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by: Hector Hudson III


Hector Hudson III
OK State
GPA 4.0

David Wright

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David Wright
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This 0 page Class Notes was uploaded by Hector Hudson III on Sunday November 1, 2015. The Class Notes belongs to MATH 2144 at Oklahoma State University taught by David Wright in Fall. Since its upload, it has received 15 views. For similar materials see /class/232784/math-2144-oklahoma-state-university in Mathematics (M) at Oklahoma State University.

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Date Created: 11/01/15
Math 2144 Final Exam Last Week Review Page 1 Step 1 First review Exams 1 2 and 3 and be sure to understand all the principles involved in solving those problems ef ciently and correctly Step 2 Next review the new sections 55 61 62 and be sure to understand the techniques used to solve the problems below Step 3 Some main charts of formulas to review and other key pages in the textbook are c Derivatives formulas Reference page 5 at end of book 0 Table of Inde nite Integrals in Section 54 o Midpoint Trapezoid Simpson s rules for numerical integration Section 77 o Substitution example problems in Section 55 Extra sample problems mostly from the last few sections 1 Each of the pictures below shows the graph of y x between z a and z b On each one sketch and shade in the region corresponding to the numerical method stated below the picture For instance the left Riemann sum with n 1 subdivision corresponds to the picture to the right i 1 l a a aw2 1 Left sum with 111 A l a ab2 b 1 193 17 a ab2 b Right sum with 111 MidPDjT 1 511 With quot1 Le sum with 112 D E F a ab2 b a ab2 b a ab2 b Trapezoid sum with quot1 Right sum with quot2 Tmpezoid sum with quot2 D Let r be a constant greater than 1 Evaluate the following integrals by substitution of a new variable u in place of z Show the function of z you use for u show the integral in terms of u only evaluate that integral and nally give the answer as a function of m a 2 2mr3m1dm b 5 tan 99sec20d0 I w tan 13m c e x1re dm 1 de Math 2144 Final Exam Last Week Review Page 2 3 Evaluate the following de nite integrals by substitution of a new variable u in place of m Show the function of z you use for u show the integral in terms of u only with the adjusted endpoints7 and then evaluate the u integral as a number 10 1 a 16 2m dm 1 467m5 dm 0 0 9 12 arcsinm 7 5 d d 7 d c 5 xvm m 0 f 7 2 m g A particle moves along a line so that its velocity at time t is 1205 t2 7 t 7 2 in meters per second a Find the displacement of the particle during 1 g t g 3 b Find the distance traveled during this time period 5 a Write down an integral that equals the area of the region of the zy plane above the curve y 2 see picture and below the curve y 17 z b Determine the exact area of this region 6 Find the area of the bounded region between the curves y 4x and y 2x2 7 2x 4 7 Sketch the bounded regions enclosed by the curves y xi y 7 z 16 Write down an integral that represents the area of the region7 and then nd the area of the region 8 Find the volume of the foptball obtained by rotating the region be 4 Jim 39139 l3939quot ts tween the curves y 17 and y 0 around the z axis The solid is shown to the right 2 9 Let R be the region above the curve y 17 mi and below the curve y 1 7 2 see picture u a a Using the washer method7 write down an integral giving the vol 72 r u ume of the solid obtained by rotating the region R about the z axis b Evaluate the volume described in part a 10 Let R be the bounded region between the curves y 12 7 2x2 and y 8 7 2 a Sketch the region R carefully and determine the points of intersection of the curves b Write down an integral for the area of R and then evaluate that area c A solid A is created by rotating R about the line y 71 Write down an integral for calculating the volume of A by means of the washer method Do not evaluate Math 2144 Final Exam Extra Review Answers a co a aw2 b a aw2 b a ab2 b M39d 39 1 39th 1 Right sum w1th nl 1 pm sum W n Left sum VVlth 112 Dr E F a ab2 b a nb2 b a ab2 b Trapezold sum wnh quot1 Right sum with quot2 Trapezoid sum with quot2 21 For any constant K the general antiderivatives are au122zr7 du2z2dz7 usdu7 z22zr4K b u5tant97 dusec2t9dt97 ugdu7 17105tan1910K c u139regc7 du39rexdz ul2d7u7 331T E32K 7 7 1 1 1 3 e 1 71 2 du7tan 317 du713z2dz 1137 6tan K 1 17T 72M 7 2x 01 7 23032 K e u sin 17 du coszdz7 1FT du7 sin z1 T K 27 du 7727 f 1147217 du72dz7 CA3 b 1 Using dz 7 fay we obtain 152 35 d au162z7 du2dz7 ll 15 2 3 1 b 1115 du5z4dz7 e ud7u7 117671 0 5 5 592 15 4 c 11177 dudz u51du7 0 d u arcsinz7 g 1 a The displacement is ff vt dt 13 t2 7 t7 2 dt 3 7 tQ 7 2t 3 337 32767lt772gt 1 b The total distance travelled is ff lvtl dt ff W 7 t7 2 dt Firsty we must determine where the velocity is positive or negative in order to remove the absolute value symbols Since vt t 7 2t 1y we see it is positive for 2 lt t g 3 and negative for 1 g t lt 21 Thusy the total distance travelled is 2 3 1 1 2 1 1 7 t27t72dt t27t72dt7 7t377t272t 7t377t272t 1 2 3 2 1 3 2 3 2 Math 2144 Final Exam Extra Review Answers 5 The curves intersect when 12 1 7 2 which leads to 212 1 and z i7 Then the area is 1 2 12 2 12 2 2 1712712dz2 172z2dz2z7 13 i 12 o 3 0 3 6 The intersection points occur at solutions to 4x 212 7 21 4 or 212 7 61 4 07which turn out to be 1 17 2 2 Then the area is 4x 7 212 7 21 4dz 13 1 7 The curves look as follows The intersection points come from 137 or z 1297 or 12 7 91 07 leading to z 07 9 The total area is then 9 16 2 1 7 73532 7 7352 0 3 6 9 2 1 2 1 2 7932779 7 7163277162 lt3 6 3 6 IE 8 We have 1 7 72 0 at z i2 Then the integral corresponding to disk cross sections is 2 2 2 2 2 4 3 5 x x x x x 1 7 7 d 2 1 7 7 7 d 2 7 7 7 127 4 x 770 2T16x W 6T80 9 a7b The curves intersect when 7 472 41 7 2 or 1 7 72 41 7 12 4 7 412 after squaring Then 412 7 72 4 71 37 and so 3 4 712 z2 which means 1 i The outer radius is 417 12 and the inner radius is 4 1 7 972 Then the volume integral is 2N3 1 2 2 2N53 15 2 5 771712277r 7 17z7 11277 77712dz7r7 gNg 2 4 0 4 16 5 10 These are two parabolas7 one opening upward and the other opening downward 9 1 16 1 2 1 0E7 zdz79 7 zdz zg27gzz This evaluates to 2732 777139 0 15 The intersection points come from 10 7 12 12 7 8 or Zzz 187 hence z i3 The area is then 3 3 3 3 4 1071271278d1 187212dx2187212dx36x7 13 73 73 O O Math 2144 Final Exam Extra Review Answers Important Formulas to Review De nition of derivative f d lim M lim M 067a z 7 a hao h Linear approximation of fz at z a fz s fd f dz 7 a L Hospital s Rule lim lim provided both functions tend to 0 or both tend to ioo and the second z7gta gz z7gta g limit converges n Sigma notation 2 di am am am2 dna an im n71 Riemann sums Left endpoints Ln Z Az i0 n Right endpoints Rn Z Az 11 271Tz 72 i n Midpoint rule Mn Az Where z 7 Trapezoid rule Tn Ax fz0 2fz1 2fz2 2fzn1 Simpson s rule Sn Ax 7f104flt11gt2fltI2gt4f132f14quot 2f1n724frn1f1n7 Where n is even and Az 12 Area integral 7 dz a 12 Volume integral by crosssection Az dz a 12 Volume of revolution integral by washer method 7rfz2 7gz2 dz where fz is the outer radius and a gz is the inner radius Important but not on nal exam 12 Volume of revolution integral by cylinder method 27rz fz dz where fz is the height of the cylindri a cal shell and z is the radius 1 Work integral Fz dz where Fz is the force exerted at position z Average value The average value of the function over the interval ab is b fltzgt dz 1 b b a bia zfzdzi


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