtion is x, wheres the lower limit is fixed t c. 2

3 Continuous functions re integrble, s discussed bove, so F turns out to be well-defined. Further, F is continuous on [, b], differentible on (, b), nd hs derivtive This is Theorem F (x) = f(x) for ll x (, b) 1.4. Concept of ntiderivtive. Suppose f is continuous on [, b]. An ntiderivtive for f, or primitive for f, or indefinite integrl for f, is function G on [, b] such tht: G is continuous on [, b]. G (x) = f(x) for ll x (, b). A little while bck, we hd seen the following result: if F, G re functions on n intervl I such tht F = G for ll points in the interior of I, then F G is constnt function on I. In other words, F nd G differ by constnt. Conversely, if F nd G re functions on n intervl I with F differentible on the interior of I, nd F G is constnt, then G is lso differentible on I nd F = G on the interior of I. Thus, the ntiderivtive of function is not unique we cn lwys dd constnt function to one ntiderivtive to obtin nother ntiderivtive. However, the ntiderivtive is unique up to differing by constnts. In other words, ny two ntiderivtives differ by constnt. We re now in position to stte the fundmentl theorem of clculus. Suppose f is continuous function on the intervl [, b]. If G is n ntiderivtive for f on [, b], then we hve: f(t) dt = G(b) G() Also, s we lredy noted, ny two ntiderivtives differ by constnt, so if we replce G by nother ntiderivtive, the right side remins the sme becuse both G() nd G(b) get shifted by the sme mount. For nottionl convenience, this is sometimes written s: f(t) dt = [G(t)] b Here, the right side is interpreted s the difference between the vlues of G(t) for t = b nd t =, which simplifies to G(b) G(). 2. Computing ntiderivtives nd integrls: esy fcts 2.1. Computing some ntiderivtives. We now compute some common expressions for ntiderivtives of functions. (1) If f(x) = x r, nd r 1, then we cn set G(x) = x r+1 /(r + 1). The fctor of 1/(r + 1) is intended to cncel the fctor of r + 1 tht ppers s coefficient when we differentite x r+1. In prticulr, if f(x) = x, G(x) = x 2 /2, nd if f(x) = x 2, G(x) = x 3 /3. Most importntly, if f(x) = 1, then G(x) = x. (2) An ntiderivtive for sin is cos nd n ntiderivtive for cos is sin. Note the sign differences between these formuls nd those for the derivtive. The derivtive of sin is cos but the ntiderivtive of sin is cos. The derivtive of cos is sin nd the ntiderivtive of cos is sin. (We will see more trigonometric ntiderivtives lter) Linerity of the integrl. The integrl is liner, in the sense of being dditive nd llowing for the fctoring out of sclrs. Specificlly: nd [f(x) + g(x)] dx = f(x) dx + 3 g(x) dx

4 αf(x) dx = α f(x) dx Thus, we cn pull out sclrs nd split sums dditively when computing integrls, just s we did for derivtives Linerity of the ntiderivtive. The linerity of the integrl turns out to be closely relted to the linerity of the ntiderivtive. Of course, it is not precise to sy the ntiderivtive, since the ntiderivtive is defined only up to differences of constnts. Wht we men is the following: (1) If F is n ntiderivtive for f nd G is n ntiderivtive for g, then F + G is n ntiderivtive for f + g. (2) If F is n ntiderivtive for f nd α is rel number, then αf is n ntiderivtive for αf. (These sttements re immedite corollries of the corresponding sttements for derivtives) Generl expression for indefinite integrl. Once we hve computed one ntiderivtive for the integrl, the generl expression for the indefinite integrl is obtined by tking tht ntiderivtive nd writing + C t the end, where C is freely vrying rel prmeter. Wht this mens is tht every specific choice of numericl vlue for C gives yet nother ntiderivtive for the originl function. Note tht the letter C is used conventionlly, but there is nothing specil bout this ltter. If the sitution t hnd lredy uses the letter C in some other context, plese use nother letter. For instnce: (x sin x) dx = (x 2 /2) + cos x + C 2.5. Getting our hnds dirty. We re now in position to do some strightforwrd computtions of integrls for polynomils nd some bsic trigonometric functions. For instnce: (x 2 x + 1) dx We cn find n ntiderivtive for this function, by finding ntiderivtives for the individul functions x 2, x, nd 1, nd then dding up. An ntiderivtive tht works is x 3 /3 x 2 /2 + x. Now, to clculte the definite integrl, we need to clculte the difference between the vlues of the ntiderivtive t the upper nd lower limit. We write this s: Next, we do the clcultion: [ x 3 3 x2 2 + x ] 1 ( ) ( + ) = 5 6 Thus, the vlue of the definite integrl is 5/6. In other words, the signed re between the grph of the function x 2 x + 1 nd the x-xis, between the x-vlues nd 1, is 5/6. Some people prefer to split the definite integrl s sum first nd then compute ntiderivtives for ech piece. The work would then pper s follows: (x 2 x + 1) dx = x 2 dx x dx + There is no substntive difference in the computtions. 1 dx = [x 3 /3] 1 [x 2 /2] 1 + [x] 1 = 1/3 1/2 + 1 = 5/6 4

5 3. Higher derivtives nd repeted integrtion 3.1. Finding ll functions with given k th derivtive. Suppose the second derivtive of function is given. Wht re ll the possibilities for the originl function? In order to nswer this question, we need to integrte twice. For instnce, suppose f (x) = cos x. Then, we know tht: f (x) = cos x dx = (sin x) + C 1 where C 1 is n rbitrry rel number. Integrting gin, we get: f(x) = f (x) dx = [(sin x) + C 1 ] dx = ( cos x) + C 1 x + C 2 Here, both C 1 nd C 2 re rbitrry rel numbers. Thus, the fmily of ll possible fs tht work is described by two prmeters, freely vrying over the rel numbers. More generlly, if the k th derivtive of function is known, then the originl function is known up to dditive difference of polynomil fo degree t most k 1. Ech coefficient of tht polynomil is freely vrying rel prmeter, nd there re k such coefficients: the constnt term, the coefficient of x, nd so on till the coefficient of x k Degree of freedom nd initil/boundry vlues. One wy of thinking of the preceding mteril is tht ech time we integrte, we introduce one more degree of freedom. Thus, integrting thrice introduces totl of three degrees of freedom. In prctice, when we re sked to find function f in the rel world, we know the k th derivtive of f, but we lso hve informtion bout the vlues of f t some points. The two typicl wys this informtion is pckged re: Initil vlue problem pckging: Here, the vlue of f nd ll its derivtives, up to the (k 1) th derivtive, t single point c re provided, long with the generl expression for the k th derivtive. For this kind of problem, we cn, t ech stge of ntidifferentition, determine the vlue of the constnt we get, nd thus we get single function t the end. Boundry vlue problem pckging: Here, the vlue of f t k distinct points is specified. To solve this kind of problem, we first find the generl expression for f with k unknown constnts, then use the vlues t k distinct points to get system of k liner equtions in k vribles, which we then proceed to solve. 4. Subtle issues/dditionl notes 4.1. Vrible of integrtion don t reuse! When writing something like f(t) dt, plese remember tht the letter t, which is used loclly s vrible of integrtion, cnnot be used outside the expression Definite integrl s size or norm of function. To completely describe function f on closed intervl [, b] requires lot of work, since it requires specifying the function vlue t infinitely mny points. On the other hnd, the vlue of the integrl of f on [, b], given by f(x) dx, is single rel number. Since numbers re esier to grsp thn functions, we often use the integrl of function on n intervl to get n pproximte estimte of its size. More generlly, we re often interested in expressions of the form f(x)g(x) dx where g(x) plys the role of weighting function. Usully, we hve bunch of two or three functions g nd we re interested in the bove integrl on [, b] for ech of those gs. We use the collection of two or three numbers we get tht wy to sy profound things bout the function f, even without knowing f directly Liner lgebr interprettion of ntiderivtive. (This mteril is not necessry for this course, but is useful for subsequent mthemtics we ll see it gin in 153 nd you ll see more of these ides if you tke Mth 196/199 or dvnced courses in the socil nd/or physicl sciences). Denote by C 1 the set of functions on R tht re continuously differentible everywhere. Denote by C the set of continuous functions on R. 5

6 First, note tht C nd C 1 re both vector spces over R. Here s wht this mens for C : the sum of two continuous functions is continuous, nd ny sclr multiple of continuous function is continuous. Here s wht this mens for C 1 : the sum of two continuously differentible functions is continuously differentible, nd ny sclr multiple of continuously differentible function is continuously differentible. Differentition is liner opertor from C 1 to C in the following sense: first, for ny f C 1, f is n element of C. Second, we hve the rules (f + g) = f + g nd (αf) = αf. In other words, differentition respects the vector spce structure. The kernel of liner opertor is the set of functions which go to zero. For ny liner opertor between vector spces, the kernel is subspce. Our bsic result is tht the kernel of the differentition opertor is the spce of constnt functions. Two elements in vector spce hve the sme imge under liner opertor iff their difference is in the kernel of tht opertor. In our context, this trnsltes to the sttement tht two functions hve the sme derivtive iff their difference is constnt function. Our second bsic result is tht ny function in C rises s the derivtive of something in C 1. This something cn be computed using definite integrl. Thus, the kernel of the differentition opertor is copy of the rel line inside C 1, given by the sclr functions. For ny element of C, the set of elements of C 1 which mp to it is line inside C 1 prllel to the line of constnt functions. Insted of looking t functions on the entire rel line, we cn lso restrict ttention to functions on n open intervl inside the rel line qulittively, ll our results hold. 6