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(x=0,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(x=0,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 u=l(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),t=0..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. 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

8 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 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

9 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 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)

10 > b := -> 0^/sqrt(!); > b(+), b(); b := 0 ( + ), ( + )! > 'b(+)/b()' = b(+)/b(); 0! 0! b ( + ) 0 ( + )! = b( ) ( + )! 0 > 'b(+)/b()' = 0*sqrt(!/(+)!); b ( + ) = 0 b( ) ( + )! > 'b(+)/b()' = simplify(0*sqrt(!/(+)!)) assumig (,posit); b ( + ) b( ) = 0! + > Limit('b(+)/b()', =ifiity) = limit(0/(+)^(/), =ifiity); lim b ( + ) b( ) = 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);

11 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 =. 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

12 > 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). 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).

13 . 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. 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).

14 II > a := ->!/(0^(00*)); > r := simplify(a(+)/a()); a :=! 0 ( 00 ) r := / \ / \ > limit(r, = ifiity); 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

15 i) I: F, II: F j) Wrog aswer Solutio: g I > a := -> ((+^)/(0+00*^+^))^; a := > r := combie(a()^(/), symbolic); + 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 + =

16 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 > 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 )

17 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 > 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 =

18 ) 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 := -; 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; R = 4 ρ := 4 The left edpoit is a diverget p-series ( p = c R, or -7. Substitutig x = ). 7 results i the series = 0 ( ), which is + > sigma := 0; σ := 0 The right edpoit is c + R, or. Substitutig x = results i the series = 0, which is a +

19 coverget alteratig series because + decreases to 0. > tau := ; Fially, τ := > c+rho+sigma+tau; 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 +! ( x ) 8 c

20 8 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 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), t=0, ); #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^, t=0, 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, t=0.. x) + `...`; t 9 t

21 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, > subs(x=/, x^4/0); 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:

22 > 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) 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

23 46080 x x x ( 4 x ) 5 ( 4 x ) 4 ( 4 x ) ( 4 x ) 6 > Aswer = (D@@5)(f)(0)/5!; Aswer = 8 The Maple way: > Maclauri := series(8*x/(4-x^),x=0,8); p := covert(maclauri, polyom); Aswer := coeff(p, x^5); 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 + ),!

24 is writte as biomial(alpha, ). > NewtoFormula := (+u)^alpha = Sum(biomial(alpha,)*u^, = 0.. ifiity); NewtoFormula := ( + u) α = biomial ( α, ) u = 0 > subs({u=x^, alpha = -/}, NewtoFormula); = 0 = ( + x ) ( / ) - biomial, ( ) x 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:

25 > 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