x = b a N. (131) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is


 Poppy Bond
 2 years ago
 Views:
Transcription
1 Jnury 28, The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According to Webster s dictionry, to integrte mens to mke complete by dding prts. This is indeed wht is done in integrtion in clculus. The integrl is the resulting totl DEFINITIONS, TERMINOLOGY, AND NOTATIONS We must begin by estblishing some terminology. Region of integrtion. The region of integrtion is segment [, b] of the rel line. Tht is, the region is the set of points x with x b. Differentil elements. The region of integrtion will be subdivided into mny smll prts, s illustrted in Fig These smll segments of the region [, b] re clled the differentil elements, or just elements. There re in generl N elements, ech of length x. Note tht N x is the length of the full region of integrtion, i.e., b, so x = b N. (131) The set of points used to subdivide the rnge [, b] (see Fig. 13.1) is {x 0, x 1, x 2, x 3,..., x N 1, x N } where x 0 = is the left endpoint nd x N = b is the right endpoint. We denote generic point in this set by x k, where the index k hs n integer vlue from 0 to N. Therefore the position on the rel line of the point x k is x k = + k ( x). (132) For exmple, x 1 = + x, x 2 = + 2 x, etc. Note tht the rightmost point is x N = + N x = + (b ) = b, i.e, the right endpoint of the full rnge. The kth differentil element is the smll segment [x k 1, x k ] whose length is x k x k 1 = x. 1
2 2 Chpter 13 The integrl will be defined s certin sum of terms, in the limit x 0 nd N. In tking the limit, N x remins constnt, equl to b. The definite integrl. The definite integrl 1 of function g(x) for the integrtion region [, b] is the sum of terms g(x k ) x for the subdivision shown in Fig. 13.1, in the limit x 0. Written s n eqution, Here the definite integrl = lim x 0 g(x k ) x. indictes tht we sum over the N elementl segments, of which the kth is [x k 1, x k ]. The mthemticl symbol for discrete sum is Σ (the Greek letter with the sound of S ). The Σ nottion for summtion is described in Appendix B Nottion for the definite integrl The mthemticl symbol for the integrl is. It resembles n elongted S, which is pproprite becuse it stnds for Summtion of mny prts. A definite integrl depends on three things, nd the nottion for the integrl must disply ll this informtion. First, there is the function tht is being integrted, which is clled the integrnd; let g(x) denote the function. Second, there is the vrible of integrtion, which is the independent vrible x of the function g(x). Third, there is the region (or rnge) of integrtion [, b]; this is the set of rel numbers x with x b. The nottion for the definite integrl is g(x)dx. (133) Note how it displys the three pieces of informtion: the function is g(x), the independent vrible (x) is indicted in dx, nd the endpoints of the rnge of integrtion re nd b. The expression in (133) is red s the integrl of g(x) from to b. The complex symbol in (133) stnds for single number. For exmple, the integrl of the function x 2 for x from 1 to 3 is denoted by 3 1 x 2 dx; 1 We define in this chpter the definite integrl, which must hve specified region of integrtion. The indefinite integrl will be defined in Chpter 14.
3 Dniel Stump 3 the vlue of this integrl, which we ll lern to clculte in Chpter 14, is 26/3. Thus we my write n eqution, 3 1 x 2 dx = 26 3 = Opertionl definition of the integrl Summrizing ll tht hs been sid, the eqution tht defines the definite integrl of function g(x) over the rnge x b is g(x)dx = lim x 0 g(x k ) x. (134) The set of points {x 0, x 1, x 2,..., x N } is specified in Fig The sum on the righthnd side of (134) is clled the Riemnn sum. Note, in (133) nd (134), the ppernce of our old friend the differentil dx. A good nottion simplifies mthemtics. The clever ide behind the nottion in (133) is tht g(x)dx is the infinitesiml contribution to the totl from the differentil element dx. The integrl is the sum of ll the elementl prts. If we think of g(x) s the density of quntity Q, i.e., the mount of Q per unit of x, then g(x)dx is the mount of Q in the infinitesiml segment dx of the x xis. The integrl (134) is the totl of Q over the integrtion region [, b]. In principle, (134) could be used to evlute ny integrl, by numericl computtion of the sum for very smll vlues of x. Indeed, this method is used to evlute integrls by computer. We ll consider some simple exmples of this opertion in the next section. However, doing the clcultion this wy is very cumbersome in prctice. We will develop more powerful techniques to clculte integrls in Chpters 14 to 17. But in ny cse (134) is the mthemticl definition of the integrl.
4 4 Chpter EXAMPLES We will evlute here some simple integrls directly from the definition (134). The sum on the righthnd side of (134) is clled Riemnn sum, nmed for the mthemticin Riemnn. 2 Evluting the Riemnn sum is very tedious wy to clculte n integrl. We will discover more powerful method in Chpter 14. But it is instructive to look t some very simple integrls by using (134) directly, to understnd the bsic ide of integrtion. Exmple 1. Consider the constnt function y(x) = 3.6. Determine the definite integrl of y(x) over the integrtion rnge 2 x 7. Solution. Subdivide the region of integrtion [2, 7] into N elements of length x = 5/N. The Riemnn sum is sum of N terms, nd the contribution from the elementl segment [x k 1, x k ] is y(x k ) x = N. (135) This elementl contribution is independent of k becuse y(x) is constnt, so the Riemnn sum is sum of equl terms, i.e., just equl to N times ny one term, y(x k ) x = N = 18. (136) N In this exmple constnt function the limit x 0 (or, equivlently, N ) is trivil becuse the Riemnn sum does not depend on x. The limit of constnt is the constnt! Hence the integrl is dx = 18. (137) Generliztion. The exmple of constnt function is so simple tht we cn immeditely generlize Exmple 1 to n rbitrry constnt function, sy g(x) = C, nd n rbitrry rnge of integrtion, x b. We ll prove tht C dx = C (b ). (138) Note tht the nswer to Exmple 1 grees with this generl formul. 2 Bernhrd Riemnn, who lived from 1826 to 1866, clrified the proper definition of the integrl. Another of his developments Riemnnin geometry is the essentil mthemtics used in Einstein s theory of generl reltivity.
5 Dniel Stump 5 To prove the generl formul, repet the nlysis in Exmple 1. In the generl cse the point x k in the subdivision is x k = + k( x) where x = (b )/N is the width of n elementl segment. Then the definition of the integrl gives C dx = lim N C x = lim C x N. (139) N Substituting N x = b we find the generl result C dx = C(b ). (1310) The limit N is trivil in this cse, becuse the Riemnn sum does not depend on N. Exmple 2. Consider the function f(ξ) = 3ξ where the independent vrible is denoted by ξ. Determine the definite integrl of f(ξ) from ξ = 0 to ξ = 10. Solution. Subdivide the integrtion region [0, 10] on the rel line of ξ, into N segments of width ξ = 10/N. The endpoints of the elements re points in the set {ξ 0, ξ 1, ξ 2,..., ξ N } where ξ k = k ( ξ). (Note tht ξ 0 = 0 nd ξ N = 10.) The contribution to the Riemnn sum from the kth segment is f(ξ k ) ξ = 3ξ k ξ = 3k ( ξ) 2. (1311) The Riemnn sum for N elements, cll it R N, is R N = 3k ( ξ) 2 = 3 ( ξ) 2 N k; (1312) in the second equlity we hve fctored out the constnt fctors 3 nd ( ξ) 2. Recll tht the sum of the first N integers is 3 Thus k = N = 1 N(N + 1). (1313) 2 R N = 3 2 N(N + 1)( ξ)2, (1314) or, substituting ξ = 10/N, R N = N(N + 1)100 = N 2 N. (1315) 3 See Appendix B.
6 6 Chpter 13 Now, to obtin the integrl we must tke the limit ξ 0, which is equivlent to N becuse ξ = 10/N. The term 150/N goes to 0 in the limit, so the integrl is ξdξ = 150. (1316) We ll generlize this result fter the next exmple. Exmple 3. Consider gin f(ξ) = 3ξ, but now integrte the function from ξ = 5 to ξ = 10. By the definition (134), ξdξ = lim ξ 0 3ξ k ( ξ). (1317) In this cse ξ is 5/N becuse the region of integrtion hs length 5 nd is subdivided into N elements. Also, ξ k is the vlue of ξ on the rel line t the right side of the kth element, ξ k = 5 + k ξ. (1318) For exmple ξ 1 is t 5+ ξ, ξ 2 is t 5+2 ξ, nd so on; ξ N is t 5+N ξ = 10 s required. The Riemnn sum for subdivision with N elements is R N = 3 = 3 (5 + k ξ) ξ [5N + 12 N(N + 1) ξ ] ξ = N(N + 1) 2N 2 (1319) where in the lst step we hve substituted ξ = 5/N. Finlly, we obtin the vlue of the integrl by tking the limit N. As in the previous exmple, N(N + 1)/N 2 1. Hence the integrl is ξdξ = = (1320) Generliztion. We my generlize the results of Exmples 2 nd 3 to ny liner function of the form f(x) = mx with m constnt, integrted between rbitrry points x = nd x = b. We ll prove tht mxdx = m 2 ( b 2 2). (1321)
7 Dniel Stump 7 Note tht the result of Exmple 3 grees with this generl formul, for m = 3, = 5 nd b = 10. To prove the generl formul, repet the nlysis in Exmple 3. In the generl cse the point x k in the subdivision is x k = + k( x) where x = (b )/N is the width of n elementl segment. Then the definition of the integrl gives mxdx = lim N = lim N m m ( + k x) x [ N x + N(N + 1) 2 Now substitute N x = b nd tke the limit N : ] mxdx = lim [(b m ) + 1 N(N+1) N 2 N (b ) 2 2 ( x) 2 ]. (1322) = m [ (b ) (b )2] = m [ (b )] (b ) ( ) b + = m (b ) = m ( b 2 2). (1323) 2 2 In this clcultion the limit N ws evluted in the second step by replcing N(N + 1)/N 2 by The integrl of n rbitrry liner function between rbitrry endpoints Equtions (138) nd (1321) re nice, generl formuls. But we cn do better. In this section we ll determine the formul for the integrl of ny liner function. A liner function of the independent vrible x hs the form C 0 + C 1 x where C 0 nd C 1 re constnts. The function y(x) = 3.6 in Exmple 1 is n exmple of liner function, with C 0 = 3.6 nd C 1 = 0. The function f(ξ) = 3ξ in Exmples 2 nd 3 is liner, with C 0 = 0 nd C 1 = 3. Now, wht is the integrl of the most generl liner function C 0 + C 1 x, over n rbitrry region of integrtion [, b]? We ll prove tht (C 0 + C 1 x) dx = C 0 (b ) C ( 1 b 2 2). (1324) After proving this result, we ll never gin need to evlute Riemnn sum for liner function! We cn just plug in the specific vlues of the constnts
8 8 Chpter 13 C 0, C 1, nd b to clculte the integrl. Eqution (1324) is just simple extension of the erlier results (138) nd (1321). The integrl of sum of functions is simply the sum of the integrls, [f 1 (x) + f 2 (x)] dx = f 1 (x)dx + f 2 (x)dx. (1325) This is true becuse the Riemnn sum of f 1 + f 2 is the sum of f 1 plus the sum of f 2, nd, fter tking the limit N, the sums become the integrls in (1325). Now note tht the two terms on the righthnd side of (1324) follow from the integrls in (138) nd (1321).
9 Dniel Stump GRAPHICAL INTERPRETATION OF THE INTEGRAL Grphicl nlysis of functions is n importnt technique in science nd engineering. For exmple, consider the function F (x) = x 2 e x/5, (1326) which might describe cusendeffect reltionship between two vribles, x nd F, in physicl system. A scientist encountering this function, e.g., in theoreticl clcultion, would immeditely sketch its grph (Fig. 13.2); the pictoril representtion is esier to comprehend thn the symbolic formul. The grph shows tht s x increses from 0 to 30, the vrible F strts t 0, then rises until F reches mximum t x = 10 nd then decreses more grdully to 0. The derivtive df/dx, or F (x), is n importnt quntity in the nlysis of function F (x). It is the rte of chnge of the vrible F with respect to chnge in x. The derivtive function hs simple grphicl interprettion, which we studied in Sec. 4.2: The derivtive F (x) is the slope of the grph of F (x) t x. For exmple, looking t Fig for the function in (1326) we cn immeditely see tht the slope strts t 0 t x = 0, becomes positive s x increses, but returns to 0 t the mximum (x = 10); s x increses beyond this point the slope becomes negtive but pproches 0 (from below) s x tends to. All of the detiled properties of F (x) nd F (x) cn be seen from the grph. The slope of grph is geometric concept, becuse it involves the shpe of the curve. The physicl ppliction of the function F (x) my be completely unrelted to geometry. But we introduce this geometric concept the slope of the grph s n bstrct mthemticl construction tht will help us to nlyze the function. The definite integrl F (x)dx is nother importnt quntity in the nlysis of function. It is the mount of quntity whose density (mount per unit of x) is F (x). This concept hs mny physicl pplictions, s illustrted in Chp. 12. Cn the grph of F (x) help us to nlyze the integrl? The integrl does hve simple grphicl interprettion. F (x)dx is certin re in the grph of F (x): It is the re bounded bove by the curve F (x), bounded below by the x xis, nd bounded on the sides by the verticl lines x = nd x = b. For exmple, the integrl of the specific function F (x) in (1326), from x = 5 to x = 20, is the re of the shded region in Fig We ll prove tht F (x)dx is equl to the shded re presently. It is importnt to understnd tht the re of the grph does not generlly refer to rel physicl re. In most pplictions, F (x)dx is not physicl re, or ny other geometric quntity. It could be mss, or kinetic
10 10 Chpter 13 energy, or n electric field, etc. (cf. the exmples in Chp. 12). We introduce the re of the grph s n bstrct mthemticl construction tht helps us to nlyze the integrl. Some pplictions of integrtion re in fct clcultions of geometricl quntities, such s res or volumes. For exmple, we considered re clcultions in Sec s elementry exmples of integrtion; see lso, Chpter 16. In n re clcultion the grphicl interprettion of the integrl is identicl to the ctul ppliction. But in other pplictions the grphicl interprettion is just n id to understnding. Then the ctul mening of the integrl is not physicl re but some other physicl quntity mss, kinetic energy, electric field, etc. The following sttements summrize the mening nd grphicl interprettion of the definite integrl. Mening: The integrl g(x)dx is the totl mount of quntity whose density is g(x), for x between nd b. Grphicl interprettion: The integrl g(x)dx is the re bounded bove by g(x) nd below by the x xis, between x = nd x = b, on grph of g(x) Proof of the grphicl interprettion Figure 13.4 illustrtes the grph of g(x). Subdivide the rnge of integrtion [, b] into N smll segments of width x. For ech segment two rectngulr strips re shown in the figure, which hve width x nd height g mx or g min. The re A under the curve my be bounded bove nd below. A is less thn the sum of the strips if we tke the mximum vlue of g(x) in the kth strip s the height of the strip; A is greter thn the sum of the strips if we tke the minimum vlue of g(x) in the kth strip s the height of the strip. The res of these lrger nd smller strips ( = width times mximum or minimum height, respectively) re gk mx x nd gk min x. Adding ll the strip res, the bounds on the re A under the curve re gk min x A g mx k x. (1327) Now tke the limit x 0 nd N, with N x = b, constnt. The sums in (1327) both tend to the sme limit, nmely the integrl, by the definition (134). Either sum serves s the Riemnn sum for the integrl. The two bounds squeeze together to the vlue of the integrl s x 0.
11 Dniel Stump 11 Therefore the re A under g(x) must be equl to the integrl, A = g(x)dx. (1328) Figure 13.4 gives n intuitive picture of the grphicl interprettion of the definite integrl. Integrtion is equivlent to subdividing the integrtion region into smll elements, dding the elementl contributions (either gk mx x or gk min x) nd tking the limit x dx. The result is the re under the curve becuse g(x)dx is the elementl re (height times width) of n infinitesiml element Negtive integrls In the discussion leding to (1328) it ws tcitly ssumed tht g(x) is positive in the integrtion region [, b]. Wht bout function f(x) tht is negtive for some prt of the region, s illustrted in Fig. 13.5? The elementl contribution f(x k ) x to the Riemnn sum is negtive for segment in the x region [c, d] where f(x) is negtive. If we interpret the integrl s n re, wht is the mening of negtive re? In the grphicl interprettion we must understnd negtive re s re below the x xis. For exmple, in Fig. 13.5, the integrl of f(x) from x = to x = b hs three prts: positive re for x from to c, where the curve is bove the x xis; negtive re for x from c to d, where the curve is below the x xis; nd positive re for x from d to b. The integrl from to b is the sum of these three res. So the bsic ide is tht f(x)dx is the re between the curve of f(x) nd the x xis. Where f(x) is positive, the curve is bove the x xis nd the re is positive; where f(x) is negtive, the curve is below the x xis nd the re is negtive. In the grphicl interprettion of the integrl ( f(x)dx = re) regions where the curve is below the x xis count s negtive re. Exmple 4. Consider the sine function. Figure 13.6 is grph of sin x, for x between 0 nd 2π. Wht is the integrl from x = 0 to x = 2π? Solution. The integrl from 0 to π is positive, nd in fct equl to The vlue +2 is the re of the region bounded by the first hlfcycle of the sine curve nd the x xis; this re is positive becuse the curve lies bove the x xis. The integrl from π to 2π is negtive, nd equl to 2. The vlue 2 is the re bounded by the second hlfcycle of sin x nd the x xis, nd it is negtive becuse the curve lies below the x xis. The integrl from 0 to 2π 4 We ll lern to evlute this integrl in Chpter 15.
12 12 Chpter 13 is 0 becuse positive nd negtive contributions cncel, 2π 0 sin x dx = 0. (1329)
13 Dniel Stump DISTANCE TRAVELED IN ACCELERATING MOTION Consider n object M moving long line 5 with vrying velocity v(t). The ccelertion is (t) = dv dt. (1330) In grph of velocity v versus time t, the slope of the curve t t is the ccelertion t tht time. Wht is the distnce trveled by the object during some time intervl, sy from time t 1 to t 2? The distnce is the integrl of the velocity, D = t2 t 1 v(t)dt. (1331) The reson is becuse v is the distnce per unit time, so tht vdt is the distnce trveled during the elementl time intervl dt. Adding ll the elementl distnces, i.e., integrting the velocity, gives the totl distnce trveled. We could mke the explntion more rigorous by describing Riemnn sum: Subdivide the intervl [t 1, t 2 ] into smll elements t, nd write D v(t k ) t; (1332) the sum pproximtes D becuse v(t k ) t pproximtes the distnce trveled during t. In the limit t 0, the pproximtion becomes exct. But the limit of the Riemnn sum is, by definition, just the integrl (1331). Positive or negtive displcements Suppose the velocity v(t) is negtive, v = dx/dt < 0. Negtive velocity mens tht the object M is moving towrd smller vlues of the coordinte x. Then vdt is negtive, nd (1331) would give negtive distnce. A better terminology is to cll D the displcement, which is positive if the object moves to lrger coordinte, or negtive if the object moves to smller coordinte. Then the distnce is defined s the bsolute vlue of D (which must be positive). 6 Exmple 5. Wht is the distnce trveled from time t = 0 to time t = T if v is constnt? Solution. We know tht t 2 t 1 Cdt = C(t 2 t 1 ) by the generl result (1310). 5 We nme the object M for moving. 6 For motion in three dimensions, displcement is vector D nd distnce is the sclr D (the length of the vector) which must be positive.
14 14 Chpter 13 Thus, for constnt velocity v 0 from t 1 = 0 to t 2 = T, D = v 0 T. (1333) The result is just velocity time, fmilir from grde school. Exmple 6. Wht is the distnce trveled from time t = 0 to time t = T if there is constnt ccelertion? Solution. The velocity, for constnt ccelertion, is v(t) = v 0 + t where v 0 is the initil velocity. Using the generl formul (1324), the displcement is D = T 0 (v 0 + t) dt = v 0 T T 2. (1334) This formul ws used in Chp. 9 for motion with constnt ccelertion. Exmple 7. Wht is the distnce trveled from time t = 0 to time t = T if the velocity is v(t) = U sin(πt/t )? Figure 13.7 shows grph of v versus t. Solution. The distnce trveled is the integrl, i.e., the re under the curve in Fig. 13.7, D = T 0 U sin πt dt. (1335) T We will lern to clculte this integrl in Chpter 15. The result is D = 2UT/π. Exmple 8. Consider the velocity function v(t) whose grph is shown in Fig The object M is locted t the origin t time t = 0. Is the coordinte positive or negtive t t = t FIN? Solution. The re under the curve from t = 0 to t FIN is clerly negtive. This re is the integrl vdt, i.e., the displcement. The displcement is negtive, so the finl coordinte is negtive. The net chnge of position during the time intervl from 0 to t FIN is to the left (with positive displcement defined to be to the right). Summry The three bsic functions of kinemtics 7 re x(t), v(t), nd (t) position, velocity, nd ccelertion. Recll from Chpter 9 the differ 7 Kinemtics is the nlysis of motion.
15 Dniel Stump 15 entil equtions obeyed by these functions, dx dt = v(t) nd dv dt = (t). We my now dd n integrl eqution, x(t) = x 0 + t 0 v(t )dt, becuse by (1331) the integrl of v(t ) from t = 0 to t = t is the displcement during the time intervl [0, t]. Similrly, relting v nd, v(t) = v 0 + t 0 (t )dt. The connection between the differentil eqution nd the corresponding integrl eqution (e.g., v is dx/dt nd x is vdt) is n exmple of the fundmentl theorem of clculus the subject of Chpter 14.
16 16 Chpter SUMMARY Two bsic mthemticl ides re contined in this chpter. The definition of the integrl is f(x)dx = lim x 0 f(x k ) x (1336) where {x 0, x 1, x 2,..., x N } is the subdivision of [, b] into N segments (elements) of length x. The righthnd side is clled the Riemnn sum. The integrl f(x)dx is equl to the re under the curve in grph of f(x). More precisely it is the re of the region bounded by the curve, the x xis, nd the lines x = nd x = b.
17 Figure 13.1: Differentil elements. The x xis from to b is subdivided into N smll segments of width x = (b )/N. The curve represents function g(x) tht is being nlyzed. Figure 13.2: The function F (x) in Eq. (1326). 17
18 Figure 13.3: The integrl of the function F (x), from x = 5 to x = 20, is equl to the re of the shded region in the grph. Figure 13.4: Proof of the geometricl interprettion of the definite integrl; the integrl equls the re A under the curve from x = to x = b. The re A is bounded bove by the sum of the strip res with the mximum g mx of g(x) in ech strip s the height of tht strip (upper sum); the re is bounded below by tking the minimum g min of g(x) in ech strip s the height of the strip (lower sum). Either sum is Riemnn sum for the integrl. In the limit x 0 (nd N, with N x = b ) the upper nd lower bounds squeeze together, nd both pproch the integrl. 18
19 Figure 13.5: A function tht is negtive in prt of the integrtion rnge [, b]. Figure 13.6: The sine function. The integrl of sin x from 0 to 2π is 0, becuse the positive re bove the x xis (from x = 0 to π) is equl but opposite to the negtive re below the x xis (from x = π to 2π). Figure 13.7: Velocity versus time in Exmple 7. The object M strts t rest (t t = 0), ccelertes to mximum velocity U, nd decelertes bck to zero velocity (t t = T ). How fr did it move? 19
20 Figure 13.8: Velocity versus time in Exmple 8. The object M strts t the origin (x = 0) t t = 0. Is the position positive or negtive t t = t FIN? 20
21 Dniel Stump 21 EXERCISES Section 2: Exmples TBA TBA Section 3: Grphicl interprettion of the integrl Consider the sine function, shown in Fig Consider the integrl from 0 to, C() = 0 sin x dx. The lower endpoint is fixed t 0. The upper endpoint is n rbitrry vlue. Sketch grph of the function C() versus. We ll lern to clculte this integrl in Chp. 15. The result is C() = 1 cos. Verify tht grph of 1 cos mtches your grph Consider n intervl [, b] on the x xis, nd point c inside the intervl, so tht < c < b. From the definition (134) of the definite integrl, prove f(x)dx = c f(x)dx + c f(x)dx. Also, explin this eqution bsed on the grphicl interprettion of the integrl Derive eqution (1324) from the grphicl interprettion of the integrl. (Hint: The grph of C 0 + C 1 x is stright line with slope C 1. Wht is the re under the curve between x = nd x = b?) Section 4: Distnce trveled in ccelerting motion Suppose n object M moves with velocity v(t) given by Figure How fr does M move? Give the nswer in meters. Describe the motion in words.
22 22 Chpter 13 Figure 13.9: Velocity versus time in Exercise For ech grph of velocity v(t) versus time t in Fig , sketch grph of position x(t) versus time t for the full time intervl. In ll cses ssume x = 0 t t = 0. [Hint: Displcement = vdt = re under the curve of v(t).] Figure 13.10: Exercise An object is initilly (t t = 0) t rest t the origin. During the time intervl from t = 0 to 10 s, its ccelertion is (t) = γt where γ = 0.3 m/s 3 ; from t = 10 s to 20 s, the ccelertion is (t) = γ (20 s t). Solve this exercise grphiclly.
23 Dniel Stump 23 () How fst is the object moving t t = 20 s? (b) Estimte how fr the object moves from t = 0 to t = 20 s. Generl Exercises TBA TBA
24 24 Chpter 13 Contents 13.1 Definitions, terminology, nd nottions Nottion for the definite integrl Opertionl definition of the integrl Exmples The integrl of n rbitrry liner function between rbitrry endpoints Grphicl interprettion of the integrl Proof of the grphicl interprettion Negtive integrls Distnce trveled in ccelerting motion Summry
Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
More information5 Accumulated Change: The Definite Integral
5 Accumulted Chnge: The Definite Integrl 5.1 Distnce nd Accumulted Chnge * How To Mesure Distnce Trveled nd Visulize Distnce on the Velocity Grph Distnce = Velocity Time Exmple 1 Suppose tht you trvel
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
More informationConservation Law. Chapter Goal. 5.2 Theory
Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationSection Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?
Section 5.  Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationMAT 168: Calculus II with Analytic Geometry. James V. Lambers
MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationFinal Exam  Review MATH Spring 2017
Finl Exm  Review MATH 5  Spring 7 Chpter, 3, nd Sections 5.5.5, 5.7 Finl Exm: Tuesdy 5/9, :37:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationMATH 115 FINAL EXAM. April 25, 2005
MATH 115 FINAL EXAM April 25, 2005 NAME: Solution Key INSTRUCTOR: SECTION NO: 1. Do not open this exm until you re told to begin. 2. This exm hs 9 pges including this cover. There re 9 questions. 3. Do
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationChapter 5. Numerical Integration
Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationAn Overview of Integration
An Overview of Integrtion S. F. Ellermeyer July 26, 2 The Definite Integrl of Function f Over n Intervl, Suppose tht f is continuous function defined on n intervl,. The definite integrl of f from to is
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationStudent Session Topic: Particle Motion
Student Session Topic: Prticle Motion Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position, velocity or ccelertion my be
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More information