Math 132 Fall 2007 Final Exam 1. Calculate cos( x ) sin( x ) dx a) 1 b) c) d) e) f) g) h) i) j)

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1 Math Fall 007 Fial Exam π. Calculate cos( x ) si( x ) dx. 0 a) b) f) g) 4 c) h) d) 4 i) 4 e) 5 j) 6 Solutio: d > J : It(cos(x)*si(x)^, x 0..Pi/); J : 0 π cos( x ) si( x ) > K : studet[chagevar](u si(x), J, u); > value(k); K : 0 4 u du dx. Let F( x ) x 5 + t 4 dt. Calculate the derivative D( F )( ) of F at. + t a) 4 b) 5 c) 6 d) 7 e) 8 f) 4 g) 5 h) 6 i) 7 j) 8 Solutio: i

2 > F : (x) -> It((5+t^4)/sqrt(+t^),t x.. ); > D(F)(x); > D(F)(); > simplify(d(f)()); F : x x 5 + x t 4 dt + t + x x. Calculate dx. ( x + ) ( x + ) 0 a) f) l 9 8 l 9 5 b) g) l 7 6 l 8 c) h) l 5 4 l 9 4 d) i) l 4 l 6 e) j) l l 6 9 Solutio: a > J : It(x/(x+)/(x+),x 0.. ); J : > R : studet[itegrad](j); 0 x dx ( x + ) ( x + ) x R : ( x + ) ( x + ) > PFE : covert(r, parfrac, x);

3 PFE : + x + > atiderivative : it(pfe, x); x + atiderivative : l ( x + ) + l ( x + ) > defiiteitegral : subs(x,atiderivative) - subs(x0,atiderivative); defiiteitegral : l( ) + l( ) + l( ) > Aswer : combie(defiiteitegral, l); 8 x 4. Calculate d + x + 6 ( + x ) ( + x x. ) 0 Aswer : l 8 9 a) l( ) b) l( ) c) l( ) d) l( ) e) l( ) 4 f) 4 l( ) g) 5 l( ) h) 6 l( ) i) 7 l( ) j) 8 l( ) Solutio: i > J : It((8*x^+*x+6)/(+x)/(+x^),x 0.. ); J : > R : studet[itegrad](j); 0 8 x + x + 6 dx ( x + ) ( + x ) 8 x + x + 6 R : ( x + ) ( + x ) > PFE : covert(r, parfrac, x); x 6 PFE : + + x x + > atiderivative : it(pfe, x); atiderivative : l ( + x ) + 6 l ( x + ) > defiiteitegral : subs(x,atiderivative) - subs(x0,atiderivative); defiiteitegral : 7 l( ) 7 l( )

4 > Aswer : combie(defiiteitegral, l); e 5. Calculate x l( x ) dx. Aswer : 7 l( ) a) e b) ( e ) c) ( e ) d) ( e ) e) ( e + ) f) ( e + ) g) ( e + ) h) ( 9 e + ) i) ( 9 e + ) j) ( 9 e + ) Solutio: h > J : It(x^*l(x), x.. exp()); J : e x l( x ) > K : studet[itparts](j, l(x)); #Itegratio by Parts with ul(x) > value(k); K : ( e ) e + 9 ( e ) 9 dx x dx x 6. What is the derivative of x with respect to x at x? a) l( ) b) f) l( ) c) l( ) d) l( ) g) + l( ) h) + l( ) i) 4 l( ) e) l( ) l( ) j) 4

5 Solutio: g > restart; > eq : f(x) x^(/x); eq : f( x) x > eq : map(z-> simplify(l(z), symbolic), eq); eq : l ( f( x )) > eq : map(z -> diff(z,x), eq); x l( x) d dx f( x ) l( x ) eq : + f( x) x x > eq4 : D(f)(x) solve(eq, diff(f(x),x)); > eq5 : subs(x /, eq4); eq5 : D( f) > eq6 : subs(x /, eq); > subs(eq6, eq5); 7. If y( 0 ) 0 ad a) si( π cos( x )) b) si ( si( x )) c) eq4 : D( f)( x) f( x ) ( l( x ) ) 4 f eq6 : f 4 x x D( f ) l( ) + l dy dx cos( x ) y, the what is y( x )? cos π cos ( x ) d) cos ( si( x )) - e) arcsi( x ) f) arcsi ( arcsi( x )) g) si ( ta( x ) ) h) ta ( si( x ) ) i) arcsi ( ta( x )) j) arcsi ( arcta( x )) Solutio: b By the Method of Separatio of Variables:

6 > eq : It(/sqrt(-t^),t 0.. y(x)) It(cos(t),t0..x); > eq : map(value, eq); y( x) x eq : dt cos( t) dt t 0 eq : arcsi ( y( x )) si( x ) > Aswer : y(x) solve(eq, y(x)); Aswer : y( x ) si ( si( x ) ) 0 For those who are iterested, here is how to get MAPLE to solve this differetial equatio without the user supplyig ay guidace: > ode : diff(y(x),x)cos(x)*sqrt(-y(x)^); d ode : dx y( x ) cos( x ) y( x ) > iitialcoditio : y(0)0; iitialcoditio : y( 0 ) 0 > IVP : {ode, iitialcoditio}; d IVP : { y( 0) 0, } dx y( x ) cos( x ) y( x ) > dsolve(ivp, y(x)); #Usig Maple's differetial equatio solver y( x ) si ( si( x )) 8. Cosider the followig three statemets about a series a with positive terms: I: The series coverges because lim a 0. a + II: The series coverges because lim. ad b coverges. b a + III: The series coverges because lim. a

7 For each statemet, determie whether the reasoig is correct or icorrect. a) I: correct, II: correct, III: correct b) I: correct, II: correct, III: icorrect c) I: correct, II: icorrect, III: correct d) I: correct, II: icorrect, III: icorrect e) I: icorrect, II: correct, III: correct f) I: icorrect, II: correct, III: icorrect g) I: icorrect, II: icorrect, III: correct h) I: icorrect, II: icorrect, III: icorrect i) Wrog aswer j) Bous wrog aswer Solutio: f I) Icorrect: Whe a the terms of the series satisfy lim a 0 but the series diverges. II) Correct: The assertio follows from the Limit Compariso Test. It is true that the Limit Compariso Test is stated with the slightly differet hypothesis lim a L. b To see why the Limit Compariso Test applies oetheless, defie the series c a +. The sum of the first N terms of 0 the series Sice N c is c, which equals 0 0 N c by a. Sice a ad c have the same partial sums, they both coverge or both diverge. 0 lim c a + lim. b b

8 ad b coverges, we coclude from the Limit Compariso Test that Therefore a also coverges. c 0 coverges. III) Icorrect: The limit coditio is the icoclusive case of the Ratio Test. (Whe a we have lim a + a but the series diverges.) 9. Cosider the followig three statemets about a series a with positive terms: I: The series coverges because a < 0 +. II: The series diverges because < a. a + III: The series coverges because lim 0. a For each statemet, determie whether the reasoig is correct ( C ) or icorrect ( F ). a) I: C, II: C, III: C b) I: C, II: C, III: F c) I: C, II: F, III: C d) I: C, II: F, III: F e) I: F, II: C, III: C f) I: F, II: C, III: F g) I: F, II: F, III: C h) I: F, II: F, III: F i) Wrog aswer j) Bous wrog aswer Solutio: g

9 I) Icorrect ( F ): The series diverget..) No iformatio about II) Icorrect ( F ): The series deduced from 0 + b < a whe b is coverget. is diverget (by compariso with the diverget p-series a ca be deduced from < a b whe b is is coverget. No iformatio about a ca be III) Correct (C): Sice the limit, amely 0, is less tha, this assertio follows from the Ratio Test. 0. Cosider the three series 5 I: 0, II: 0 ad the statemets 0!, ad III: l( ) ( C ) The series coverges ( D ) The series diverges For each series, decide which of statemets (C), (D) a) I: C, II: C, III: C b) I: C, II: C, III: D c) I: C, II: D, III: C d) I: C, II: D, III: D e) I: D, II: C, III: C f) I: D, II: C, III: D g) I: D, II: D, III: C h) I: D, II: D, III: D i) Wrog aswer j) Bous wrog aswer is correct. Solutio: b

10 I) > a : -> ^5/^; > a(+), a(); > 'a(+)/a()' a(+)/a(); a ( + ) 5 a : ( + ) 5, ( + ) 5 ( + ) 5 a( ) ( + ) 5 > 'a(+)/a()' simplify(a(+)/a()); a ( + ) ( + ) 5 a( ) 5 > Limit('a(+)/a()', ifiity) limit(/*(+)^5/^5, ifiity); lim a ( + ) a( ) Sice this limit is less tha, the Ratio Test gives covergece (C). II) > b : -> 0^/sqrt(!); > b(+), b(); b : 0 ( + ), ( + )! > 'b(+)/b()' b(+)/b(); 0! 0! b ( + ) 0 ( + )! b( ) ( + )! 0 > 'b(+)/b()' 0*sqrt(!/(+)!);

11 b ( + ) 0 b( ) ( + )! > 'b(+)/b()' simplify(0*sqrt(!/(+)!)) assumig (,posit); b ( + ) b( ) + > Limit('b(+)/b()', ifiity) limit(0/(+)^(/), ifiity); lim b ( + ) b( ) 0! 0 Sice this limit is less tha, the Ratio Test gives covergece (C). III) > f : x -> /x/l(x); f : x x l( x ) > It(/x/l(x),x.. N) it(/x/l(x),x.. N); N x l( x ) dx l ( l( N ) ) l ( l( )) > It(/(x*l(x)),x.. ifiity) Limit( l(l(n))-l(l()), N ifiity); x l( x ) dx lim N l ( l( N ) ) l ( l( )) > It(/(x*l(x)),x.. ifiity) limit( l(l(n))-l(l()), N ifiity); x l( x) The give series diverges (D) by the Itegral Test. dx

12 . Cosider the two series I: ( ) ad the statemets + ( ) ad II: ( AC ) The series coverges absolutely ( CC ) The series coverges coditioally ( D ) The series diverges For each series, decide which of statemets (AC), (CC), (D) is correct. a) I: AC, II: AC b) I: AC, II: CC c) I: AC, II: D d) I: CC, II: AC e) I: CC, II: CC f) I: CC, II: D g) I: D, II: AC h) I: D, II: CC i) I: D, II: D j) Wrog aswer Solutio: h > a : -> (/(+))^; a : + We studied the limit of this sequece i cojuctio with cotiuous compoudig. > limit(a(), ifiity); e (- ) Sice this umber is ot 0, it follows from the divergece test that the series diverges (D).

13 II) > a : -> ^(/)/(+^(4/)); a : ( / ) + ( 4 / ) > b : -> /^(/); #Captures the size of a() b : ( / ) > limit(a()/b(), ifiity); Sice this umber is ot 0 ad ot, we coclude from the Limit Compariso Test that the series + 4 has the same behavior as the series, which is diverget sice it is a p-series with p. As a result, we coclude that the give series is ot absolutely coverget. Sice the give series coverges by the Alteratig Series Test, we coclude that it is coditioally coverget (CC).. Cosider the two series I: 0 π ad the statemets ad II: 0! 0 ( 00 ) ( C ) The Ratio Test establishes covergece ( D ) The Ratio Test establishes divergece ( F ) The Ratio Test is ot coclusive. Apply the Ratio Test to series I ad II ad for each, decide which of statemets (C), (D), (F) is correct.

14 a) I: C, II: C b) I: C, II: D c) I: C, II: F d) I: D, II: C e) I: D, II: D f) I: D, II: F g) I: F, II: C h) I: F, II: D i) I: F, II: F j) Wrog aswer Solutio: h I > a : -> /^Pi; a : π > Limit(a(+)/a(), ifiity) limit(a(+)/a(), ifiity); lim π ( + ) π Because this is equal to, the Ratio Test is ot coclusive (F). II > a : ->!/(0^(00*)); > r : simplify(a(+)/a()); a :! 0 ( 00 ) r : / \ / \ > limit(r, ifiity);

15 Because this limit is greater tha, the Ratio Test establishes divergece (D).. Cosider the two series + I: ad the statemets ad II: + ( C ) The Root Test establishes covergece ( D ) The Root Test establishes divergece ( F ) The Root Test is ot coclusive. Apply the Root Test to series I ad II ad for each, decide which of statemets (C), (D), (F) is correct. a) I: C, II: C b) I: C, II: D c) I: C, II: F d) I: D, II: C e) I: D, II: D f) I: D, II: F g) I: F, II: C h) I: F, II: D i) I: F, II: F j) Wrog aswer Solutio: g I > a : -> ((+^)/(0+00*^+^))^; a : > r : combie(a()^(/), symbolic);

16 + r : > Limit(a()^(/), ifiity) limit(r, ifiity); lim Because this is equal to, the Root Test is ot coclusive (F). II > a : -> ((+)//)^; + a : > r : combie(a()^(/), symbolic); + r : > Limit(a()^(/), ifiity) limit(r, ifiity); lim + Because this is less tha, the Root Test establishes covergece (C). 4. Let f( x) 7 x e ( x ). What is f ( 7 ) ( 0 )? a) 0 b) 40 c) 60 d) 80 e) 00 f) 0 g) 40 h) 60 i) 80 j) 00 Solutio: g

17 > f : x -> x^*exp(*x^)/7; f : x 7 x e ( x ) > Maclauri : series(f(x), x 0, 0); Maclauri : x 6 x5 6 x7 54 x9 O( x ) > p : covert(maclauri, polyom); p : > Aswer 7!*coeff(p, x^7); x x 5 x 7 Aswer 40 x 9 By direct calculatio (ot recommeded!): > (D@@7)(f)(x); 40 e ( x ) x e ( x ) x 4 e ( x ) 96 + x 6 e ( x ) x 8 e ( x ) > (D@@7)(f)(0); 40 x 0 e ( x ) 0 si( x 5 ) 5. Calculate L lim x 0 x cos( 5 x by fidig the Maclauri series of the umerator ) x ad the Maclauri series of the deomiator. These two Maclauri series begi with the same degree p moomial. (I other words, for the Maclauri series for both the umerator ad deomiator, the coefficiets of x are 0 for < p ad the coefficiets of x p are ozero.) What is the value of the product p L? a) - 6 b) - 48 c) - 60 d) - 7 e) - 84 f) - 96 g) -08 h) -0 i) - j) - 44 Solutio: f

18 > ratio : x -> 0*si(*x^5)/(x*cos(5*x^)-x); 0 si( x 5 ) ratio : x x cos( 5 x ) x > Limit(ratio(x), x 0) limit(ratio(x), x 0); lim x 0 0 si( x 5 ) -96 x cos( 5 x ) x 5 > series(umer(ratio(x)), x 0, 0); series(deom(ratio(x)), x 0, 0); > Aswer 5*(-96/5); 40 x 5 + O( x 0 ) x5 4 x9 O( x ) Aswer -96 ( ) ( x + ) 6. Calculate the iterval of covergece of. Let R be the radius of covergece. You will eed to calculate the sum of four itegers ad it might help to record them as you go. Let c be the base poit of the power series. ( c ) Set ρ R if R is a iteger ad - otherwise. ( ρ ) Set σ if the left edpoit belogs to the iterval of covergece ad 0 otherwise. ( σ ) Set τ if the right edpoit belogs to the iterval of covergece ad 0 otherwise. ( τ ) What is the value of c + ρ + σ + τ? a) -4 b) - c) - d) e) f) 4 g) 7 h) 8 i) 0 j) Solutio: f > c : -;

19 c : - > a : -> (-)^/sqrt(+)/4^; a : (-) + 4 > eq : R Limit(abs(a()/a(+)), ifiity); (-) + 4 ( + ) eq : R lim + 4 ( -) ( + ) > R : limit(sqrt(+)*4^(+)/sqrt(+)/4^, ifiity); > rho : 4; The left edpoit is 0, which is a + diverget p-series ( p R 4 ρ : 4 c R, or -7. Substitutig x ). ( ) 7 results i the series 0 ( ), or + > sigma : 0; σ : 0 The right edpoit is c + R, or. Substitutig x results i the series 0 ( ), which is a + coverget alteratig series because + decreases to 0. > tau : ; Fially, τ : > c+rho+sigma+tau;

20 4 7. Let T( x ) be the degree Taylor polyomial of l( x ) with base poit. What is T( ) l( )? 5 a) b) c) d) e) f) g) h) i) j) Solutio: c > c : : f : l: > T : x -> sum((d@@)(f)(c)/!*(x-c)^, 0.. ); T : x 0 ( D ( ) )( f )( c ) ( x ) > T(x); #The degree Taylor polyomial of l(x) with base poit > T()-l(); l( ) x + 8! ( x ) 8 c arcta( x ) x 8. To approximate d x x, the Maclauri series of arcta( x ) (ad, from that, 0 the Maclauri series of the itegrad) is used. A alteratig series for the (exact) value S of the defiite itegral results. A approximatio to S is obtaied by usig the miimum

21 umber of terms that, by the Alteratig Series Test, guaratee a absolute error less tha What is the approximatio? a) f) b) g) 60 c) h) d) 6 i) 80 e) j) Solutio: j > series(arcta(t), t0, ); #The startig poit. Subtract t the divide by t^. t t 5 t5 7 t7 9 t9 O( t ) > Maclauri : series((arcta(t)-t)/t^, t0, 5); #More terms tha eeded Maclauri : t 5 t 7 t5 9 t7 t9 t O( t ) > p : covert(maclauri, polyom); t t t 5 t 7 p : > eq : It((arcta(t)-t)/(t^),t 0.. x) it(p, t0.. x) + `...`; arcta( t ) t x x 4 x 6 x 8 x 0 x eq : dt t x We eed to idetify the first summad o the right side of this equatio that evaluates to less tha 0.00 whe x. The term x 6 clearly does ot do the job. Next, t 9 t > subs(x/, x^4/0);

22 0 is also ot small eough. Next, > subs(x/, x^6/4); 688 shows that this is the first term that is less tha the acceptable error. We therefore add the terms up to but ot icludig this oe. > Aswer : subs(x/, -/6*x^+/0*x^4); -7 Aswer : 960 As a verificatio, MAPLE's (exact) itegratio is: > defiiteitegral : it((arcta(x)-x)/(x^),x 0.. /); defiiteitegral : arcta + + l( 5 ) l( ) The absolute error is: > abs(evalf(aswer-defiiteitegral)); What is the coefficiet of x 5 i the Maclauri series of 8 x 4 x? a) 6 f) 4 b) g) 6 c) 8 h) d) e) 8 4 i) j)

23 Solutio: c Method : The geometric series way (as iteded): 8 x We write x 4 x Thus u where u 8 x x ( + u + u + u +... ) 4 x x 4. 8 x x x x 4 4 x x x 5... x The coefficiet of x 5 is 8. Method : The brute force computatioal way (ot recommeded!): > f : x -> 8*x/(4-x^); > (D@@5)(f)(x); f : x 8 x 4 x x x ( 4 x ) 5 ( 4 x ) 4 ( 4 x ) > Aswer (D@@5)(f)(0)/5!; Aswer x 6 ( 4 x ) 6 The Maple way: > Maclauri : series(8*x/(4-x^),x0,8); p : covert(maclauri, polyom); Aswer : coeff(p, x^5);

24 Maclauri : x x 8 x5 x7 O( x 9 ) x x 5 x 7 p : x Aswer : 8 0. What is the coefficiet of x 4 i the Maclauri series of ( + x )? a) 9 b) 9 c) 6 d) 6 e) 9 f) 9 g) h) i) j) Solutio: f Method : The Newto way (as iteded): I MAPLE, the umber "α choose ", amely α ( α )... ( α + ),! is writte as biomial(alpha, ). > NewtoFormula : (+u)^alpha Sum(biomial(alpha,)*u^, 0.. ifiity); NewtoFormula : ( + u) α biomial ( α, ) u 0 > subs({ux^, alpha -/}, NewtoFormula); 0 ( + x ) ( / ) - biomial, ( ) x

25 The coefficiet of x 4 ( or ( x ) for ) is: > biomial(-/,); 9 Method : The brute force computatioal way (ot recommeded!): > f : x -> (+x^)^(-/); > (D@@4)(f)(x); f : x ( + x ) ( / ) 4480 x x + 8 ( + x ) ( / ) 9 ( + x ) ( / ) > (D@@4)(f)(0); > (D@@4)(f)(0)/4!; ( + x ) ( 7 / ) The Maple way: > Maclauri : series((+x^)^(-/), x 0, 5); p : covert(maclauri, polyom); Aswer : coeff(p, x^4); Maclauri : + + x 9 x4 O( x 6 ) x x 4 p : + 9 Aswer : 9