Math- 106, FALL 2005 Solutions to Final Exam Problems

1-a) Using the identity ( )2 1sin 1 cos22

θ θ= − , we find

( )/ 4/ 4 / 4

2

00 0

1 1 1 1 1 1sin 1 cos2 sin 22 2 2 2 4 2 8 4

d dππ π π πθ θ θ θ θ θ = − = − = − = − ∫ ∫ .

b) Using the substitution 29u x= − , with 2du x dx= − and 12

x dx du= − , we find

( ) ( )1/ 2 3/ 22 1/ 2 3/ 2 21 1 2 19 92 2 3 3

x x dx u du u C x C− = − = − + = − − +∫ ∫ .

c) We can break the integral up: 1/ 2 1/ 2 1/ 2

2 2 21/ 2 1/ 2 1/ 2

1 2 1 21 4 1 4 1 4

x xdx dx dxx x x− − −

+= +

+ + +∫ ∫ ∫ .

Since 2

21 4

xx+

is an odd function, 1/ 2

21/ 2

2 01 4

x dxx−

=+∫ .

To evaluate1/ 2

21/ 2

11 4

dxx− +∫ , we make the substitution 2u x= , with 2du dx= .

Now1/ 2

21/ 2

11 4

dxx− +∫ [ ]

11

2 11

1 1 1arctan2 1 2 2 4 4 4

du uu

π π π−

−

= = = − − = + ∫ .

Thus 1/ 2

21/ 2

1 21 4 4

x dxx

π

−

+=

+∫ .

d) / 2 / 2

7 6

0 0

cos cos cost dt t t dtπ π

=∫ ∫ . Now use parts, with 6cosu t= , 56cos sindu t t= − ,

cos , and sindv t dt v t= = . Thus / 2 / 2

7 6

0 0

cos cos cost dt t t dtπ π

=∫ ∫

( )/ 2 / 2

/ 26 5 2 5 2

00 0

cos sin 6 cos sin 6 cos 1 cost t t t dt t t dtπ π

π = + = − ∫ ∫

/ 2 / 2

5 7

0 0

6 cos 6 costdt tdtπ π

= −∫ ∫ . It follows that / 2 / 2

7 5

0 0

87 cos 6 cos 615

tdt tdtπ π

= =∫ ∫ , so

/ 2

7

0

6 8 48cos7 15 105

tdtπ

= =∫ .

e) First we compute 21

lnb xdxx∫ . We will use parts, with ln , dxu x du

x= = ,

2 1, anddv x dx vx

−= = − . Now 2 21 11 1

ln ln ln 1 ln 1 1b bb bx x dx b bdx

x x x b x b b = − + = − − = − − + ∫ ∫ .

Then 2 21 1

ln ln ln 1lim lim 1 1b

b b

x x bdx dxx x b b

∞

→∞ →∞

= = − − + = ∫ ∫ .

2- a) ( )

( ) ( )∑∑∞

=

∞

=

⋅−=+

−

111

1ln1

nn

n

n

n

un

where ( )1ln1+

=n

un for 1≥n .

(1) ( )1ln1+

=n

un are all positive.

(2) 1+≥ nn uu , since ( ) ( )2ln1

1ln1

+≥

+ nn for all 1≥n .

(3) ( ) 01ln

1→

+=

nun as ∞→n .

Then by the Alternating Series Test, the series converges.

b) Here, 12

12

2

−++

=nn

nan and 021

121limlim 2

2

≠=−+

+=

∞→∞→ nnna

nnn. Therefore by the n-th

Term Test, the series diverges.

c) First, notice that !1

nnan+

= are all positive and ( )( ) ( ) ( )2

1

12

1!

!12

++

=+

⋅++

=+

nn

nn

nn

aa

n

n .

So ( )

.1012limlim 2

1 <=++

=∞→

+

∞→ nn

aa

nn

n

n Hence, by the Ratio Test, the series converges.

d) 2

110 2

2

nee

nn ≤=≤ − for 1≥n and∑

∞

=12

1n n

converges (this is a p-series with 12 >=p ).

Therefore ( by the Direct Comparison Test) ∑∞

=

−

1

2

n

ne converges, too.

3) Domain of =y Ñ\{1} and 0→y only when −∞→x but y is never equal to 0. So, x-intercept does not exist.

At ⇒−=−

== 110

,00eyx ( )1,0 is the y-intercept.

∞=−∞→ 1

limxe x

x and 0

1lim =

−−∞→ xe x

x. So 0=y is the horizontal asymptote.

Since, ∞=−+→ 1

lim1 x

e x

x and −∞=

−−→ 1lim

1 xe x

x, 1=x is the vertical asymptote.

( )( )

( )( )

01

211

22 =′⇒−−⋅

=−

−−⋅=′ y

xxe

xexey

xxx

when 2=x (critical point at 2=x ).

Note: y′ is undefined at 1=x , but 1 is not in the domain of 1−

=xey

x

.

Next, ( )( ) ( ) ( ) ( )( )

( ) ( )( )( )3

2

4

2

1221

121212

−−−−⋅

=−

−⋅⋅−⋅−−⋅+−⋅=′′

xxxe

xxexxexey

xxxx

.

( ) ( ) 05402210 22 =+−⇒=−−−⇒=′′ xxxxy . This equation has no real roots , since 042016 <−=−=∆ (i.e. The discriminant is negative). This implies that, there is no inflection point. We have y ′′ is undefined at 1=x , which is not in the domain of y . So we have

Combining our results;

So the graph of y will be as follows ;

4- a)

( )222 10 xxh −=+

( ) xxxxxxh 21002010010 22222 −=−+−=−−= xh 2100 −=⇒ .

So, ( ) xxxV 210031 2 −⋅⋅= where 50 ≤≤ x .

b) ( ) ( ) ( )20210021

34201002

32

21

2 −⋅−⋅⋅+−⋅⋅=′ −xxxxxV

x

xxx20100

134020100

38 2

−⋅−−⋅= .

( )x

xxxxV20100

13402100

380 2

−⋅=−⋅⇒=′

42510052100 =⇒=⇒=−⇒ xxxx . Since ( ) ( ) 050 ==VV , ( )4V is the maximum volume. So the length side .4mx = and the height .522080100 mh ⋅==−= 5-a) Use the ratio test (or the root test) to find the radius of convergence.

11

1 3 1 1lim lim 3 lim 322 3

nnn

nnn n nn

a x n nx xa nn x

+++

→∞ →∞ →∞

+ += = = ++

. If 3 1x < , or 13

x < , then

the power series converges; if 3 1x > , or 13

x > , then the power series

diverges. Thus the radius of convergence is 13

R = .

To find the interval of convergence, we need to examine the cases 13

x = and

13

x = − . In the case 13

x = the power series takes the form ( )0

11

n

n n

∞

=

−

+∑ ; this

Leibniz series converges. For 13

x = − we have 0

11n n

∞

= +∑ , a divergent p-series .

Thus the interval of convergence is 1 1,3 3

− .

b)2 3 5 3 3

2sin 1 ... ... ...2 6 120 6 2

x x x x x xe x x x x x = + + + − + − = − + + +

=

3

2 ...3xx x

+ + +

.

c) 22

1 3 2 (1 2 )(1 ) 1 2 1 (1 2 )(1 )x x A B A Ax B Bx

x x x x x x x x− + −

= = + =− + − − − − − −

. We must have

0A B+ = and 2 1A B− − = , so 1A = and 1B = − . Now 21 3 2x

x x− +

( ) ( )2 21 1 1 2 4 ... 2 ... 1 ... ...1 2 1

n n nx x x x x xx x

= − = + + + + + − + + + + +− −

( ) ( )2

1

3 ... 2 1 ... 2 1n n n n

nx x x x

∞

=

= + + + − + = −∑ .

6- a) ( ) ( ) ( ) ( ) ( ) ( ) ( ).1

11

cos1

1cos1lnsinlim

sin1lncoslim1lncotlim

0'

00==+

++⋅−=

+⋅=+⋅

→↑→→ xx

xxx

xxxxx

xHL

xx

b) Let x

xy

+=

31 , then

+⋅=

xxy 31lnln and

03

3lim31

3lim1

31

/3

lim1

31lnlimlnlim

0

2

20

2

2

0'

00=

+=−⋅

+⋅

−=

−

+

−

=

+

=+++++ →→→↑→→ x

xx

xx

x

x

x

x

xyxxx

HLxx

.

So, 1lim

0=

+→y

x.

c) ( )1lnln 1 +⋅= + xey x and by taking derivatives of both sides with respect to x, we have

( )1

11ln1 11

+⋅++⋅=⋅ ++

xexe

dxdy

yxx and ( )

111ln 11

+⋅++⋅⋅= ++

xexey

dxdy xx

So, ( ) ( )

+++⋅+= ++

111ln1 11

xxex

dxdy xex

.

d) ( ) ( ) ( ) ( ) ( ) 2lnsin2sincos2

cos12 ⋅−⋅+−⋅⋅=′ − θθθθθ

f .

e) ( )2

1

22lncosh

1lnln xxxxeex

xx +=

+=

+=

−−

, so

( ) ( )( ) ( ) ( ) ( )

+=

+⋅+=

+⋅+=⋅+

xx

dxd

xxx

dxdx

xx

dxdxx

dxd

21

211

2

1

1lncosh1222

222

( ) ( ) ( ) ( ) ( )2

2424

2

222

2

2222

21244

2112

421212

xxxxx

xxx

xxxx −−−+

=+

−+=⋅+−⋅+

=

−+=

−+= 2

22

24 12321

2123

xx

xxx .