1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

Size: px

Start display at page:

Download "1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute"

Osborne Marsh
5 years ago
Views:

1 Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral, we have a ad b, so j a + j + 4 j. Thus the limit defiitio of the defiite itegral gives 4 ad 4 d j j j f j ) ) ) 4 j 4 4 j + j 4 j 4 j j 4 j lim j j ) 4 j ) + ) 4 ) + ) + ) ) + ) 4 ) + ) + ) lim ) ). Now, usig the Evaluatio Theorem to check our work, we fid 4 d 4 4) ) 4 ) ) poits) Suppose f is the cotiuous fuctio give by f) e t dt. i a) What is f)? Solutio: By defiitio of f, f) e t dt, so f) because the defiite itegral of ay fuctio from a to a is always zero. b) Is f) positive or egative? Support your aswer. Solutio: By defiitio of f, f) e t dt, so f) is positive because e t > e o the iterval,, which implies that f) e e >., i i + ), i i + ) + ).

2 Math, Calculus II Fial Eam Solutios c) Fid f ). Solutio: By the Fudametal Theorem of Calculus, we have f ) d e t dt e. d d) If F ) fsi ), fid F ). Solutio: Usig the Chai Rule together with our aswer i part c), we have F ) f si ) d d si e si ) cos e si cos. 5 poits) Cosider the fuctio f) o the iterval,. a) Compute d. Solutio: By the Evaluatio Theorem, we have d l l l l l b) Use the Trapezoidal Rule with 4 to approimate d. Solutio: With 4 subitervals, we have 4, ad j + j for j,,,, 4. Thus the Trapezoidal Approimatio is d T 4 f) + f.5) + f) + f.5) + f) / + /) + /) + /5) + /) 7. c) Accordig to the error formula for the Trapezoidal Rule what is the maimum possible error for this approimatio i part b)? Compare this to the actual error. Solutio: First, with f), we have f ) ad f ). So, o the iterval,, we have f ) /, so take M i the error formula: E 4 ) 4) ).8 So the maimum possible error i part b) is.8, but the actual error is E 4 7 l which is substatially smaller tha the maimum possible error. If f ) M o a, b, the the error formula for the Trapezoidal Rule is E Mb a).

3 Math, Calculus II Fial Eam Solutios 4. poits) The gamma fuctio Γ) is defied for all > by Γ) e t t dt. a) Evaluate Γ). Solutio: To evaluate Γ), we substitute i for withi the itegrad, so the epoet of t is ow ) ad the itegral simplifies to Γ) e t t dt e t dt. Thus we have Γ) e t dt e t dt e t b e b + e + b) For >, show that Γ) )Γ ). Assume that all these improper itegrals eist ad use itegratio by parts. Solutio: Followig the suggestio, we use itegratio by parts with u t, du )t dt; dv e t dt, v e t Therefore, recallig the formula u dv uv v du, we have Γ) e t t dt t ) e t dt) t e t ) ) b e t ) )t dt b e b e ) + e t )t dt + ) e t t dt )Γ ) c) Use parts a) ad b) to fid Γ), Γ), Γ4). What is the patter? Solutio: By part b), we fid that Γ) )Γ ) Γ), Γ) )Γ ) Γ), Γ4) 4 )Γ4 ) Γ) ) From just these few computatios, we ca already see the patter emerge that Γ) )Γ ) ) )Γ ) )! 5. poits) Cosider the regio R i the first quadrat bouded o the left by the parabola y ad o the right by the lie y. a) Sketch the curves ad shade the regio below: y y

4 Math, Calculus II Fial Eam Solutios b) Set up, but do ot evaluate!, itegrals that compute i. the area of R; Solutio: Settig it up with respect to, the parabola is o top ad the lie is the bottom, so the area is equal to A ) ) d If, o the other had, we itegrate with respect to y, the the lie is o top while the parabola is the bottom, so the area betwee them is A y y) y ) dy ii. the volume of the solid obtaied by revolvig R about the y-ais; Solutio: If we revolve about the y-ais, we should set the itegral up with respect to y. As i part a), the lie y is o top ad the parabola y is o bottom, so the volume we obtai is give by V y πy) πy ) dt y πy y 4 ) dy iii. the volume of the solid obtaied by revolvig R about the -ais. Solutio: If we ow revolve the regio R about the -ais, we should write our itegral with respect to, i which case y is o top ad y is o bottom, so we have V π ) π) d π ) d. 5 poits) Phoebe Small is caught speedig. The fie is $. per miute for each mile per hour above the speed limit. Sice she was clocked at speeds as much as 4 MPH over a -miute period, the judge fies her $.)umber of miutes)mph over 55) $.))4 55) $. Phoebe believes that the fie is too large sice she was goig 55 MPH at times t ad t miutes, ad was goig 4 MPH oly at t. She reckos, i fact, that her speed v was give by v 55 + t t. a) Show that Phoebe s equatio does give the correct speed at times i. t : v) 55 + ) ) 55 ii. t : v) 55 + ) ) 4 iii. t : v) 55 + ) ) 55 b) Phoebe argues that sice her speed varied, the fie should be determied by her average speed rather tha her maimum speed. What was Phoebe s average speed?

5 Math, Calculus II Fial Eam Solutios Solutio: Assumig that Phoebe s speed is give by vt) 55 + t t, as we will, her average speed is v ave vt) dt ) ) 55t + t t 55 + t t dt 55) + ) ) ) ) Thus Phoebe s average speed is mph, which is still over the speed limit, so she should still pay a fie. c) What should Phoebe propose to the judge as a reasoable fie? Solutio: Based o Phoebe s average speed, her fie should be $.)umber of miutes)average MPH over 55) $.)) 55) $8. Thus, if the judge goes alog with Phoebe s argumet, her fie would oly be $ poits) The Riema zeta-fuctio ζ is defied by ζ) ad is used i umber theory to study the distributio of prime umbers. a) What is the domai of ζ? That is, for what values of does coverge?) Solutio: Lookig at the right-had side of the Riema zeta-fuctio, we recogize it to be a p-series with p. We kow that p-series oly coverge whe p > ad that they diverge wheever p, so the domai of the Riema zeta-fuctio is, ) { : > }. b) Approimate the value of ζ) by usig the sum of the first terms, s. Solutio: We use the approimatio ζ) s c) Estimate the error ζ) s usig the error boud R Solutio: The error i part b) is, at worst, ζ) s R d f) d. d b b d) How large must we take to isure the error R ζ) s is less tha.? Solutio: To make sure R <., we ll fid large eough so that d <.. Now d d b b so we wat <.. R R is less tha.. Hece we eed <, so take to esure that

6 Math, Calculus II Fial Eam Solutios 8. 5 poits) Cosider the series ). a) Fid the series radius of covergece. Solutio: To fid the radius of covergece, we use the Ratio Test: lim a + a ) + / + ) ) / ) + Accordig to the Ratio Test, the power series coverges absolutely) wheever < ad diverges whe >. Now < < < < < < < has width /, so the radius of covergece is R /. b) Fid the iterval of covergece of the series. Solutio: So far, we kow that the power series coverges for / < < ad diverges whe < / or >. It oly remais to check the two edpoits: / ) ) i. /: The series at this edpoit is, which is the Alteratig Harmoic Series that we already kow to coverge by the Alteratig Series Test, i case you do t remember). So / is i the iterval of covergece. ii. : The series at this edpoit is ), which is the Harmoic Series that we already kow to diverge or you may thik of it as p-series with p ). Thus is ot i the iterval of covergece. Therefore, the iterval of covergece is /, ) { : / < }. c) Where does the series coverge absolutely? Where does it coverge coditioally? Solutio: Accordig to the Ratio Test, we kow that the series coverges absolutely o the iterval /, ) { : / < < }. By part b), we also kow that the series coverges at / but ot at, so it is oly coditioally coverget at /. Thus the power series coverges absolutely o /, ) coverges coditioally at / diverges wheever < / or 9. 5 poits) Use substitutio, multiplicatio, differetiatio, ad/or itegratio to fid Taylor series cetered at a for the followig fuctios: a) + Solutio: b) cos) + ) ) ) ) +

7 Math, Calculus II Fial Eam Solutios Solutio: Sice si + ) + ) ) + )! ) c) ta )! d + )!, cos d ) )!. 4) )!. Solutio: We have ta + ) +. d) ta Solutio: ta e) e Solutio: e )! d + si ) d + )! Thus cos) ) ) )! + t dt t ) dt t + ) + ta + ) + ) +.! +. 5 poits) Fid the Taylor polyomial of order 4 i.e., oly up to the a) 4 -th term) geerated by f) si cetered at a π/. Solutio: I order to fid the Taylor polyomial of order 4 cetered at a π/ for f) si, which is 4 f ) a) T 4 ) a)! we eed to kow the first 4 derivatives of f) ad their values at a π/. So we fid f) si fπ/) siπ/) / f ) cos f π/) cosπ/) / f ) si f π/) siπ/) / f ) cos f π/) cosπ/) / f 4) ) si f 4) π/) siπ/) / Thus, the Taylor polyomial of order 4 geerated by f) si cetered at a π/ is T 4 ) 4 f ) π/)! π ) π ) f π/) +! fπ/) + f π/)! + / π ) + / + π ) π 4 π! π ) f π/) +! ) / + ) π π ) π ) f 4) π/) + 4! ) / + 4 π ) 4 π ) 4 π ) 4

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this