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


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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
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