tapia (jat4858) – HW03 – clark – (52990) 1This print-out should have 18 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsIf the points(0, 7),12, 4, (1, 2),32, 3, (2, 5)lie on the graph of a continuous function y =f(x), use the trapezoidal rule and all thesepoints to estimate the integralI =Z20f(x) dx .1. I ≈ 72. I ≈152correct3. I ≈ 84. I ≈2945. I ≈314Explanation:The trapezoidal rule estimates the integralI ash2f(0) + 2f12+ 2f(1) + 2f32+ f(2).With h =12and the given values of f , there-fore, the area is estimated byI ≈152.002 10.0 pointsIf the points(0, 1),12, 5, (1, 6),32, 3, (2, 5)lie on the graph of a continuous function y =f(x), use Simpson’s Rule and al l these pointsto estimate the integralI =Z20f(x) dx.1. I ≈2632. I ≈253correct3. I ≈4964. I ≈ 85. I ≈172Explanation:Simpson’s Rule estimates the integral I ash3f(0) + 4f12+ 2f(1) + 4f32+ f(2).With h =12and the given values of f, there-fore, the area is estimated byI ≈253.keywords: Simpson’s Rule, integral, graph003 10.0 pointsBelow is the graph of a function f .-7-6-5-4-3-2-10123456789−1−2−31 2 3246tapia (jat4858) – HW03 – clark – (52990) 2Estimate the integralI =Z3−3f(x) dxusing the Midpoint Rule with six equal subin-tervals.1. I ≈ 92. I ≈ 73. I ≈ 64. I ≈ 8 correct5. I ≈ 5Explanation:Since [−3, 3] is subdivided into six equalsubintervals, each of these will have length1 and the six corresponding rectangles areshown a s the gray-shaded areas in-7-6-5-4-3-2-1012345678−1−2−31 2 3246The heights of the rectangles are midpointsample values of f that can be read off fromthe graph. Thus, with midpoints,I ≈ 4 − 2 − 3 + 1 + 3 + 5 = 8 .keywords: midpoint rule, subint erval, inte-gral, graph004 10.0 pointsA farmer wishes to estimate the area of ariver-front pasture. He knows some calculus,so he takes measurements every 100 feet anddraws the following graph40 60 50 80 60What value does he obt ain if he uses thetrapezoidal rule and all the data to estimatethe area?1. Area ≈ 24050 sq. feet2. Area ≈ 24000 sq. feet correct3. Area ≈ 24150 sq. feet4. Area ≈ 24100 sq. feet5. Area ≈ 24200 sq. feetExplanation:By the trapezoidal ruleArea ≈12100(y0+ 2y1+ 2y2+ 2y3+ y4)= 50(40 + 120 + 100 + 160 + 60).Consequently,Area ≈ 24000 sq.feet .keywords: word problem, estimate area,trapezoidal rule005 10.0 pointstapia (jat4858) – HW03 – clark – (52990) 3A radar gun was used t o record the speedof a runner during the first 5 seconds of a raceas shown in the tablet (secs) vel (m/ sec)0 00.5 41.0 7.31.5 82.0 9.72.5 10.13.0 10.63.5 10.74.0 10.84.5 10.885.0 10.99Use Simpson’s Rule and all the given datato estimate the distance the runner coveredduring those 5 seconds.1. dist ≈ 43.75 meters correct2. dist ≈ 43.77 meters3. dist ≈ 43.81 meters4. dist ≈ 43.73 meters5. dist ≈ 43.79 metersExplanation:The distance covered during those 5 sec-onds is given by the integralI =Z50v(t) dtwhere v(t) is the veloci ty of the runner at t imet. Simpson’s Rule estimates this integral asI ≈16nv(0) + 4v12+ 2v(1) + 4v32+ 2v(2) + 4v52+ 2v(3) + 4v72+ 2v(4) + 4v92+ v(5)o.Reading off the values of v(t) from the tablewe thus see thatI ≈ 43.75 meters .006 10.0 pointsWhich o f the following integrals ar e im-proper?I1=Z100111 + x2dx ,I2=Z20x − 2x − 1dx ,I3=Z101√xdx .1. I2only2. I1and I2only3. none of them4. I1and I3only5. all of them6. I1only7. I3only8. I2and I3only correctExplanation:An integralI =Zbaf(x) dxis improper when one or more of the followingconditions are satisfied:tapia (jat4858) – HW03 – clark – (52990) 4(i) the interval of integration is infinite, i.e.,when a = −∞ or b = ∞, or both;(ii) f has a vertical asympt ote at one ormore of x = a, x = b or x = c for somea < c < b.Consequently,Z100111 + x2dxis not improper;Z20x − 2x − 1dxis improper;Z101√xdxis improper.007 10.0 pointsDetermine if the improper integralI =Z∞06(x + 4)2dxis convergent, and if it is, find its value.1. I is not convergent2. I = 23. I =524. I =945. I =746. I =32correctExplanation:The integral is improper b ecause the in-terval of integration is infinite. To test forconvergence we thus have to determine iflimt → ∞It, It=Zt06(x + 4)2dx ,exists.To evaluate It, set u = x + 4, Thendu = dx ,whilex = 0 =⇒ u = 4x = t =⇒ u = t + 4 .In this case,It=Zt+446u2du =h−6uit+44,so thatlimt → ∞It= limt → ∞32−6t + 4=32.Consequently, the integral isconvergent and I =32..008 10.0 pointsDetermine if the improper integralI =Z∞22xe−x/3dxconverges, and if it does, compute its value.1. I = 30e232. I = 6e−233. I does not converge4. I = 6e235. I = 30e−23correctExplanation:The integralZ∞22xe−x/3dxtapia (jat4858) – HW03 – clark – (52990) 5is improper because of the infinite range ofintegration. To overcome this we restrict to afinite interval of integration a nd consider thelimitlimt → ∞It, It=Zt22xe−x/3dx.To evaluate Itwe integrate by parts. ThenIt=h−6xe−x/3it2+ 6Zt2e−x/3dx= −6h(x + 3)e−x/3it2= 30e−23− 6(t + 3)e−t/3.On the other hand,limx → ∞xme−x= 0for any power m of x. Thuslimt → ∞(t + 3)e−t/3= 0.Consequently, I converges a ndI = 30e−23.009 10.0 pointsDetermine if the improper integralI =Z∞16 tan−1(x)1 + x2dxconverges, and if it does, find its value.1. I =9162. I =34π23. I does not converge4. I =345. I =98π26. I =916π2correctExplanation:The integral I is improper because the in-terval of integration is infinite. Thus theintegral will converge iflimt → ∞It, It=Zt16 tan−1(x)1 + x2dx ,exists, and its value will then be the value ofthe limit.The substitution u = tan−1(x) is sug-gested. For thendu =11 + x2dx ,whilex = 1 =⇒ u =π4,andx = t =⇒ u = tan−1(t) .In this caseIt=Ztan−1(t)π/46u du = 3hu2itan−1(t)π/4.ThusIt= 3tan−1(t)2−π216.On the other hand,limt → ∞tan−1(t) =π2.Consequently, I converges and has val ueI = 314−116π2=916π2.010 10.0 pointsDetermine if the integralI =Z202(x − 1)2/3dxtapia (jat4858) – HW03 – clark – (52990) 6is convergent or divergent; and if convergent,find its value.1. I = 42. I = 63. I = 12 correct4. I is divergent5. I = 0Explanation:The integral is improper because the func-tionf(x) =2(x − 1)2/3has a vertical asymptot e at x = 1 in the i nter-val of integration. Thus we have