Section 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


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1 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, Since t =., we ll need to use the following vlues: t v(t) To find n upper estimte, use the lefthnd endpoint of ech intervl since v(t) is decresing. We get U = (. hr)( ) m hr =.75 m. Similrly, The verge is L = (. hr)( ) m hr =.65 m. (U + L)/ =.7 m. 5.. Since f (t) = 5t + 8 is incresing, the formul on p. 45 gives the difference between the upper nd lower estimtes. U L = t f () f (b) = 3 f (3) f () = ( ) (5 + 8) = /5. 5
2 5..6 () The first hlf hour consists of the first two intervls: t 5 nd 5 t 3. Since Roger s speed is never incresing, use the left endpoints for n upper estimte nd right endpoints for lower estimte. To mke the units cncel we ll use t =.5 hr rther thn t = 5 min. (b) This prt is very similr. U = (.5 hr)( + ) mi hr L = (.5 hr)( + ) mi hr = 5.75 mi = 5.5 mi. U = (.5 hr)( ) mi hr L = (.5 hr)( ) mi hr = 7 mi = 4 mi (c) This prt is sking: How smll does t hve to be in order to gurntee tht U L.? To nswer this question, use the formul on p. 45 which reltes these two quntities: U L = t f (b) f (). mi t. mi mi/hr =.833 hr = 3 sec. (It s fine if you used equtions insted of inequlities; just relize tht t needs to be 3 seconds or smller.) 5.. For t 3, the region between the xxis nd the grph is tringulrshped, with re / b h = / (3 sec) (6 cm/sec) = 9 cm. For 3 t 4, it is gin tringulrshped region, with re / ( sec) ( cm/sec) = cm, but it counts s negtive becuse it lies below the xxis. Therefore the net distnce trveled by the prticle when t 4 is 8 cm.
3 5..5 () Looking t the grph, cr A ttins the lrger mximum velocity becuse it hs the higher pek. (b) (c) Cr A stops first becuse its grph is the first to rech the xxis. Cr B trvels frther becuse the re under its velocity curve is greter We use the methods of section 5. gin. The tble vlues indicte tht t = 3 nd f ppers to be decresing, so nd our best estimte is the verge: U = 3( ) = 4, L = 3( ) = 7, (U + L)/ = Using clcultor, x dx The re under the curve y = cos x for x is the sme s the following integrl (which you cn find using clcultor): cos x dx.6. 3
4 5.. () 6 (x + ) dx = 78. This is the re bounded between the xxis, the verticl lines x = nd x = 6, nd the prbol y = x +. (b) A lefthnd sum with n = 3 gives L = ( f () + f () + f (4)) = 46. Evidently this is n underestimte. In fct, we could hve known beforehnd tht lefthnd sum would give n underestimte becuse the function is incresing. (c) The corresponding righthnd sum is n overestimte. R = ( f () + f (4) + f (6)) = 8, Note: the point of this problem is to see these vlues represented on grph. If you hd trouble drwing the grphs, you cn sk during section or office hours Using Figure 5.9, () b f (x) dx = 3 (b) c b f (x) dx =  (It s below the xxis). (c) c (d) c f (x) dx = f (x) dx = If lefthnd sum underestimtes definite integrl by certin mount, then the corresponding righthnd sum will overestimte the integrl by the sme mount. This sttement is flse in generl. Problem is counterexmple, since the lefthnd sum underestimted the integrl by 3, but the righthnd sum overestimted it by 4. (Remember tht one counterexmple disproves clim!) However, the sttement is true when the function being integrted is line: f (x) = mx + b. 4
5 5.3.4 v(t) dt, where v(t) is the velocity in meters/sec nd t is the time in seconds. Recll tht v(t) = x (t), where x(t) is the position t time t. According to the Fundmentl Theorem of Clculus, this integrl gives x(3) x(), which represents the chnge in position (in meters) between t = nd t = 3 seconds f (t) dt, where f (t) is the rte t which the world s popultion is growing in yer t, in billions of people per yer: If P(t) is the world s popultion in yer t, this integrl gives P(4) P(), which represents the chnge in the world s popultion (in billions) between nd The formul for verge vlue is given by the boxed eqution on p. 6: b g(t) dt = ( + t) dt =. b The first hour is the intervl t 6, since t is in minutes. The integrl representing the oil leked during this time is 6 f (t) dt () Since t is mesured from the strt of 99, the desired integrl is f (t) dt. (b) A lefthnd sum with five subdivisions gives f (t) dt 3(e + e.5 + e. + e.5 + e. ) 77.3 billion brrels. (c) The first term in the sum, 3 e = 3 billion brrels, represents the oil consumed during 99, ssuming tht the rte of oil consumption did not increse during the yer. The other terms represent the sme thing for the yers 99, 99, 993, 994. However, the rte of oil consumption is continuously incresing, which is why the sum of these terms will give n underestimte. 5
6 5.3.8 () There re mny wys to see why this verge must be bigger thn /. One wy is to drw line t y = / nd observe tht the re lying bove the line thn is greter thn the re missing from below it. Therefore y = / is not the verge vlue, nd the line must be moved upwrds to compenste. On the other hnd, the verge vlue must be less thn becuse sin x lwys. (b) vg. vlue = π π using clcultor. The exct vlue is /π. sin x dx The integrl tht computes the chnge during 993 is (nnul income) = 4(.) t dt dollrs On x, the grph of x /3 lies bove x /. The re between these curves is. (x /3 x / ) dx Using Theorem 5.3 nd the informtion given in the problem, = b b ( c g(x) + (c f (x)) ) dx ( c g(x) + c f (x)) dx b = c g(x) dx + c = c + c. b f (x) dx 5.4. Since f (x) is odd, f (x) dx =. Using Theorem 5. (prt ), f (x) dx + f (x) dx = f (x) dx + f (x) dx = 8 f (x) dx = 8. 6
7 5.4.3 Use the informtion given nd solve for f (x) dx s though it were vrible (3 f (x) + 4) dx = 8 f (x) dx + 4 dx = 8 3 f (x) dx + = 8 f (x) dx = (8 )/3 =. () Using grph (or mth knowledge), we see tht the mximum vlue of f (x) = e x / is. It follows tht so the integrl is less thn. e x / < (b) Using clcultor, e x / dx =, () f (x) = e x is positive everywhere. This mens tht whenever < b, it must be true tht b ex dx >. (b) Agin, f (x) = is positive lmost everywhere, except for when cos(x+) +tn x cos(x + ) =. This occurs only for specific, isolted vlues of x, so the sme resoning pplies. 7
8 5.4.4 For convenience we ll write f (x) = π e x /. f (x). This is where Theo () We need f (x), nd the tble gives us f (x) nd rem 5. comes in hndy: f (x) dx f (x) dx = f (x) dx =.4987 f (x) dx =.574 f (x) dx (b) Since the grph of f (x) is symmetric bout the yxis (tht is, f is n even function), we know tht f (x) dx = f (x) dx =.477. Now use Theorem 5. gin: f (x) dx = f (x) dx + f (x) dx = =
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