Math 156 Applied Honors Calculus II Final Exam Review Sheet Solutions Fall 2016 1. True or

Math 156 Applied Honors Calculus II Final Exam Review Sheet Solutions Fall 20181. True or False

a) TRUE. 1n + 2

n + 3n + · · ·+ n

n =nP

i=1

in = 1

n ·nP

i=1i = 1

n · n(n+1)2 = n+1

2

b) TRUE. limn!1

nPi=1

f0(xi)�x =

R ba f

0(x)dx = f(b)� f(a) by FTC

c) FALSE. If n is doubled, then �x decreases by a factor of 12 and the error in the right-hand Riemann sum decreases by

a factor of 12 , not

14 , as we saw in examples in class.

d) TRUE. Apply integration by parts. TO BE COMPLETED

e) FALSE. SinceR 10

dxx2 diverges by p�test )

R10

dxx2 diverges.

f) FALSE. When the spring is stretched from length 20 cm to 30 cm, the work done isR 30�L20�L kxdx =

R 30�2020�20 kxdx =

R 100 kxdx = 1

2kx2��100

= 50k = 2Joule. Then when the spring is stretched from length 30 cm to 40 cm, the work done isR 40�L30�L kxdx =

R 40�2030�20 kxdx =

R 2010 kxdx = 1

2kx2��2010

= 12k(400� 100) = 150k = 3 · 50k = 6Joule.

g) FALSE. TO BE COMPLETED

h) FALSE. A counterexample is an exponential distribution, f(x) attains its maximum value at x = 0 rather than µ = 1c .

(However the statement is true for a normal distribution.)

i) TRUE.R1�1(x� µ)f(x)dx =

R1�1 xf(x)dx�

R1�1 µf(x)dx = µ� µ · 1 = 0

j) FALSE. After 100 years the sample has mass 12kg, and after 400 years the sample has mass

�12

�4= 1

16 kg.

k) TRUE. continuous compounding ) y(t) = y0ert = 1000e0.05t ) y(2) = 1000e0.05·2 = 1000e0.1; hence we need to show

that 1105 < 1000e0.1 < 11121st part : we need to show that 1105 < 1000e0.1; recall that ex = 1 + x + 1

2x2 + · · · , so e

0.1 = 1 + 0.1 + 12 (0.1)

2 + · · · =1 + 0.1 + 0.005 + · · · ; then summing the first three terms yields e

0.1 = 1.105 + · · · , and since the remainder is positive,we have e

0.1> 1.105, thus 1000e0.1 > 1000 · 1.105 = 1105

2nd part : we need to show that 1000e0.1 < 1112; this is equivalent to showing that e0.1

< 1.112; in this case letus consider e

�x; it is fairly easy to see that e�x

> 1 � x for x 6= 0; for example, consider the graph of e�x and

1 � x; then let us set x = 0.1, so that we have e�0.1

> 1 � 0.1; this implies that e�0.1

> 0.9; and this implies thate0.1

<10.9 = 1

910

= 109 = 1.1111 · · · < 1.112; so we’re done

l) FALSE. y(t) = 0 is a constant solution of the di↵erential equation, but it is unstable by looking at the phase plane.

m) FALSE. This is only true if the constant solution is stable.

n) FALSE. If the step size �t decreases, then the error also decreases.

o) FALSE. A counterexample is an = 1n , bn = n

2.

p) FALSE. A counterexample is an = 1n2 , bn = 1

n , so that1P

n=0an converges, but

1Pn=0

bn diverges.

q) FALSE.

Method 1: draw a graph of y = 1x2 for x � 1 and note that

1Pn=1

1n2 is a left-hand Riemann sum for

R11

dxx2 , and hence the

sum is larger than the integral.

Method 2:1P

n=1

1n2 = ⇡2

6 ,R11

dxx2 = �1

x

��11

= 1

r) FALSE.Method 1: The error bound for a convergent alternating series says that |s � sn| < an+1. If we set n = 1, then|s� s1| < a2 ) |s� 1| < 1

2 ) 12 < s <

32 , so s 6= 0.

Method 2: In class we showed that 1� 12 + 1

3 � 14 + · · · = ln 2 6= 0.

s) FALSE.1P

n=1

1n(n+1) =

1Pn�1

( 1n � 1n+1 ) =

1Pn=1

1n �

1Pn=1

1n+1 = 1�1 = 0

The 1st step is correct, but the 2nd step is incorrect; the original sum is equal to 1 (as shown on homework), so it isincorrect to write 1 = 1�1.

t) FALSE. The AST cannot be used to show that a series diverges.

u) FALSE. limn!1

��an+1

an

�� = limn!1

n2

(n+1)2 = 1 ) the ratio test is inconclusive

v) FALSE. A counterexample isP

n=1(�1)nxn, which converges for x = 1 by the AST, but diverges for x = �1 since itis the harmonic series in that case.

1

w1) TRUE. The radius of convergence is at least R = 1, so the interval of convergence is at least 0 < x 2, whichcontains x = 1

2 .

w2) FALSE. geometric series, r = 2(x�1), converges for�1 < r < 1 , �1 < 2(x�1) < 1 , � 12 < x�1 <

12 , 1

2 < x <32 ,

diverges at end points x = 12 ,

32

x) TRUE. 11+x = 1

1�(�x) =1P

n=0(�1)nxn, then di↵erentiating both sides yields � 1

(1+x)2 =1P

n=1(�1)nnxn�1 ) 1

(1+x)2 =

1Pn=1

(�1)n+1nx

n�1 =1P

n=0(�1)n(n+ 1)xn, and we can check that the ioc is �1 < x < 1

y) FALSE. Don’t try to find f(3)(0), f (6)(0) directly. Instead use the Taylor series to derive them, since the general form

of a Taylor series is f(x) =1P

n=0cnx

n, where cn = f(n)(0)n! ) f

(n)(0) = n! · cn. Since ex = 1 + x + 1

2x2 + 1

3!x3 + · · · , then

e�x2

= 1 � x2 + 1

2x4 � 1

6x6 + · · · . Then c3 = 0, since there is no x

3 term, and hence f(3)(0) = 0. Then c6 = � 1

6 , thus

f(6)(0) = 6! · c6 = 720 ·

�� 1

6

�= �120. Thus the statement is false.

z) TRUE.part 1 : Since e

x = 1 + x + 12x

2 + 13!x

3 + · · · , then setting x = 1, we have e = 1 + 1 + 12 + 1

3! + · · · , and it is clear thate > 2.part 2 : Setting x = �1 in the power series for ex, we have e

�1 = 1� 1 + 12 � 1

3! + · · · , then using the error bound for aconvergent alternating series we have |e�1 � 1

2 | <16 ) 1

3 < e�1

<23 , then e

�1>

13 ) e < 3.

aa) TRUE. ex = 1 + x + 12x

2 + · · · ) e�x2

= 1 + (�x2) + 1

2 (�x2)2 + · · · = 1 � x

2 + 12x

4 � · · · Next we note that

e�x2

> 1 � x2 for x > 0; to show this we can sketch the graph of e�x2

and 1 � x2. To confirm the sketch, consider the

function f(x) = e�x2 � (1� x

2), note that f(0) = 0 and f0(x) = �2xe�x2

+ 2x = 2x(1� e�x2

) > 0 for x > 0, so f(x) is

an increasing function, and this implies f(x) > 0 for x > 0, and hence we have e�x2

> 1 � x2 for x > 0. Then we haveR 1

0 e�x2

dx >R 10 (1� x

2)dx = (x� x3

3 )��10= 1� 1

3 = 23 .

bb) TRUE. Since the series is sin ⇡2 (note that sinx = x� 1

3!x3 + 1

5!x5 � 1

7!x7 + · · · ) and sin ⇡

2 = 1.

cc) TRUE. cosh2x� sinh2x =�ex+e�x

2

�2 ��ex�e�x

2

�2= e2x+2+e�2x

4 � e2x�2+e�2x

4 = 1.

dd) FALSE.Rtanhxdx =

Rsinh xcosh xdx =

R1

cosh xd coshx = ln(coshx) 6= sech2h. Note that (tanhx)0 = sech2x.

ee) TRUE. Note that T1(x) = f(a) + f0(a)(x� a), thus T 0

1(a) = f0(a).

↵) FALSE. In class we showed that the function defined by f(x) = e�x2

for x 6= 0 and f(0) = 0 has the property thatf(n)(0) = 0 for all n, yet f(x) 6= 0 for x 6= 0.

gg) TRUE. ex = 1+x+ 12x

2+ · · · , so e�0.1 = 1+(�0.1)+ 1

2 (�0.1)2+ · · · = 1� 0.1+0.005� · · · , so |e�0.1� 0.9| 0.005,so 0.895 e

�0.1 0.905.

hh) FALSE. Using the binomial series (1 + x)k = 1 + kx + · · · , replace x with x2, then (1 + x

2)k = 1 + kx2 + · · · , and

then setting k = 12 , we have

p1 + x2 = (1 + x

2)12 = 1 + 1

2x2 + · · · .

ii) TRUE. Since cosh ix = eix+e�ix

2 = cos x+i sin x+cos x�i sin x2 = cosx

jj) TRUE.

part 1 :R ⇡/20 sin2 ✓d✓ =

R ⇡/20 ( 12 � 1

2 cos 2✓)d✓ = ( 12✓ �12 · 1

2 sin 2✓)��⇡/20

= ⇡4

part 2 :R ⇡/20 cos2 ✓d✓ =

R ⇡/20 ( 12 + 1

2 cos 2✓)d✓ = ( 12✓ +12 · 1

2 sin 2✓)��⇡/20

= ⇡4

kk) TRUE. Since e⇡i = cos⇡ + i sin⇡ = �1 ) ⇡i = log(�1). (Actually, more rigorously, log(�1) = (2k + 1)⇡i, where

k is an integer.)

ll) TRUE.�83

�= 8!

3!5! =8·7·63·2·1 = 8 · 7 = 56

mm) FALSE.�63

�= 6!

3!3! =6·5·43·2·1 = 20

nn) TRUE.�102

�= 10!

2!·8! =10!8!·2! =

�108

�

oo) TRUE.�73

�+�74

�= 7!

3!·4! +7!

4!·3! =2·7!3!·4! =

8·7!4·3!·4! =

8!4!·4! =

�84

�

pp) TRUE. (a+ b)k =kP

n=0

�kn

�ak�n

bn, then setting a = 1, b = �1 yields

kPn=0

�kn

�(�1)n = (1 + (�1))k = 0k = 0

Question 2 Solutiona) geometric series : sum = 1

1� 20162017

= 20172017�2016 = 2017

b) It equalsR 10 (1 + x)dx =

⇣x+ x2

2

⌘��1

0= 3

2 (change in ! x, 1

n ! dx)

c) It equalsR 10

11+xdx = ln(1 + x)|10 = ln 2

2

d) By L’Hospital’s rule, limx!0

sin xx = lim

x!0

(sin x)0

(x)0 = limx!0

cos x1 = cos 0 = 1.

e) By L’Hospital’s rule, limx!0

1�cos xx2 = lim

x!0

(1�cos x)0

(x2)0 = limx!0

sin x2x = 1

2 .

f) Note that limn!1

�1 + x

n

�2n= ( lim

n!1

�1 + x

n

�n)2 = (ex)2 = e

2x

g) By L’Hospital’s rule limx!0

p1+x�1x = lim

x!0

(p1+x�1)0

x0 = limx!0

12p

1+x

1 = 12

h) limh!0

(x+h)4�x4

h = limh!0

x4+4x3h+6x2h2+4xh3+h4�x4

h = limh!0

4x3h+6x2h2+4xh3+h4

h = limh!0

(4x3 + 6x2h+ 4xh2 + h

3) = 4x3 (one

can use L’hospital rule, note that ‘h’ is variable here, regard x as constant)

i) By L’Hospital’s rule, limh!0f(x+h)�f(x)

h = limh!0f 0(x+h)

1 = f0(x).

j) Using L’Hospital’s rule twice, limh!0f(x+h)�2f(x)+f(x�h)

h2 = limh!0f 0(x+h)�f 0(x�h)

2h = limt!0f 00(x+h)+f 00(x�h)

2 = f00(x).

k) limh!0

R h0 f(x)dx

h

L’Hospital Rule��! limh!0

(R h0 f(x)dx)0

(h)0 = limh!0

f(h)1 = f(0)

l) limh!0

R h0 xf(x)dx

h2

L’Hospital Rule��! limh!0

(R h0 xf(x)dx)0

(h2)0 = limh!0

hf(h)2h = lim

h!0

f(h)2 = f(0)

2

integration

Question 3 Solution

a) TO BE COMPLETED

b) TO BE COMPLETED

c) ex =1P

n=0

1n!x

n ) e�x2

=1P

n=0

1n! (�x

2)n =1P

n=0

1n! (�1)nx2n )

Re�x2

dx =R 1P

n=0

1n! (�1)nx2n

dx =1P

n=0

1n! (�1)n 1

2n+1x2n+1 =

x� 13x

3 + 110x

5 � 142x

7 + · · · .d) TO BE COMPLETED

e) integration by parts :Rx sinxdx =

Rx(�1)d cosx = �

Rxd cosx = �x cosx +

Rcosxdx = �x cosx + sinx =

sinx� x cosx

f) use integration by parts twiceonce :

Re�x sinxdx =

Re�x(�1)d cosx = �e

�x cosx+Rcosxde�x = �e

�x cosx�Re�x cosxdx

twice :Re�x cosxdx =

Re�x

d sinx = e�x sinx�

Rsinxde�x = e

�x sinx+Re�x sinxdx

)Re�x sinxdx = �e

�x cosx� e�x sinx�

Re�x sinxdx ) 2

Re�x sinx = �e

�x cosx� e�x sinx

)Re�x sinx = 1

2 (�e�x cosx� e

�x sinx) = � 12e

�x(cosx+ sinx)

g)R

dx4x2 = � 1

4x

h)R

x4+x2 dx =

R 12

4+x2 dx2 =

R 12

4+x2 d(4 + x2) = 1

2 ln(4 + x2)

If you do not like the above way, instead use a change of variable u = 4 + x2, du = 2xdx ) dx = 1

2xdu.Rx

4+x2 dx =R

xu · 1

2xdu =R

12udu = 1

2 lnu+ C = 12 ln(4 + x

2)

i) For this type of problem, one needs to use a trig substituion, x = 2 tan ✓, dx = 2(1 + tan2 ✓)d✓Rdx

4+x2 =R 2(1+tan2 ✓)

4+4 tan2 ✓ d✓ =R

12d✓ = 1

2✓ = 12 arctan

x2

j) One can use the formula 1 + sinh2 ✓ = cosh2 ✓, and the change of variable x = 2 sinh ✓, dx = 2 cosh ✓d✓.Rdxp4+x2 =

R2 cosh ✓p4+4 sinh2 ✓

d✓ =R

2 cosh ✓p4 cosh2 ✓

d✓ =R

2 cosh ✓2 cosh ✓d✓ =

Rd✓ = ✓ = arcsinhx

2 = sinh�1 x2

k) Using partial fractionsR

dx4�x2 =

Rdx

(2+x)(2�x) =R ⇣

1/42+x + 1/4

2�x

⌘dx =

R 1/42+xdx+

R 1/42�xdx = 1

4 ln |2 + x|� 14 ln |2� x|

l) partial fractions againR

dx4x�x2 =

Rdx

x(4�x) =R ⇣

1/4x + 1/4

4�x

⌘dx =

R 1/4x dx+

R 1/44�xdx = 1

4 ln |x|�14 ln |4� x|

m) TO BE COMPLETED

n) using integration by parts once and using the equality sin2 x+ cos2 x = 1Rsin2 xdx =

Rsinx · sinxdx =

Rsinx(�1)d cosx = � sinx · cosx+

Rcosxd sinx = � sinx · cosx+

Rcos2 xdx = � sinx ·

cosx +R(1 � sin2 x)dx = � sinx · cosx +

R1dx �

Rsin2 xdx ) 2

Rsin2 xdx = � sinx · cosx + x )

Rsin2 xdx =

� 12 sinx cosx+ x

2 = x2 � 1

4 sin 2xA simple way, using sin2 x = 1�cos 2x

2 , sin 2x = 2 cosx sinx, (cos 2x)2 = (1 + cos 4x)/2thus

Rsin2 xdx =

R1�cos 2x

2 dx = x2 � sin 2x

4

o) using the equality sin2 x+ cos2 x = 1Rsin3 xdx =

Rsin2 x · sinxdx =

R(1 � cos2 x)(�1)d cosx = �

R(1 � cos2 x)d cosx = �

R1d cosx +

Rcos2 xd cosx =

� cosx+ 13 cos

3x

3

p) using integration by parts will be too complicated, using sin2 x = 1�cos 2x2 , sin 2x = 2 cosx sinx, (cos 2x)2 = (1 +

cos 4x)/2

ThusRsin4 x =

R �1�cos 2x

2

�2dx =

R1�2 cos 2x+cos2 2x

4 dx =R

1�2 cos 2x+1�sin2 2x4 dx

=R �

12 � 1

2 cos 2x� 14 sin

2 2x�dx = x

2 � 14 sin 2x� 1

8

Rsin2 2xd(2x)

using 3 i)x!2x��! x2 � 1

4 sin 2x� 18

�2x2 � sin 4x

4

�= 3

8x� 14 sin 2x+ 1

32 sin 4x

Question 4 Solution

a)R 2

p3

0x3

p16�x2 dx TO BE COMPLETED

b)R1�1 x

2 1p2⇡

e�x2

2 dx TO BE COMPLETED

c)R1�1(x� 1)2 1

2p2⇡

e�(x�1)2

8 dx TO BE COMPLETED

Question 5 SolutionUsing variable substitution u = ⇡

2 � x, du = �dxR 0⇡/2

sin(⇡2 �u)

sin(⇡2 �u)+cos(⇡

2 �u) (�1)du = �R 0⇡/2

cosucosu+sinudu =

R ⇡/20

cosucosu+sinudu

both u and x are integral variables changing from 0 to 12 , one may replace them with ✓, thus

R ⇡/20

sin ✓sin ✓+cos ✓d✓ =

R ⇡/20

cos ✓sin ✓+cos ✓d✓ on one hand, one the other hand

R ⇡/20

sin ✓sin ✓+cos ✓d✓+

R ⇡/20

cos ✓sin ✓+cos ✓d✓ =

R ⇡/20

⇣sin ✓

sin ✓+cos ✓ + cos ✓sin ✓+cos ✓

⌘d✓ =

R ⇡/20 1d✓ = ⇡

2 .

ThusR ⇡/20

sin ✓sin ✓+cos ✓d✓ =

R ⇡/20

cos ✓sin ✓+cos ✓d✓ = ⇡

4

Question 6 Solutiona) convergent by p�test

R11

1x2 dx =

R10

�� 1

x

�0dx = � 1

x

��11

= 1

b) divergent by p�testR11

dxx = lnx|11 = 1

c)R11

dxx�1 =

R10

dyy divergent by p�test

d) divergent by p-testR 10

dxx2 = � 1

x

��10= 1

e) convergent by p-testR 10

dxpx= 2

px|10 = 2

f) divergent by p-test Since bothR 0�1

dxx and

R 10

dxx diverges.

Question 7 Solutionsubstitute u = r

2 � 2rx+ a2, so du = �2rdx

limits x = �a ) u = r2 � 2r(�a) + a

2 = r2 + 2ra+ a

2 = (r + a)2, x = a ) u = r2 � 2ra+ a

2 = (r � a)2

V (r) = q2a

R a�a

dxpr2�2rx+a2 = q

2a

R (r�a)2

(r+a)2� 1

2r dupu

= � q4ar · 2

pu��(r�a)2

(r+a)2= � q

2ar (|r � a|� |r + a|) =(

qr if r � a

qa if 0 r a

Question 8a Solution

Assume the aquarium has length l, width w, and height h, let x be the vertical coordinate with x = 0 at the bottom ofthe aquarium (positive x is up), divide the water into layers of width �x = h

n , each layer is a rectangular box with volumelw�x, the force on a layer is ⇢glw�x, the layer at level xi = i�x is raised a distance h�xi, after letting n ! 1,�x ! 0,

the work done is W =R h0 ⇢glw(h� x)dx = ⇢glw(hx� x2

2 )��h0= 1

2⇢glwh2, substituting the given values, the work done is

W = 12⇢g · 2 · 0.5 · 1

2 = 12⇢g Joule. From the formula W = 1

2glwh2, we see that if the width w is doubled, then the work

is also doubled, and if the height h is doubled, then the work is multiplied by 4.

Question 8b Solution TO BE COMPLETED

Question 9 Solution

a) On the x-axis, one ion is held fixed at x = 0, so the distance between the ions is x, so replace r in F = � q2

4⇡✏0r2by x.

Work =R ba F (x)dx =

R 23 � q2

4⇡✏0x2 dx = q2

4⇡✏01x

��23= q2

4⇡✏0( 12 � 1

3 ) =q2

24⇡✏0

b) On the x-axis, one ion is held fixed at x = 1, so the distance between the ions is x� 1, so replace r in F = � q2

4⇡✏0r2by

x� 1.

Work =R ba F (x)dx =

R 23 � q2

4⇡✏0(x�1)2 dx = q2

4⇡✏01

x�1

��2

3= q2

4⇡✏0( 11 � 1

2 ) =q2

8⇡✏0

c) The work can be calculated separately for each ion then added together. Work = q2

24⇡✏0+ q2

8⇡✏0= q2

6⇡✏0.

d) Let w be a coordinate along the rod, so 0 w 1. Divide the rod into small pieces of width �w. The charge in eachpiece is q�w, and the force on the piece at coordinate w is F (x) = � q·q�w

4⇡✏0(x�w)2 . The work contributed by that piece is

4

Work =R 23 F (x)dx =

R 23 � q2�w

4⇡✏0(x�w)2 dx = q2�w4⇡✏0(x�w)

��2

3= q2�w

4⇡✏0

⇣1

2�w � 13�w

⌘.

We need a second integral for w from 0 to 1 to sum the work due all the pieces, and in this process we have �w ! dw.

Total Work = q2

4⇡✏0

R 10 (

12�w � 1

3�w )dw = q2

4⇡✏0(� ln(2� w) + ln(3� w))|10 = q2

4⇡✏0(� ln 1 + ln 2� (� ln 2 + ln 3)) = q2

4⇡✏0ln 4

3

Note that Work (a) = 0.04167 q2

⇡✏0, Work (b) = 0.1667 q2

⇡✏0, Work (d) = 0.07192 q2

⇡✏0, so Work (a) < Work (d) < Work (b).

This is reasonable because case (a) and case (b) are two extreme cases (all the charge is at one end of the rod or theother), and the total charge is the same in all three cases.

Question 10 Solution

f(x) = coshx ) f0(x) = sinhx ) 1 + (f 0(x))2 = 1 + sinh2 x = cosh2 x )

p1 + (f 0(x))2 = coshx

a) arclength =R 1�1

p1 + (f 0(x))2dx =

R 1�1 coshxdx = sinhx|1�1 = sinh 1� sinh(�1) = 2 sinh 1

b) surface area =R 1�1 2⇡f(x)

p1 + (f 0(x))2dx = 2⇡

R 1�1 cosh

2xdx = 2⇡

R 1�1

⇣ex+e�x

2

⌘2dx = 2⇡

R 1�1

⇣e2x+2+e�2x

4

⌘dx

= 2⇡4

⇣e2x

2 + 2x+ e�2x

�2

⌘ ��1�1

= ⇡2

⇣e2

2 + 2 + e�2

�2 �⇣

e�2

2 � 2 + e2

�2

⌘⌘= ⇡

2 (e2+4�e

�2) = ⇡2 (2 sinh 2+4) = ⇡(sinh 2+2)

You can save some time using the fact that coshx is an even function, soR 1�1 · · · = 2

R 10 · · · .


x = My

m , y = Mxm

case 1: m =R ba f(x)dx, Mx = 1

2

R ba f

2(x)dx, My =R ba xf(x)dx

case 2: m =R ba (f(x)� g(x))dx, Mx = 1

2

R ba (f

2(x)� g2(x))dx, My =

R ba x(f(x)� g(x))dx

answers : (x, y) = (a)�32 ,

65

�, (b)

�34 ,

125

�, (c)

�0, 2

5

�, (d)

�1,

14

�

(b) m =R 20 (4� x

2)dx = (4x� x3

3 )��20= 8� 8

3 = 163

Mx = 12

R 20 (16� x

4)dx = 12 (16x� x5

5 )��20= 1

2 (32�325 ) = 64

5 ) y = Mxm = 64

5 · 316 = 12

5

My =R 20 (4x� x

3)dx = (4x2

2 � x4

4 )��20= 8� 16

4 = 4 ) x = My

m = 4 · 316 = 3

4

(d) m =R10

11+x2 dx ( = arctanx

��10= ⇡

2 )

change variable: x = tan ✓, dx = (1 + tan2 ✓)d✓, x = 0 ) ✓ = 0, x = 1 ) ✓ = ⇡2 , since tan 0 = 0, tan ⇡

2 = 1m =

R10

11+x2 dx =

R ⇡2

01+tan2 ✓1+tan2 ✓d✓ =

R ⇡2

0 1d✓ = arctanx��10= ⇡

2

x =R 10 xf(x)dx

m =R 10

x1+x2 dx

m =R 10

12dx2

1+x2⇡2

=12 ln(1+x2)

��10

⇡2

= 1 !!!

The area is finite, but it has an infinitely large x.

y =R 10

12 f

2(x)dxm = 1

m ·R10

12f

2(x)dx = 1⇡2

R10

12

⇣1

1+x2

⌘2dx = 2

⇡

R10

12

(1+x2)2 dx = 2⇡

R10

12+

12x

2� 12x

2

(1+x2)2 dx

= 2⇡

⇣R10

12 (1+x2)(1+x2)2 dx�

R10

12x

2

(1+x2)2 dx

⌘= 2

⇡

⇣R10

12

1+x2 dx�R10

12x·x

(1+x)2 dx

⌘

= 2⇡

⇣12 arctanx

��10

�R10

12x·

12 dx2

(1+x2)2

⌘= 2

⇡

h12 · ⇡

2 +R10

x4 d

⇣1

1+x2

⌘i= 2

⇡

⇣⇡4 + x

4 · 11+x2

��1

0� 1

4

R10

11+x2 dx

⌘

= 2⇡

⇣⇡4 + x

4 · 11+x2

��1

0� 1

4

R10

11+x2 dx

⌘= 2

⇡

⇣⇡4 + x

4 · 11+x2

��1

0� 1

4 arctanx��10

⌘

= 2⇡

�⇡4 + 0� 1

4 · ⇡2

�= 2

⇡ · ⇡8 = 1

4

Question 12 Solutionf(t) = ce

�ct, where c = 11000 and t � 0

a) Prob(0 t 200) =R 2000 ce

�ctdt = �e

�ct|2000 = 1� e� 1

5 ⇡ 0.18

b) Prob(t � 800) =R1800 ce

�ctdt = �e

�ct|1800 = e� 4

5 ⇡ 0.45

Question 13 SolutionFirst notice that f(x) � 0 for all x. Now we need only show that

R 10 f(x)dx = 1

R 10 f(x)dx =

R 10

1

⇡p

x(1�x)dx

x=u+ 12��!

R 12

� 12

1

⇡p

(u+ 12 )(1�u� 1

2 )du =

R 12

� 12

1

⇡p

( 12+u)( 1

2�u)du

=R 1

2

� 12

1

⇡p

14�u2

duu= 1

2 sin ✓��!

R ⇡2

�⇡2

1⇡· 12 cos ✓

· 12 cos ✓d✓ = 1

⇡ ✓��⇡2

�⇡2= 1

An alternative approach to the integral is to make the substitution w =px, x = w

2, dw = dx/(2

px). This givesR 1

0 f(x)dx =R 10

1

⇡p

x(1�x)dx =

R 10

2dw⇡p1�w2 = 2

⇡ arcsinw]10 = 2⇡ (⇡/2� 0) = 1. The alternative is more e�cient but requires

some ingenuity to think of.

5

Di↵erential EquationsQuestion 14 Solution

a) y0 = �2y, y0 = 1 (standard exponential decay) ) y = C · e�2t.y0 = 1 ) C = 1 ) y(t) = e

�2t and limt!1

y(t) = 0

b) y0 = 1�2y ) dydt = 1�2y separation of variables ) dy

1�2y = dt integrate both sides ) � 12 ln |1� 2y| = t+C

) 1� 2y = C · e�2t ) y = 1�C·e�2t

2

y0 = 0 ) C = 1 ) y = 1�e�2t

2 and limt!1

y(t) = 12

c) y0 = 1�y2 ) dy

dt = (1+y)(1�y) separation of variables ) dy(1+y)(1�y) = dt partial fraction )

12

1+ydy+12

1�ydy =

dt integrate both sides ) 12 ln |1 + y| � 1

2 ln |1� y| = t + C ) ln�� 1+y1�y

�� = 2t + C ) 1+y1�y = C · e2t where C is

constant may be positive or negative ) y(t) = C·e2t�1C·e2t+1

y0 = 0 ) C = 1 ) y(t) = e2t�1e2t+1 . Furthermore y(t) = (e2t�1)e�t

(e2t+1)e�t = et�e�t

et+e�t = tanh t

Check... y(t) = tanh t is the solution.limt!1

y(t) = 1

d) y0 = �ty ) dydt = �ty separation of variables ) dy

y = �tdt integrate both sides ) ln |y| = � 12 t

2 + C )y = Ce

� 12 t

2

y0 = 1 ) C = 1 ) y(t) = e� 1

2 t2

limt!1

y(t) = 0

Question 15 Solutiona) y = c1e

t + c2e�t ) y

0 = c1et � c2e

�t ) y00 = c1e

t + c2e�t = y, thus it is a solution of y00 = y for any constants

c1, c2.b) y(0) = 1, y0(0) = 0 ) c1 + c2 = 1, c1 � c2 = 0 ) c1 = c2 = 1

2 ) y(t) = 12e

t + 12e

�t = coshxc) y(0) = 0, y0(0) = 1 ) c1 + c2 = 0, c1 � c2 = 1 ) c1 = 1

2 , c2 = � 12 ) y(t) = 1

2et � 1

2e�t = sinhx

Question 16 Solution a) y0 = ky. Solve the equation, we have y(t) = y0e

kt. 200 = y(30) = y0e30k and 800 = y(90) =

y0e90k. Therefore, y0 = 102 = 100 cells.b)200 = 100e30k, so k = ln 2

30 . Therefore, y(t) = 100 · 2t/30. Solve the equation 6400 = 100 · 2t/30. Then t =30 ln 64/ ln 2 = 30 · 6 = 180 hours.

Question 17 Solutiony(t) = y0e

�kt

y(t) = 40 ·�12

� t1.4⇥10�4

30 = 40 ·�12

� t1.4⇥10�4

t = 1.4⇥ 10�4 ln 34

ln 12= 0.581⇥ 10�4s

Question 18 Solutiony(t) : tiger body mass (kg) as a function of time t (day)y0 = rate in� rate out = 2500 cal

day · 1 kg10000 cal � 20 cal · y kg

10000 cal ·1

day = 2500�20y10000

kgday

y0 = � 1

500 (y � 125) Newton’s heating/cooling y0 = k(y � T )

y(t) = T + (y0 � T )ekt = 125 + (y0 � 125)e�t

500 ) limt!1

y(t) = 125 kg

Question 19 Solutiony(t) = T + (y0 � T )e�kt , note that the patient’s temperature is T , y0 = 70�Fy(1) = 95 = T + (70� T )e�k

y(2) = 100 = T + (70� T )e�2k

) ( 95�T70�T )

2 = 100�T70�T ) (95� T )2 = (100� T )(70� T ) ) T

2 � 190T + 952 = T2 � 170T + 70000 ) 20T = 2025

) T = 101.25� F

Question 20 Solutiony0 = ky(M � y)

y(t) = My0

y0+(M�y0)e�kMt

y0 = 10, M = 4000y(t) = 40000

10+(4000�10)e�4000kt

measure time in days20 = 40000

10+(4000�10)e�4000·7k

6

e�k =

�199399

� 128000

y(t) = 40000

10+3990( 199399 )

t7

let y(t) = 12 · 4000 = 2000, solve t = 7 ln 399

ln 399�ln 199 ⇡ 60 days.

Question 21 SolutionThis equation, y0 = f(y) with f(y) = 2y, is particularly simple to study with Euler’s Method. The recursion for Euler’s

Method in this case is uk+1 = uk + (�t)f(uk) = (�t)(2uk). After factoring, uk+1 = (1 + 2�t)(uk). This means that, inthis case, we don’t actually need to do each iteration of the process; we can foresee that uk = (1+2�t)ku0 = (1+2�t)k.So un = (1 + 2/n)n, where n = 1/�t is the number of steps.

�t un

1 (1 + 2)1 = 31/2 (1 + 1)2 = 41/4 (1 + 1/2)4 = 5.0625Remark. The general solution of this di↵erential equation is y = Ce

2t, and the solution which fits our initial conditionis y = e

2t. Thus the true answer is y(1) = e2 ⇡ 7.389. We need to use smaller steps to get a satisfactory approximation.

With 1000 steps, we would still only have un ⇡ 7.374.

SeriesQuestion 22 Solution

a) divergent1P

n=1

12n = 1

2

1Pn=1

1n by p-test of series, p = 1.

b) convergent since1P

n=1

12n =

1Pn=1

�12

�n= 1

2

1Pn=0

�12

�n= 1

2 · 11� 1

2= 1 < 1.

c) convergent by p-test of series, p = 2.d) convergent by Alternating Series Test, lim

n!1an = lim

n!11n2 = 0, an+1 < an and the sign is alternating.

e) divergent by Ratio Test, L = limn!1

��an+1

an

�� = limn!1

2n+1

(n+1)2 · n2

2n = 2 > 1, (L > 1 divergent)


a) 0.111111.... = 0.1+0.01+0.001+0.0001+ · · · = 110 +

1100 +

11000 +

11000 + · · · =

1Pn=1

110n = 1

10

1Pn=0

�110

�n= 1

10 ·1

1� 110

= 19

b) 0.1212121212... = 12100 + 12

10000 + 121000000 + · · · =

1Pn=1

12100n = 12

100

1Pn=0

1100n = 12

100 · 11� 1

100= 12

99

c) 0.4999999... = 0.45 + 0.045 + 0.0045 + 0.00045 + · · · = 45100 + 45

1000 + 4510000 + · · ·

= 45100

�1 + 1

10 + 1100 + 1

1000 + · · ·�= 45

100

1Pn=0

110n = 45

100 · 11� 1

10= 1

2 (ie, 0.49999999... = 0.5)


a) recall :1P

n=0

xn

n! = ex for all x, so setting x = 2 we obtain

1Pn=0

2n

n! = e2

b) recall :1P

n=0xn = 1

1�x for �1 < x < 1, so setting x = 13 we obtain

1Pn=0

13n = 1

1� 13= 3

2 )1P

n=1

13n =

1Pn=0

13n �1 = 3

2 �1 = 12

c) di↵erentiating the first equation in part (b) with respect to x we obtain1P

n=1nx

n�1 = 1(1�x)2 for �1 < x < 1, so setting

x = 13 we obtain

1Pn=1

n( 13 )n�1 = 1

(1� 13 )

2 = 94 )

1Pn=1

n( 13 )n · ( 13 )

�1 = 94 )

1Pn=1

n( 13 )n = 9

4 · 13 = 3

4


Given that1P

n=1

1n2 = ⇡2

6 , note that1P

n=0

1(2n+1)2 includes all the odd terms.

1Pn=1

1n2 =odd terms+even terms )

1Pn=1

1n2 =

1Pn=0

1(2n+1)2 +

1Pn=1

1(2n)2 )

1Pn=1

1n2 =

1Pn=0

1(2n+1)2 + 1

4

1Pn=1

1n2

) ⇡2

6 =1P

n=0

1(2n+1)2 + 1

4⇡2

6 )1P

n=0

1(2n+1)2 = ⇡2

6 � ⇡2

24 = ⇡2

8 .

Question 26 Solutiona) Use |s� s10|

R1n f(x)dx since all terms are positive, where f(x) = 1

x2

|s� s10| R110 f(x)dx =

R110

1x2 dx = � 1

x

��110

= 0.1

b) Use |s� s10| an+1 where an+1 = 1(n+1)2 since the series is an alternating series.

|s� s10| an+1 = 1112 = 1

121

7

Question 27 SolutionAssume the dog starts with student A and runs to student B; if the time interval has duration t0, then we have 10t0+2t0 =20 (the distance the dog runs plus the distance student B walks is equal to the distance between them when they start).This implies that 12t0 = 20 or t0 = 20

12 = 53 .

Note that the at the end of the first interval, the distance between the students is 20�4t0 = 20�4· 53 = 20� 203 = 40

3 = 20· 23 ;in other words the distance between the students decreased by a factor of 2

3 .In the next time interval of duration t1, the dog runs back to A; then we have 10t1 +2t1 = 40

3 (the distance the dog runsplus the distance A walks is equal to the distance between the students when they start).This implies that 12t1 = 40

3 or t1 = 4012·3 = 10

9 = 53 · 2

3 .Note that at the end of the second interval, the distance between the students is 40

3 � 4 · t1 = 403 � 4 · 10

9 = 809 = 40

3 · 23 .

In the next time interval of duration t2, the dog runs back to B; then we have 10t2 +2t2 = 809 (the distance the dog runs

plus the distance B walks is equal to the distance between them when they start).This implies that 12t2 = 80

9 or t2 = 8012·9 = 20

27 = 53 · ( 23 )

2.The pattern repeats.

The total time is T = t0 + t1 + t2 + · · · = 53 + 5

3 · 23 + 5

3 ·�23

�2+ · · · = 5

3 ·⇣1 + 2

3 +�23

�2+ · · ·

⌘= 5

3 · 11� 2

3= 5.

Hence the dog travels a distance D = 10 mph ⇥ 5 hours = 50 miles.The question asks us to express D as an infinite series, but we could also find the distance directly as follows. The studentsmeet in the middle after walking a distance of 10 miles; since they walk at 2 mph, this must have taken 5 hours. This isalso how long the dog runs, so the dog must have run a total distance of 10 mph ⇥ 5 hours = 50 miles.

Question 28 SolutionConsider the remaining points as happening in pairs. In any pair of points, there are 3 possible outcomes; (a) youwin both, with probability p

2, (b) you lose both, with probability (1 � p)2, or (c) you win one and lose one, withprobability 2p(1 � p). In the first case, the game ends and you win. In the second case, the game ends and you lose.In the third case, the game continues. So any scenario in which you win has the following form; you split pairs ofpoints with your opponent a certain number of times and then you win a pair. Thus the probability of winning is

p2 + (2p(1� p))p2 + (2p(1� p))2p2 + · · · =

1Pn=0

[2p(1� p)]np2 = p2

1Pn=0

[2p(1� p)]n = p2 · 1

1�2p(1�p) =p2

1�2p+2p2 .

p = 12 ) p2

1�2p+2p2 = 12 (no surprise)

p = 14 ) p2

1�2p+2p2 = 110

p = 34 ) p2

1�2p+2p2 = 910 (no surprise that this answer and the previous would sum to 1)


a) The total length removed = 13 + 2 · 1

3 · 13 + 4 · 1

3 · 13 · 1

3 + · · · = 13

h1 + 2

3 +�23

�2+ · · ·

i= 1

3 · 11� 2

3= 1

b) Notice that any point which is the endpoint of a remaining interval will never be removed. Every stage doubles thenumber of remaining intervals, so this accounts for infinitely many points.

Power Series, Taylor SeriesQuestion 30 Solution

a) L = limn!1

��an+1

an

�� = limn!1

��xn+1

xn

�� = |x| < 1 ) the radius of convergence is 1; since at two end points x = ±1, the

series diverges, the interval of convergence is �1 < x < 1. The sum is 11�x for �1 < x < 1.

b) L = limn!1

��an+1

an

�� = limn!1

�� 2nxn+1

2n+1xn

�� =��x2

�� < 1 ) |x| < 2 ) the radius of convergence is 2; since at two end points

x = ±2, the series diverges, the interval of convergence is �2 < x < 2. The sum is1P

n=0

�x2

�n= 1

1� x2= 2

2�x for �2 < x < 2.

c) L = limn!1

��an+1

an

�� = limn!1

�� (x�1)n+1

(x�1)n

�� = |x� 1| < 1 ) � 1 < x � 1 < 1 ) 0 < x < 2 ) the radius of convergence

is 1 (the length of the interval divided by 2); since at two end points x = 0 and 2, the series diverges, the interval ofconvergence is 0 < x < 2. The sum is 1

1�(x�1) =1

2�x for 0 < x < 2.

d) L = limn!1

��an+1

an

�� = limn!1

�� nxn+1

(n+1)xn

�� = |x| < 1 ) � 1 < x < 1 ) the radius of convergence is 1 ; since at x = 1, the

series is harmonic series thus diverges, while at x = �1, the series converges by AST (alternating series test), the interval

of convergence is �1 x < 1. Note that xn =Rnx

n�1dx ) xn

n =Rxn�1

dx the sum is1P

n=1

Rxn�1

dx =R 1P

n=1xn�1

dx =

R 1Pn=0

xndx =

R1

1�xdx = � ln(1� x) = ln 11�x for �1 x < 1.

ln 11�x =

1Pn=1

xn

n

8

e) L = limn!1

��an+1

an

�� = limn!1

�� (n+1)xn+1

nxn

�� = |x| < 1 ) � 1 < x < 1 ) the radius of convergence is 1; since at x = ±1,

the series diverges, the interval of convergence is �1 < x < 1.

Note that (xn)0 = nxn�1,

1Pn=1

nxn = x

1Pn=1

nxn�1 = x

1Pn=1

(xn)0 = x ·✓ 1P

n=0xn

◆0= x ·

⇣1

1�x

⌘0= x · 1

(1�x)2 = x(1�x)2

x(1�x)2 =

1Pn=1

nxn


Namely find cn, such that f(x) = 11�x =

1Pn=0

cn(x� 12 )

n.

11�x = 1

1�(x� 12+

12 )

= 11�(x� 1

2 )�12= 1

12�(x� 1

2 )= 1

2 · 11�2(x� 1

2 )= 1

2 ·1P

n=0(2(x� 1

2 ))n =

1Pn=0

2n�1(x� 12 )

n

Question 32 Solutionf(x) = sinhx ) f(0) = sinh 0 = 0f0(x) = coshx ) f

0(0) = cosh 0 = 1f00(x) = sinhx ) f

00(0) = sinh 0 = 0f000(x) = coshx ) f

000(0) = cosh 0 = 1...

sinhx = f(0) + f0(0)x+ f 00(0)

2 x2 + f 000(0)

3! x3 + · · · = x+ 1

3!x3 + 1

5!x5 · · · =

1Pn=0

x2n+1

(2n+1)!

coshx = (sinhx)0 =

✓ 1Pn=0

x2n+1

(2n+1)!

◆0=

1Pn=0

⇣x2n+1

(2n+1)!

⌘0=

1Pn=0

x2n

(2n)!


a) sinx =1P

n=0(�1)n x2n+1

(2n+1)! = x� x3

3! +x5

5! � · · · , cosx =1P

n=0(�1)n x2n

(2n)! = 1� x2

2! +x4

4! �x6

6! + · · ·

sin2 x+ cos2 x =⇣x� x3

3! +x5

5! � · · ·⌘2

+⇣1� x2

2! +x4

4! �x6

6! + · · ·⌘2

=⇣x2 � 2x4

3! +x6

36 + 2x6

5! + · · ·⌘+⇣1� 2x2

2 + x4

4 + 2x4

4! � 2 x6

2·4! � 2x6

6! + · · ·⌘= · · · = 1 +O(x8)

b) In fact we know that all terms in the power series for sin2 x+ cos2 x vanish after the first term 1, though proving it isa nice fairly involved exercise.


Because ex =

1Pn=0

xn

n! , e�x2

=1P

n=0

(�x2)n

n! =1P

n=0(�1)n x2n

n! = 1� x2 + x4

2 � x6

6 + x8

24 � x10

120 + · · ·

T1(x) = T0(x) = 1 and T2(x) = 1� x2. See Figure below for sketch.

Figure 1: Graph for Problem 34.

Question 35 SolutionShow that 0 f(x) < 1; lim

x!1f(x) = 1; lim

x!0+f(n)(x) = lim

x!0+P ( 1x )e

�1/x, where P ( 1x ) is a polynomial of 1x ; when x ! 0+

e�1/x ! 0 exponentially (faster than any polynomial) thus f (n)(x) ! 0 regardless of the form of P ( 1x ).

9

Question 36 Solutionmethod 1 It can be shown using Taylor series for f(x) =

px about x = a, that

px =

pa+ 1

2pa(x�a)� 1

8a� 3

2 (x�a)2+· · · .Setting a = 9 yields

px = 3 + 1

6 (x� 9)� 1216 (x� 9)2 + · · · .

This is a convergent alternating series (why?), so |s� sn| < an+1, i.e.��px�

�3 + 1

6 (x� 9)�� < 1

216 (x� 9)2.

Setting x = 10 yields��p10� 3.16666

�� < 1216 <

1200 = 0.005.

The approximate value isp10 ⇡ 3.16666.

method 2 Use the binomial series, (1 + x)k = 1 + kx+ k(k�1)2 x

2 + · · · for �1 < x < 1.p10 =

p9 + 1 =

q9(1 + 1

9 ) =p9q1 + 1

9 = 3(1 + 19 )

1/2; we have x = 19 and k = 1

2p10 = 3(1 + 1

2 · 19 � 1

8 · 192 + · · · ) = 3 + 1

6 � 1216 + · · · )

��p10� 3.16666�� < 1

216 <1

200 = 0.005


Since f(x) = ln(1+x) = x� x2

2 + x3

3 � x4

4 + x5

5 + · · · =1P

n=1(�1)n+1 xn

n is an alternating series, we can use the error bound

for alternating series: |s� sn| an+1 where an+1 = xn+1

n+1 . We are interested in estimating ln 32 = ln(1 + 1

2 ); hence, we

take x = 12 . Let us find the smallest n such that an+1 = 1

(n+1)2n+1 0.001. We find that n = 6 and s6 ⇡ 0.4047; the

exact value is s = ln 32 ⇡ 0.4055, so the error is less than 10�3, as desired.


Find the first two nonzero terms in the Taylor series for f(x) about x = 0.

a) e�x sinx = x� x2 + · · ·

recall that ex = 1 + x+ 12x

2 + · · · , so we have e�x = 1� x+ 1

2x2 + · · · , also recall that sinx = x� 1

6x3 + · · ·

then we have e�x sinx = (1� x+ 1

2x2 + · · · )(x� 1

6x3 + · · · ) = x� x

2 + · · · ok

- - - - - - - - - - - - - - - - - - - -

b) 1�cos xx = 1

2x� 124x

3 + · · ·recall that cosx = 1� 1

2x2 + 1

24x4 + · · · , then 1�cos x

x =1�(1� 1

2x2+ 1

24x4+··· )

x = 12x� 1

24x3 + · · · ok

- - - - - - - - - - - - - - - - - - - -

c) tanx = x+ x3

3 + · · ·recall that sinx = x� 1

6x3 + · · · , cosx = 1� 1

2x2 + · · ·

then tanx = sin xcos x =

x� 16x

3+···1� 1

2x2+··· = (x� 1

6x3 + · · · ) · 1

1�( 12x

2+··· ) : write the 2nd factor as a geometric series

= (x� 16x

3 + · · · ) · (1 + ( 12x2 + · · · ) + · · · ) = x+ x

3( 12 � 16 ) + · · · = x� 1

3x3 + · · · ok

- - - - - - - - - - - - - - - - - - - -

d) tan�1x = x� 1

3x3 + · · ·

tan�1x = c0 + c1x+ c2x

2 + c3x3 + · · · , note that tan�1 0 = 0, so we must have c0 = 0

) tan�1x = c1x+ c2x

2 + c3x3 + · · · = y

) x = tan y = y + 13y

3 + · · · : this uses part (c)

= (c1x+ c2x2 + · · · ) + 1

3 (c1x+ c2x2 + · · · )3 + · · · = c1x+ c2x

2 + c3x3 + · · ·+ 1

3 (c31x

3 + · · · ) + · · ·we’ve shown that x = c1x+ c2x

2 + c3x3 + · · ·+ 1

3 (c31x

3 + · · · ) + · · ·the coe�cients of each power of x must match on the left and right

x1 ) 1 = c1

x2 ) 0 = c2

x3 ) 0 = c3 +

13c

31 ) c3 = � 1

3c31 = �1 ok

Question 39 SolutionB0 = f(0) = 1; B1 = f

0(0) = � 12 ; B2 = f

00(0) = 16 (using L’Hospital rule).

Question 40 Solutiona) f(x) = x, f(0) = 0, f 0(0) = 1b) f(x) = sinx, f(0) = 0, f 0(x) = cosx, f 0(0) = 1c) f(x) = ln(1 + x), f(0) = 0, f 0(x) = 1

1+x , f0(0) = 1

b) f(x) = ex � 1, f(0) = 0, f 0(x) = e

x, f

0(0) = 1

10

If the functions are sketched in a neighborhood of x = 0, the order they appear (from top to bottom), consider theirTaylor approximations

x = x, sinx = x� 16x

3 + · · · , ln(1 + x) = x� 12x

2 + · · · , ex � 1 = x+ 12x

2

Thus on right hand side of 0, from top to bottom, ex � 1, x, sinx and ln(1 + x); on left hand side of 0, from top tobottom, ex � 1, sinx, x, and ln(1 + x).

Question 41 Solutiona) J0(x) = 1� x2

4 + x4

64 � · · · it is alternating series.R 10

⇣1� x2

4

⌘dx = 11

12

error boundR 10

x4

64dx = x5

5·64 = 1320

b) J0(x)0 =1P

n=1

(�1)n2nx2n�1

22n(n!)2 =1P

n=0

(�1)n+1(2n+2)x2n+1

22n+2((n+1)!)2

J0(x)00 =1P

n=1

(�1)n2n(2n�1)x2n�2

22n(n!)2

xJ0(x)00 =1P

n=1

(�1)n2n(2n�1)x2n�1

22n(n!)2 =1P

n=0

(�1)n+1(2n+2)(2n+1)x2n+1

22n+2((n+1)!)2

xJ0(x) =1P

n=0

(�1)nx2n+1

22n(n!)2

xJ0(x)00 + J0(x)0 + xJ0(x) =1P

n=0

(�1)n+1x2n+1

22n(n!)2

h2n+2

22(n+1)2 + (2n+1)(2n+2)22(n+1)2 � 1

i= 0

Thus J0(x) satisfies xy00 + y0 + xy = 0


f(t) =1P

n=0tn = 1 + t+ t

2 + t3 + t

4 + · · ·

a) step 1 : di↵erentiate the series, f 0(t) = 1 + 2t+ 3t2 + 4t3 + · · ·step 2 : square the series, f2(t) = (1 + t+ t

2 + t3 + t

4 + · · · ) · (1 + t+ t2 + t

3 + t4 + · · · ) = 1 + 2t+ 3t2 + 4t3 + · · ·

Thus f 0(t) = f2(t), so f(t) satisfies the di↵erential equation y

0 = y2 and we can check that the initial condition is f(0) = 1.

b) y0 = y

2 ) dydt = y

2 ) dyy2 = dt )

R dyy2 =

Rdt ) � 1

y = t + C ) y = 1�C�t , now apply initial condition

t = 0, y = 1 ) C = �1 ) y = 11�t

Since we showed in part (a) that f(t) satisfes the di↵erential equation and initial condition, we obtain f(t) = 11�t . This

is a round-about way of deriving the sum of a geomertic series, f(t) =1P

n=0tn = 1 + t+ t

2 + t3 ++t

4 + · · · = 11�t .

Question 43 Solutiona1)

R10

sin xx dx converges

pf:R10

sin xx dx =

1Pn=0

R (n+1)⇡n⇡

sin xx dx =

R ⇡0

sin xx dx +

1Pn=1

R (n+1)⇡n⇡

sin xx dx

R ⇡0

sin xx dx +

1Pn=1

R (n+1)⇡n⇡

sin xn⇡ dx; the last step

follows because x � n⇡ in each interval; then we haveR10

sin xx dx

R ⇡0

sin xx dx + 1

⇡

1Pn=1

1n

R (n+1)⇡n⇡ sinxdx =

R ⇡0

sin xx dx +

1⇡

1Pn=1

� cos xn

��(n+1)⇡

n⇡=

R ⇡0

sin xx dx + 1

⇡

1Pn=1

cosn⇡�cos(n+1)⇡n =

R ⇡0

sin xx dx + 1

⇡

1Pn=1

(�1)n 2n =

R ⇡0

sin xx dx + 2

⇡

1Pn=1

(�1)n

n , which

converges since the integral is proper (note that limx!0

sin xx = 1) and the series converges by the AST ok

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -a2)

R10

�� sin xx

�� dx diverges

pf:R10

�� sin xx

�� dx =1P

n=0

R (n+1)⇡n⇡

|sin x|x dx =

R ⇡0

|sin x|x dx+

1Pn=1

R (n+1)⇡n⇡

|sin x|x dx �

R ⇡0

sin xx dx+

1Pn=1

R (n+1)⇡n⇡

|sin x|(n+1)⇡dx; the last

step follows because x n(+1)⇡ in each interval; then we haveR10

sin xx dx �

R ⇡0

sin xx dx + 1

⇡

1Pn=1

1n+1

R (n+1)⇡n⇡ |sinx| dx =

R ⇡0

sin xx dx+ 1

⇡

1Pn=1

2n+1 =

R ⇡0

sin xx dx+ 2

⇡

1Pn=1

1n+1 , which diverges because the series is the harmonic series ok

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -b)

R10

sin xx dx = ⇡

2

pf: set f(a) =R10

sin xx e

�axdx; then note that f(0) =

R10

sin xx dx; so we need to evaluate f(0); to do that we will evaluate

f(a) and then set a = 0; to solve for f(a), we will find an expression for f0(a) and then integrate; hence we have

f0(a) =

R10

sin xx

dda [e

�ax]dx =R10

sin xx ·�xe

�axdx = �

R10 sinxe�ax

dx

substitute: u = sinx, dv = e�ax

dx ) du = cosxdx, v = e�ax

�a

11

) f0(a) = �

hsinx · e�ax

�a

��10

�R10 cosx e�ax

�a dx

i= � 1

a

R10 cosxe�ax

dx

substitute: u = cosx, dv = e�ax

dx ) du = � sinxdx, v = e�ax

�a

) f0(a) = � 1

a

hcosx · e�ax

�a

��10

�R10 sinx e�ax

a dx

i= � 1

a

⇥1a + 1

af0(a)

⇤) f

0(a) = � 1a

⇥1a + 1

af0(a)

⇤) f

0(a)⇥1 + 1

a2

⇤=

� 1a2 ) f

0(a)⇥a2 + 1

⇤= �1 ) f

0(a) = � 11+a2 ) f(a) = � tan�1(a) + C; to evaluate the constant note that lim

a!1f(a) =

lima!1

R10

sin xx e

�axdx =

R10

sin xx lim

a!1[e�ax]dx =

R10

sin xx ·0dx =

R10 0dx = 0; hence we have lim

a!1f(a) = lim

a!1� tan�1(a)+

C = � tan�1(1) + C = �⇡2 + C = 0 ) C = ⇡

2 ) f(a) = � tan�1(a) + ⇡2 ) f(0) = � tan�1(0) + ⇡

2 = ⇡2 ok

Question 44 Solutioncosx = 1� 1

2x2 + 1

4!x4 � · · · ; it is a convergent alternating series, so |cosx� 1| 1

2x2, and

��cosx� (1� 12x

2)�� 1

4!x4;

setting x = ⇡5 gives the result

Question 45 Solutionerf(x) = 2p

⇡

R x0 e

�t2dt = 2p

⇡

R x0 (1� t

2 + t4

2 + ...)dt = 2p⇡(x� x3

3 + x5

10 + ...)


a) aa+b = a

b · 11+ a

b= a

b · 11�(� a

b )= a

b ·1P

n=0

��a

b

�n

aa+b = a

b

⇣1� a

b + a2

b2 + · · ·⌘= a

b � a2

b2 + a3

b3 + · · ·b) Using the Binomial Series Theorem

(1 + x)k = 1 + kx+ k(k�1)2 x

2 + · · · for �1 < x < 1pR2 � r2 = R

q1� r2

R2 = R

⇣1� r2

R2

⌘ 12= R

1� 1

2 · r2

R2 +( 12 )(

12�1)2

⇣� r2

R2

⌘2+ · · ·

�= R� r

2 · rR � r

8 · r3

R3 + · · ·

Question 47 SolutionStarting from the Taylor series of f(x) about x = a:

f(x) = f(a) + f0(a)(x� a) +

f00(a)

2!(x� a)2 + · · · ,

replace x ! x+ h, a ! x. It follows that x� a ! (x+ h)� x = h, and the Taylor series becomes

f(x+ h) = f(x) + f0(x)h+

f00(x)

2!h2 + · · ·

Question 48 Solutiona) let y = 0 ) x = ±(1 + ✏), let x = 0 ) y = ±1

b) Solve for y: y = f(x) = ±r1�

⇣x

1+✏

⌘2

A(✏) = 2R 1+✏�(1+✏)

r1�

⇣x

1+✏

⌘2dx = 4

R 1+✏0

r1�

⇣x

1+✏

⌘2dx

c)

A(✏) = 4R 1+✏0

r1�

⇣x

1+✏

⌘2dx = 4(1 + ✏)

R 1+✏0

r1�

⇣x

1+✏

⌘2d

x1+✏ = 4(1 + ✏)

R 10

p1� u2du = 4(1 + ✏)⇡4 = (1 + ✏)⇡

The first two nonzero terms are ⇡ + ⇡✏.

Question 49 SolutionV (x) = Gm1

|x�x1| +Gm2|x�x2| for x ! 1 ie., x > x1 and x > x2 ) V (x) = Gm1

x�x1+ Gm2

x�x2

Using the hint set y = 1/x, ie., x = 1/y and expand the potential in powers of yV (1/y) = Gm1

1/y�x1+ Gm2

1/y�x2= Gm1y

1�x1y+ Gm2y

1�x2y

Using Geometric Series Formula 11�xiy

=1P

n=0(xiy)

n where i = 1, 2

V (1/y) = Gm1y

1Pn=0

(x1y)n +Gm2y

1Pn=0

(x2y)n = Gm1

1Pn=0

xn1y

n+1 +Gm2

1Pn=0

xn2y

n+1

= (Gm1 +Gm2)y + (Gm1x1 +Gm2x2)y2 + (Gm1x22 +Gm2x

22)y

3 + · · ·change y = 1

x backV (x) = (Gm1 +Gm2)

1x + (Gm1x1 +Gm2x2)

1x2 + (Gm1x

22 +Gm2x

22)

1x3 + · · ·

Thus a = (Gm1 +Gm2), b = (Gm1x1 +Gm2x2), and c = (Gm1x22 +Gm2x

22)

12


a) limr!0

V (r) = limr!0

V0((r0r )

12 � 2( r0r )6) = V0 lim

r!0

r120 �2r60r6

r12 = V0r1200+ = 1

limr!1

V (r) = limr!1

V0((r0r )

12 � 2( r0r )6) = V0 lim

r!1r120 �2r60r

6

r12 = V0 · 0 = 0

b) V 0(r) = V0((�12) r120

r13 � 2(�6) r60

r7 ) = V0(�12r120

r13 + 12r60r7 ) = V0

�12r120 +12r60r6

r13

V0(r) = 0 , �12r120 + 12r60r

6 = 0 , 12r60r6 = 12r120 , r

6 = r60 , r = r0 .

d) Find the quadratic Taylor approximation at x = x0, ie., c0 + c1(x� x0) + c2(x� x0)2

using the Theorem(1 + x)k = 1 + kx+ k(k�1)

2 x2 + · · · for �1 < x < 1

V (x) = V0

h�x0x

�12 � 2�x0x

�6i= V0

⇣xx0

⌘�12� 2

⇣xx0

⌘�6�= V0

⇣x�x0+x0

x0

⌘�12� 2

⇣x�x0+x0

x0

⌘�6�

= V0

⇣1 + x�x0

x0

⌘�12� 2

⇣1 + x�x0

x0

⌘�6�

using the above Theorem⇣1 + x�x0

x0

⌘�12= 1� 12x�x0

x0+ (�12)·(�12�1)

2

�x�x0

x

�2+ · · · = 1� 12

x0(x� x0) +

78x20(x� x0)2 + · · ·

⇣1 + x�x0

x0

⌘�6= 1� 6x�x0

x0+ (�6)·(�6�1)

2

�x�x0

x

�2+ · · · = 1� 6

x0(x� x0) +

21x20(x� x0)2 + · · ·

V (x) = V0

h⇣1� 12

x0(x� x0) +

78x20(x� x0)2 + · · ·

⌘� 2

⇣1� 6

x0(x� x0) +

21x20(x� x0)2

⌘i+ · · ·

= V0

h�1 + 36

x20(x� x0)2

i+ · · · = �V0 + 36V0

x20(x� x0)2 + · · ·

T2(x) = �V0 + 36V0

x20(x� x0)2

e) We know that work equals the integral of the force:

W =1Rr0

f(r)dr =1Rr0

�V0(r)dr = �V (r)|1r0 = �0� (�V (r0)) = V (r0) = V0(1� 2) = �V0

binomial seriesQuestion 51 SolutionUsing the Binomial Series Theorem,(1 + x)k = 1 + kx+ k(k�1)

2 x2 + k(k�1)(k�2)

3! x3 + k(k�1)(k�2)(k�3)

4! x4 + · · · for �1 < x < 1

(1 + x2)k = 1 + kx

2 + k(k�1)2 x

4 + k(k�1)(k�2)3! x

6 + k(k�1)(k�2)(k�3)4! x

8 + · · ·(1 + x

2)12 = 1 + 1

2x2 +

12 (

12�1)2 x

4 +12 (

12�1)( 1

2�2)3! x

6 +12 (

12�1)( 1

2�2)( 12�3)

4! x8 + · · · = 1 + 1

2x2 � 1

8x4 + 1

16x6 � 5

128x8 + · · ·

This is an alternating series and the assumptions in the AST apply.Using the second order Taylor approximation, T2(x) = 1 + 1

2x2, we have |s� T2| < 1

8x4, where s is the exact value.

R 10

p1 + x2dx ⇡

R 10 (1 +

12x

2)dx = x+ 16x

3��10= 7

6 and the error is less thanR

18x

4dx = 1

40x5��10= 1

40

Question 52 SolutionConsider the expansion 1p

1�2ax+x2 = c0 + c1x+ c2x2 + · · · , where a is a constant. Find c0, c1, c2 in terms of a.

We can apply the binomial expansion.1p

1�2ax+x2 = (1� 2ax+ x2)�1/2 = (1 + (�2ax+ x

2))�1/2 = 1 + (� 12 )(�2ax+ x

2) +(� 1

2 )(�32 )

2 (�2ax+ x2)2 + · · ·

= 1 + ax� 12x

2 + 38 (4a

2x2 � 4ax3 + x

4) + · · · = 1 + ax+ (� 12 + 3

2a2)x2 + · · · , hence c0 = 1, c1 = a, c2 = 1

2 (3a2 � 1)

Question 53 Solutiona) Show that

�k+1n+1

�=

�kn

�+� kn+1

�

This is true since the left hand side is number of ways of choosing n+ 1 objects from a set of k+ 1 objects (disregardingthe order in which the objects are chosen).The right hand side means that: assume all k + 1 objects are white, one may randomly pick one object from the k + 1objects, coloring it red, then put it back. Now choose n + 1 objects from these k + 1 objects, there are two di↵erentsituations: one situation is that the red one is chosen, the number of ways is

�kn

�(it is equivalent to choosing n from k

objects); the other situation is the red one is not chosen, the number of ways is� kn+1

�(it is equivalent to choosing n+ 1

objects from k objects).The left hand side equals the right hand side, since it is the same thing, choosing n+ 1 objects from k + 1 objects.�k+1n+1

�= (k+1)!

(n+1)!(k�n)! =k!·(k+1)

(n+1)!(k�n)! =k!·(k�n+n+1)(n+1)!(k�n)! == k!·(k�n)+k!·(n+1)

(n+1)!(k�n)! = k!·(k�n)(n+1)!(k�n)! +

k!·(n+1)(n+1)!(k�n)! =

k!(n+1)!(k�n�1)! +

k!n!(k�n)! =

� kn+1

�+

�kn

�

b)�00

�1�1

0

� �11

�1 1�2

0

� �21

� �22

�1 2 1

13

�30

� �31

� �32

� �33

�1 3 3 1�4

0

� �41

� �42

� �43

� �44

�1 4 6 4 1�5

0

� �51

� �52

� �53

� �54

� �55

�1 5 10 10 5 1�6

0

� �61

� �62

� �63

� �64

� �65

� �66

�1 6 15 20 15 6 1

Denote elements in the triangle by ak,n, nth element on kth row. Each row is obtained by adding the two entries diagonallyabove, ak+1,n+1 = ak,n + ak,n+1 which is equivalent to

�k+1n+1

�=

�kn

�+� kn+1

�.

c) (a+b)6 =�60

�a6+

�61

�a5b+

�62

�a4b2+

�63

�a3b3+

�64

�a2b4+

�65

�ab

5+�66

�b6 = a

6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6

complex numbersQuestion 54 Solutiona) 1 + i =

p2e

⇡4 i (x = 1, y = 1 already in Cartesian form)

b) (1 + i)2 = 1 + 2i+ i2 = 1 + 2i� 1 = 2i = 2e

⇡2 i (x = 0, y = 2)

c) (1 + i)3 = (1 + i)2(1 + i) = 2i(1 + i) = �2 + 2i = 2p2e

3⇡4 i (x = �2, y = 2)

d) 11+i =

1·(1�i)(1+i)·(1�i) =

1�i1�i2 = 1�i

1�(�1) =1�i2 = 1

2 � 12 i =

1p2e

�⇡4 i (x = 1

2 , y = � 12 )

e)p1 + i =

�p2e

⇡4 i� 1

2 = 214 e

⇡8 i = 2

14

�cos ⇡

8 + i sin ⇡8

�= 2

14 cos ⇡

8 + i 214 sin ⇡

8 (x = 214 cos ⇡

8 , y = 214 sin ⇡

8 )

Question 55 Solutiona) See 49 c) (1 + i)6 = 1 + 6 · i+ 15 · i2 + 20 · i3 + 15 · i4 + 6 · i5 + i

6 = 1 + 6i� 15� 20i+ 15 + 6i� 1 = �8ib) 1 + i =

p2e

⇡4 i, since x = 1, y = 1, r =

px2 + y2 =

p2, ✓ = arctan y

x = arctan 1 = ⇡4 , re

✓i =p2e

⇡4 i

(1 + i)6 = (p2e

⇡4 i)6 = 2

62 e

64⇡i = 23e

32⇡i = 8(cos 3

2⇡ + i sin 32⇡) = �8i


a) z2 + 2z � 2 = 0 ) a = 1, b = 2, c = �2, z1,2 = �b±pb2�4ac2a = �2±

p4+8

2 = �1±p3 two real roots

b) z2 +2z+2 = 0 ) a = 1, b = 2, c = 2, z1,2 = �b±pb2�4ac2a = �2±

p4�8

2 = �2±p�4

2 = �2±2p�1

2 = �1±p�1 = �1± i

c) z2 = (re✓i)2 = r2e2✓i = 1 ) r = 1, ✓ = ⇡k, where k is any integer ) z1,2 = ±1

d) z3 = r3e3✓i = 1 ) r = 1, ✓ = 2⇡

3 k, where k is any integer ) three roots: z1 = 1 and z2,3 = � 12 ±

p32 i

e) z4 = r4e4✓i = 1 ) z

2 = ±1 ) four roots: z1,2 = ±1, z3,4 = ±i

f) ez = 1 on real axis there is on root z = 0, but in complex plane there are infinite roots, let z = x + yi, ez = ex+yi =

ex(cos y + i sin y) = 1 ) x = 0 and y = 2k⇡, where k is any integer, roots are zk = 2k⇡ i.

Question 57 Solution(cos ✓ + i sin ✓)n =

�ei✓�n

= ein✓ = cosn✓ + i sinn✓

Question 58 Solutionei(a+b) = e

ia ·eib ) cos(a+b)+i sin(a+b) = (cos a+i sin a)·(cos b+i sin b) = cos a cos b�sin a sin b+i(sin a cos b+cos a sin b)take real and imaginary parts : cos(a+ b) = cos a cos b� sin a sin b , sin(a+ b) = sin a cos b+ cos a sin b


a1)Reax cos bx dx =

Reax 1

b d sin bx = 1b e

ax · sin bx �R

1b sin bx de

ax = 1b e

ax · sin bx �R

ab e

ax sin bx dx = 1b e

ax · sin bx �Rab e

ax�� 1

b

�d cos bx = 1

b eax ·sin bx+

Rab2 e

axd cos bx = 1

b eax ·sin bx+ a

b2 eax ·cos bx� a

b2

Rcos bx deax = 1

b eax ·sin bx+ a

b2 eax ·

cos bx�a2

b2

Reax cos bx dx )

Reax cos bx dx= 1

b eax·sin bx+ a

b2 eax·cos bx�a2

b2

Reax cos bx dx )

⇣1 + a2

b2

⌘ Reax cos bx dx =

1b e

ax · sin bx+ ab2 e

ax · cos bx )Reax cos bx dx = eax

a2+b2 (b · sin bx+ a · cos bx)- - - - - - - - - - - - - - - - - - - -

a2)Reax sin bx dx =

Reax

�� 1

b

�d cos bx = � 1

b eax · cos bx +

R1b cos bx de

ax = � 1b e

ax · cos bx +R

ab e

ax cos bx dx = � 1b e

ax ·cos bx+

Rab e

ax · 1b d sin bx = � 1

b eax · cos bx+

Rab2 e

axd sin bx = � 1

b eax · cos bx+ a

b2 eax · sin bx� a

b2

Rsin bx deax = � 1

b eax ·

cos bx + ab2 e

ax · sin bx � a2

b2

Reax sin bx dx )

Reax sin bx dx = � 1

b eax · cos bx + a

b2 eax · sin bx � a2

b2

Reax sin bx dx )⇣

1 + a2

b2

⌘ Reax sin bx dx = � 1

b eax · cos bx+ a

b2 eax · sin bx )

Reax sin bx dx = eax

a2+b2 (a · sin bx� b · sin bx)- - - - - - - - - - - - - - - - - - - -

b) e(a+ib)x = eax+ibx = e

ax · eibx = eax (cos bx+ i sin bx)

Re(a+ib)x

dx = 1a+ibe

(a+ib)x = 1·(a�ib)(a+ib)·(a�ib)e

(a+ib)x = a�iba2+b2 e

(a+ib)x

- - - - - - - - - - - - - - - - - - - -

c) SinceRe(a+ib)x

dx =R(eax cos bx+ i e

ax sin bx) dx =Reax cos bx dx+ i

Reax sin bx dx = a�ib

a2+b2 e(a+ib)x, and

a�iba2+b2 e

(a+ib)x = a�iba2+b2 (e

ax cos bx+ i eax sin bx) = eax

a2+b2 (a cos bx+ b sin bx) + i

heax

a2+b2 (a sin bx� b sin bx)i, we have

14

Reax cos bx dx = eax

a2+b2 (a cos bx+ b sin bx) andR

eax sin bx dx = eax

a2+b2 (a sin bx� b sin bx)


a) since eix = cosx+ i sinx and e

�ix = cosx� i sinx ) eix + e

�ix = 2 cosx ) cosx = eix+e�ix

2

b) eix � e�ix = 2i sinx ) sinx = eix�e�ix

2i

c) ddx cosx = d

dx

⇣eix+e�ix

2

⌘= ieix�ie�ix

2 =(ieix�ie�ix)i

2i = �eix+e�ix

2i = � sinx

d) ddx sinx = d

dx

⇣eix�e�ix

2i

⌘= ieix+ie�ix

2i = eix+e�ix

2 = cosx

e) TO BE COMPLETEDf) TO BE COMPLETEDg) TO BE COMPLETEDh) TO BE COMPLETED

15

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Math 156 Applied Honors Calculus II Final Exam Review Sheet Solutions Fall 2016 1. True or