Math 142, Final Exam. 5/2/11.

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1 Math 4, Fial Exam 5// No otes, calculator, or text There are poits total Partial credit may be give Write your full ame i the upper right corer of page Number the pages i the upper right corer Do problem o page, problem o page, etc Circle or otherwise clearly idetify your fial aswer Some idetities: cos θ + cos θ, si θ cos θ, si θ si θ cos θ, 64 poits: Evaluate the followig itegrals: a e x cos x dx Name: Solutio: We itegrate by parts twice To begi, we have We obtai: u e x, du e x dx; dv cos x dx, I ex si x For the secod applicatio of parts, we have We compute: I ex si x ex si x u e x, dv si x dx, Addig /9 I to both sides gives v du e x dx; si x e x si x dx v ex cos x + + ex cos x 9 9 I ex 9 I si x + To coclude, we multiply both sides by 9/: si x + I ex cos x cos x cos x x + a dx x a ta + C a e x cos x dx + C ex si x + cos x + C

2 b 6 poits: π/ si θ cos 5 θ dθ Solutio: This is a trigoometric itegral We look for a u-substitutio, writig I π/ si θ cos 4 θcos θ dθ π/ si θ si θ cos θ dθ Hece, we set u si θ It follows that du cos θ dθ We therefore obtai I u u du u u5 5 + u7 7 ] u u + u 4 du u u 4 + u 6 du c 6 poits: t t 9 dt Some of the idetities at the top of the page may be useful; if you make a substitutio, covert the ati-derivative back to the variable t Solutio: We require a trigoometric substitutio Sice the itegral cotais t 9, a suitable substitutio is t sec θ To see this, oe ca draw for referece a right triagle with base agle θ, hypoteuse t, side opposite of legth t 9, ad side adjacet of legth We the compute dt sec θ ta θ dθ ad t 9 sec θ 9 9sec θ 9 ta θ ta θ Alteratively, the referece triagle gives ta θ opposite adjacet t 9 We substitute i the itegral to obtai sec θ ta θ I sec θ ta θ dθ 7 sec θ dθ cos θ dθ + cos θ dθ cos θ dθ si θ θ + + C θ + si θ cos θ + C t sec + t 9 + C 54 t We ote that i the above calculatio, we used: cos + cos θ θ, si θ si θ cos θ, cos θ t, si θ t 9, t t t sec θ sec θ

3 x + x + x + d 6 poits: x + x + dx Solutio: We use partial fractios to evaluate the itegral We have x + x + x + x + x + Ax + B x + + Cx + D x + We multiply by x + x + to clear deomiators, obtaiig x + x + x + Ax + Bx + + Cx + Dx + To fid the costats, we compare coefficiets : Ax + Bx + Ax + B + Cx + Dx + Cx + D A + Cx + B + Dx + A + Cx + B + D x : A + C, x : B + D, x : A + C, x : B + D Subtractig first equatio from the third yields A ; subtractig the secod equatio from the fourth yields B Hece, we also fid that C ad D To coclude, we compute I x + dx + x x + dx ta x + u du ta x + lx + + C We ote that i the secod itegral, we made the substitutio u x + 6 poits: Determie whether the itegral dx is coverget or diverget x + If it coverges, fid its value At some poit i the calculatio, you will eed to make a substitutio Solutio: This is a improper itegral We compute t I lim 6 lim x x + dx lim t The itegral coverges to the value /4 t + x u du lim ] u t + 5 poits: Use a compariso test your choice to determie whether the series coverges or diverges Solutio: For large, we have a 7 7 Therefore, we apply the limit compariso test LCT with b We have 5/ / / lim / lim / / 5/ lim > + +

4 Sice b / it is a p-series with p / <, the LCT implies that a diverges 4 poits: Determie whether the series coverges coditioally, or diverges! 6 9 coverges absolutely, Solutio: We apply the ratio test We have L lim a + a lim + +! ! + lim + < Hece, the series coverges absolutely 5 poits: Fid the radius of covergece ad the iterval of covergece of the power x series Test the edpoits of the iterval, if ecessary Solutio: We apply the ratio test: L lim x x x lim + x x < lim + x < < x < < x < 4 < x < Therefore, the radius is R / it is half of the legth of the iterval of covergece It remais to test the edpoits x The series is It is a p-series with p / < ; therefore, it diverges x The series is We observe that the sequece of absolute values of the terms is decreasig ad teds to zero Therefore, the alteratig series test AST implies that this series coverges We coclude that the iterval of covergece is I, ] 4

5 6 poits: Set up a itegral or sum of itegrals which gives the volume obtaied by revolvig the give regio R aroud the give lie L You may use disks, washers, or shells Do ot compute the itegrals Sketch the give regio ad a sample rectagle or rectagles i the regio to help justify your reasoig Label the sketch as ecessary What are the poits of itersectio of the equatios givig R? What is the thickess of a sample rectagle? a 5 poits: R : y the x-axis, y x, x + y ; L : y the x-axis Solutio: The itersectio of y x ad x + y occurs at, To fid the volume usig a sigle itegral, a sample rectagle i R should be orieted horizotally ad have thickess y Sice the axis is also horizotal, we use shells ad itegrate with respect to y The radius is ry y, ad the height legth of the rectagle is hy y y y Therefore, we have Alteratively, oe ca use disks: Volume The commo value is π/ b 5 poits: Volume πx dx + 4πy y dy π x dx R : x the y-axis, y x +, y x + ; L : y Solutio: The itersectio of y x + ad y x + occurs at, To fid the volume usig a sigle itegral, a sample rectagle i R should be orieted vertically ad have thickess x Sice the axis is horizotal, we use washers ad itegrate with respect to x The outer radius is r x x + x; the ier radius is r x x + x Therefore, we have Volume Alteratively, oe ca use shells: Volume The commo value is π πy dy + π x x dx πy y dy 5

6 7 poits: Fid a power series represetatio of the form What is the radius of covergece? a poits: Solutio: + x + x x c x for each fuctio x + x, x < x b poits: + x Differetiate the power series from part a Suitably adjust the resultig derivative Solutio: We first observe that d dx + x + x Hece, we have x + x d x x dx + x + x + + d x x dx + x, x < x + 6

7 8 5 poits: Let fx cos x a 5 poits: Write dow the Taylor series for fx at a This is sometimes called the MacLauri series for fx What is its radius of covergece? You do ot have to give justificatio You ca simply write it dow from memory, or you ca derive it from scratch Solutio: cos x x! x + x4 4 x6 7 +, x R cos x + x b 5 poits: Use your aswer from a to compute the limit lim x4 4 x x 6 You are ot allowed to use L Hôpital s rule Solutio: We compute x lim + x4 x6 + + x x x x 6 c 5 poits: Use part a to evaluate the sum Give a exact umerical value π Solutio: From part a, the sum is cos 4 lim x x6 π 4! + 7 x 6 7 6! 7

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME

MH1101 AY1617 Sem 2. Question 1. NOT TESTED THIS TIME MH AY67 Sem Questio. NOT TESTED THIS TIME ( marks Let R be the regio bouded by the curve y 4x x 3 ad the x axis i the first quadrat (see figure below. Usig the cylidrical shell method, fid the volume of