172

Calculus 2

Fall 2019

CRN 40769

Section 002

Time: Saturday 12:30 – 4:35 pm

Room BR-11

172 Calculus 2

Syllabus

Catalog description

Fundamental theorem of calculus, integration by substitution, areas and volumes, techniques of integration, arc length, improper integrals, polar coordinates and parametric equations, conic sections, sequences, infinite series, power series, convergence tests, alternating series, Taylor and Maclaurin series. Prerequisite: MATH 166 with a grade of “C” or better or MATH 171 with a grade of “C” or better.

Learning outcomes

Upon successful completion of this course, students will be able to:

1. Evaluate integrals using a variety of techniques 2. Solve problems involving applications of integrals such as finding areas,

volumes, arc length, work, etc. 3. Differentiate and integrate functions defined by parametric equations or in

polar form 4. Test infinite series for convergence and represent functions using power

series

Book: James Stuart. Calculus. Early Transcendentals. MATH 172, Community College of Philadelphia Edition 8 CENGAGE Learning ISBN: 978-1-327-05226-9 Instructor: Dr. Arkady Kitover Office: NE campus 327, Main – B2-25J

Office hours: Main campus, Saturday 4:40 – 5:40 pm (by appointment only), NE campus, TW 4:30 – 6:25 pm Email: (the best way to contact me) akitover@ccp.edu or

akitover@hotmail.com Web Page: http://faculty.ccp.edu/FACULTY/akitover The web page contains the syllabus, the reviews with detailed solutions, and a Maple attachment. Contents:

I. Indefinite integrals and technique of integration. Section 4.7 – Antiderivatives. This is a review of the corresponding section in the Calculus 1 book. For your convenience a copy of this section is provided in the package you received. Section 5.5 – The Substitution Rule. Sections 7.1 – 7.6

II. Definite integrals and their applications. Sections 5.3 – 5.5. Sections 6.1 – 6.4. Sections 8.1 and 8.2 Section 10.2

III. Improper integrals and power series. Section 7.8.

Sections 11.6 and 11.8 - 11.11 Tests: Three tests in class, with possibilities for extra credit.

Test 1. Trigonometric integrals and trigonometric substitutions. Integration by parts, integration of rational functions, rationalizing substitutions. (60 points) Test 2. Areas, volumes, and other applications of definite integrals. (60 points) Test 3. Improper integrals. Power series, Taylor and McLaurin series, and their applications. (80 points)

mailto:akitover@ccp.edumailto:akitover@hotmail.comhttp://faculty.ccp.edu/FACULTY/akitover

Cumulative final 100 points. Grading: Average of all tests. A 90%-100% (270 – 300 points) B 80%-89% (240 – 269 points) C 70%-79% (210 – 239 points) D 60%-69% (180 – 209 points) F 0% - 59% (less than 180 points) No matter what your average is, you will not get an “A” or a “B” if you get less than 50% or 40%, respectively, on the final. If you miss a test, I will give you a possibility to take it only if you have a valid and documented excuse. No food is allowed in the classroom. Put your cell phones in vibration mode before the class starts. You may not use any electronic devices for texting, web surfing, and other activities not related to the class work. I will subtract 10 points from your total sum for every violation of this rule. You may use your cell phone during a test only as a calculator

Recommended Homework

The numbers of sections and problems are given from Stewart, Calculus, Early Transcendentals, 8

MATH 172 Community College of Philadelphia

(For section 4.9 see the package you received)

Section Problems 4.9 3, 11, 37, 61, 73 5.5 11,13,19, 23, 33, 41, 49 7.1 13, 19, 21, 33, 37 7.2 5, 15, 29, 41 7.3 11, 17, 19, 25 7.4 9, 19, 27, 37, 45 7.5 27, 31, 47, 71 7.6 7, 15, 25, 27 5.3 21, 29, 37, 41, 55 5.4 35, 39, 43 5.5 55, 57, 59, 61, 63, 65 6.1 3, 5, 11, 15, 27, 31, 51 6.2 1, 9, 15, 33, 51 6.3 5, 13, 21, 25, 35 6.4 15, 21, 33 8.1 9, 13, 15, 29, 41 8.2 7, 15, 21 10.2 41, 45, 54, 65 7.8 9, 13, 21, 33, 37 11.6 3, 5, 15, 25, 35 11.7 15, 25, 33 11.8 3, 17, 27, 35, 41 11.9 7, 9, 17, 23, 29, 31

11.10 5, 23, 27, 47, 53, 59, 65, 77

11.11 9, 17, 21, 25, 29, 37

172 Calculus 2 Review 1 In problems 1 - 8 find the antiderivatives.

5

64

31.2 1

x dxx +∫

. We make the following substitution 62u x 1= + . Then and

therefore

512du x dx=

5 134

x dx du= . After the substitution we get the following integral 4

1 14

duu∫

and by Power Rule (1/ 4) 3/ 4 3/ 4 6 3/ 41 1 4 1 1 (2 1)4 4 3 3 3

u du u C u C x− = × + = + = + +∫ C

dt

t

.

2 42. tan sect t∫ . Because the power of secant is even it is convenient to write the

integral as and, recalling that2 2 2tan sec sect t td∫ 2 2sec tan 1t t= + , as . After performing the substitution 2 2 2tan (tan 1)sect t t+∫ dt ttanu = (recall

that ) we have2secdu tdt=5 3 5 3

2 2 4 2 tan tan( 1) ( )5 3 5 3u u t tu u du u u du C C+ = + = + + = + +∫ ∫ .

2 23.

5dx

x x −∫. We will use here a trigonometric substitution 5 secx t= . Then

2 25 5sec 5 5 tan2x t t− = − = and 2 5 5 tanx t− = . Also 5 tan secdx t tdt= .We plug in these expressions and obtain the following

integral2

5 tan sec 1 1 1 1cos sin5 sec 5 55sec 5 tan

t tdt dt tdt t Ctt t

= = =∫ ∫ ∫ + . Using the identity

tansinsec

ttt

= and recalling that 2 5tan and sec5 5

x xt −= t = we can express our result

as2

2 2

555

dx x Cxx x−

= +−

∫ .

24. 2 2 3x x d+ +∫ x . First we will complete the square.

2 2 21 1 12 2 2( ) 2[( ) ] 2( )2 4 2

x x x x x x+ = + = + − = + −2 12

. The integral becomes

21 52( )2 2

x dx+ +∫ which is a little bit more convenient to write

as21 14 5

22x dx⎛ ⎞+ +⎜ ⎟

⎝ ⎠∫ . Let 12

u x= + then du dx= and the integral

becomes 21 4 52

u d+∫ u . Now we perform a trigonometric substitution. 5 tan

2u t= .

Then and2 24 5 5 tan 5 5secu t 2+ = + = t 25 sec2

tdt=du . Thus our integral

becomes ( )35 5sec sec tan ln | sec tan |2 2 4 2

tdt t t t t C= + +∫ + . But 2 2tan

5 5u xt += = 1and

2 24 5 4 4sec5 5

u x xt + += = 6+ whence the answer can be written

as2

21 2 1 4(2 1) 4 4 6 ln4 2 5

x x x4 6x x x⎛ ⎞+ + + +

+ + + + +⎜⎜⎝ ⎠

C⎟⎟ . We can drop the sign of

absolute value because the expression 22 1 4 4 6x x x+ + + + is always positive. Finally, because we can include ln 5 in the constant of integration, we have

2 212 2 3 (2 1) 4 4 6 ln(2 1 4 4 6)4 2

2x x dx x x x x x x C+ + = + + + + + + + + +∫ .

25. cos(3 )x x dx∫ . We have to integrate by parts twice. First time we take . Then2, cos(3 )u x dv x dx= = 12 and sin(3 )

3du xdx v x= = . Plugging these expressions

into the formula we getudv uv vdu= −∫ ∫ 2 21 2cos(3 ) sin(3 ) sin(3 )3 3

x x dx x x x x dx= −∫ ∫ . To the integral in the right part we apply integration by parts once again taking . Then and sin(3 )u x dv x dx= = 1and cos(3 )

3du dx v x dx= = − . Thus we have

2 2

2

1 2 1 1cos(3 ) sin(3 ) cos(3 ) cos(3 )3 3 3 3

1 2 2sin(3 ) cos(3 ) sin(3 )3 9 27

x x dx x x x x x dx

x x x x x C

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

= + − +

∫ ∫ =

2

3

16.( 1)

x x dxx x+ ++∫ . We integrate a proper rational fraction so we can start with its

decomposition into partial fractions. 2

3 2 3

1( 1) 1

x x A B C Dx x x x x x+ +

= + + ++ +

. Multiplying both parts of this identity by the common

denominator we get the identity 3( 1x x + )*)2 2 31 ( 1) ( 1) ( 1) (x x Ax x Bx x C x Dx+ + = + + + + + +

Two of the coefficients in (*) can be found quite easily. If we plug in we get , and if we plug in we get

0x =1C = 1x = − 1D = − . Next, let us compare coefficients by

3x in both parts of (*). In the left part the coefficient is 0, in the right it is A D+ .

Thus, 1A = . Finally, to find B we compare coefficients by 2x in both parts of (*). We get 1 A B= + whence . Now we can finish the integration. 0B =

2

3 3 2

1 1 1 1 1 1ln | | ln | 1| ln( 1) 1 2 1 2

x x xdx dx x x C Cx x x x x x x x+ + ⎛ ⎞= + − = − − + + = − +⎜ ⎟+ + +⎝ ⎠∫ ∫ 2 .

2

2 2 2

3 5 67.( 1)( 4)

x x dxx x

− +− +∫ . Again we integrate a proper rational function. The

corresponding decomposition into partial fractions is

( )2

22 2 2 2 2

3 5 6( 1)( 4) 1 1 4 4

x x A B Cx D Exx x x x x x

− + + += + + +

− + − + + +

F

2 )

; After multiplying both parts by the

common denominator we have

. 2 2 2 2 2 2 23 5 6 ( 1)( 4) ( 1)( 4) ( )( 1)( 4) ( )( 1x x A x x B x x Cx D x x Ex F x− + = + + + − + + + − + + + −

If we plug in 1x = we get whence4 50A= 225

A = . Respectively, taking gives us

and

1x = −

14 50B= − 725

B = − . Comparing coefficients by 5x provides the equation

whence0 A B C= + + 15

C = . On the other hand if we compare coefficients by 4x we

get and0 A B D= − + 925

D = − . To find E we will compare coefficients by x,

, whence 5 16 16 4A B C− = + − − E 32 112 45 16 16 4 5 125 25 5

E A B C= + + − = + − − = . Finally,

comparing the constant terms we get whence6 16 16 4A B D= − − − F 32 112 36 30 616 16 4 6 6

25 25 25 25 5F A B D= − − − = + + − = = .

We got that

2

2 2 2 2 2 2 2 2 2

3 5 6 2 1 7 1 1 9 1 6 1( 1)( 4) 25 1 25 1 5 4 25 4 ( 4) 5 ( 4)

x x x xdx dx dx dx dx dx dxx x x x x x x x

− += − + − + +

− + − + + + + +∫ ∫ ∫ ∫ ∫ ∫ ∫

The computation of the first two integrals is immediate, 2

2

1 ln( 4)4 2

x dx xx

= ++∫ because the numerator equals to ½ of the derivative of the

denominator, and 21 1 arctan

4 2 2xdx

x⎛ ⎞= ⎜ ⎟+ ⎝ ⎠∫ according to the

formula 2 21 1 arctan , 0xdx a

x a a a⎛ ⎞= >⎜ ⎟+ ⎝ ⎠∫ .

To compute the integral 2 2( 4)x dx

x +∫ we make the substitution . It

results in

2 4u x= +

2 2 2 2

1 1 1 1( 4) 2 2 2( 4

x dx dux u u

= = − = −+∫ ∫ )x + . Finally, to compute the integral

2 2

1( 4)

dxx +∫ we use the reduction

formula 2 2 2 2 2 1 2 2 11 1 1 2 3 1

( ) 2 2 ( ) 2 2 ( )n nx ndx

x a a n x a n x a−⎡ ⎤−

= ⋅ +⎢ ⎥+ − + − +⎣ ⎦∫ ∫ n− . Applying this

formula in case when we

get

2 and 2n a= =

2 2 2 2 2

1 1 1 1 1 1 arctan( 4) 8 4 8 4 8 4 16 2

x xdx dxx x x x

⎛ ⎞= + = + ⎜ ⎟+ + + + ⎝ ⎠∫ ∫x . Plugging these

expressions into the formula above and combining like terms we obtain the answer

22

2 2 2 2 2

3 5 6 2 7 1 21 1 3ln | 1| ln | 1| ln( 4) arctan( 1)( 4) 25 25 10 200 2 2( 4) 20 4

x x x xdx x x x Cx x x x

− + ⎛ ⎞= − − + + + − − +⎜ ⎟− + + +⎝ ⎠∫ +

sin 8.sin cos

x dxx x−∫ . We will compute the integral with the help of a rationalizing

substitution. First we will divide both the numerator and the denominator by cos x . Then sin tan

sin cos tan 1x dx dxx

x x x=

−∫ ∫ − . Next we perform the substitution .

Then

tanu x=

arctanx u= , 21

1dx du

u=

+,and 2

sin sin cos ( 1)( 1)

x udx dux x u u

=− −∫ ∫ + .The

decomposition into partials provides 2 2( 1)( 1) 1 1u A Bu

u u u uC+

= +− + − +

whence

. Plugging in 2( 1) ( )( 1u A u Bu C u= + + + − ) 1u = we obtain 12

A = . Comparing

coefficients by provides whence2u 0A B+ = 12

B = − . Finally, comparing the constant

terms we get whence0A C− = 12

C = .Therefore

22 2 2

1 1 1 1 1 1ln | 1| ln( 1) arctan( 1)( 1) 2 1 1 1 2 4 2

u udu du du du u u uu u u u u

⎡ ⎤= − + = − − + +⎢ ⎥− + − + +⎣ ⎦∫ ∫ ∫ ∫ . Recall that whencetanu = x tan u x= and . Now, we can write the answer as

2 21 tan 1 secu x+ = + = 2 x

sin 1 1 ln | tan 1| ln | sec |sin cos 2 2 2

x xdx x x Cx x

= − − + +−∫ .

Calculus 2 172 Review 2 Consider the region between curves cosh , sinh , 0, 1y x y x x x= = = = . 1. Find the area of the region.

Solution. Recall that cosh2

x xe ex−+= andsinh

2

x xe ex−−= , whence sinh coshx x≤ .

Therefore the area is given by the integral

11

00

1 1

(cosh sinh ) (sinh cosh ) sinh1 cosh1 sinh 0 cosh 0

1 10 1 1 0.63.2 2

A x x dx x x

e e e e ee e

− −

= − = − = − − + =

− + −= − − + = − = ≈

∫

2. Find the volume of the solid of revolution when the region is revolved about the x-axis. Solution. We will compute the volume using the washers’ method and the identity 2 2cosh sinh 1x x− = .

1 12 2

0 0

(cosh sinh ) 1 .xV x x dx dxπ π π= − = =∫ ∫

3. Find the volume of the solid of revolution when the region is revolved about the y-axis. Solution. We will use the cylindrical shells’ method.

1

0

2 (cosh sinh )V x x x dxπ= −∫ . We will integrate by parts taking u x= and (cosh sinh )dv x x dx= − . Then du dx= and sinh coshv x x= − whence

11

00

1

0

2 (sinh cosh ) 2 (sinh cosh )

2 (sinh1 cosh1) 2 (cosh sinh ) .

V x x x x x dx

x x dx

π π

π π

= − − − =

= − + −

∫

∫

The last integral we have already computed in Problem 1 and therefore 1 1 22 2 2 1.66y

e eVe e e

π π π− −= − + = ≈

4. Find the coordinates of the geometric center of the region. Solution. By the Pappus’s Centroid Theorem we have

2 0.422 1

yc

V exA eπ

−= = ≈−

, and

0.80.2 2( 1)

xc

V eyA eπ

= = ≈−

5. Find the area of the surface of revolution when the region is revolved about the x-axis. Solution. The area in question consists of four parts. (a) The area of the surface generated by the rotation of the segment of the y-axis from 0 to 1about the x-axis. This surface is a disk with the radius 1 and its area is equal toπ . (b) The area of the surface generated by the rotation of the segment of the vertical line 1x = from sinh1 to cosh1 about the x-axis. This surface is an annulus with the radii sinh1 and cosh1 and its area is equal to 2 2(cosh 1 sinh 1)π π− = . (c) The area of the surface generated by the rotation of the curve coshy x= about the x-axis. This area is given by the formula

21 12

10 0

1 1 12 2

0 0 0

2 2 4 21

20

2 1 2 cosh 1 sinh

2 cosh cosh 2 cosh (cosh 2 1)

4 1sinh 2 1 .2 4 4

dyS y dx x xdxdx

x xdx xdx x dx

e e e exe

π π

π π π

π π π π−

⎛ ⎞= + = + =⎜ ⎟⎝ ⎠

= = = + =

⎛ ⎞− + −= + = + =⎜ ⎟⎝ ⎠

∫ ∫

∫ ∫ ∫

(d) The area of the surface generated by the rotation of the curve sinhy x= about the x-axis. This area is given by the formula

12

20

2 sinh 1 coshS x xdxπ= +∫ . Let coshu x= , then sinhdu xdx= and therefore cosh1cosh1 21

2 2 22

1 1

2 2

12 1 2 1 ln( 12 2

cosh1 1 cosh 1 2 ln(1 2) ln(cosh1 1 cosh 1) .

formula uS u du u u uπ π

π

⎛ ⎞= + = + + + + =⎜ ⎟⎝ ⎠⎡ ⎤= + − − + + + +⎣ ⎦

∫

Adding all four areas from (a) – (d) we get that the surface area in question is approximately 20.65 6. Find the area of the surface of revolution when the parametric curve

cosh , sinh , 0 1x t y t t= = ≤ ≤ , is revolved about the x-axis and about the y-axis. Solution. (a) To find the surface area in case when the curve is revolved about the x-axis we use the formula

2 21 1 12 2 2

0 0 0

2 ( ) 2 sinh sinh cosh 2 sinh 2cosh 1 .xdx dyS y t t t tdt t t dtdt dt

π π π⎛ ⎞ ⎛ ⎞= + = + = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ ∫

Taking coshu t= we have sinhdu tdt= andcosh1

2

1

2 2 1xS u duπ= −∫ . Let 2v u= then

( )2 cosh1 2 cosh1formula 39

2 2 2

22

2

2

22 1 1 ln( 12

2 2 1(cosh1 2cosh 1 1 1) ln 5.072 2 cosh1 2cosh 1 1

xS v dv v v v vππ

ππ

= − = − − + − =

+= − − + ≈+ −

∫

(b) If the curve is revolved about the y-axis we have

2 21 1 12 2 2

0 0 0

2 ( ) 2 cosh sinh cosh 2 cosh 2sinh 1 .ydx dyS x t t t tdt t t dtdt dt

π π π⎛ ⎞ ⎛ ⎞= + = + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ ∫

Taking sinhu t= we have coshdu tdt= andsinh1

2

0

2 2 1yS u duπ= +∫ . Let 2v u= then 2 sinh1 2 sinh1formula 21

2 2 2

00

2 2

22 1 1 ln( 12

2sinh1 2sinh 1 1 ln( 2 sinh1 2sinh 1 1) 10.002

yS v dv v v v vππ

ππ

⎡ ⎤= + = + + + + =⎣ ⎦

= + + + + ≈

∫

7. Find the arc length of the curve 3 , 0 1.xy x= ≤ ≤ Solution. According to the formula for arc length we have

21 1 12 2

0 0 0

1 1 (3 ln3) 1 9 ln 3 .x xdyL dx dx dxdx

⎛ ⎞= + = + = +⎜ ⎟⎝ ⎠∫ ∫ ∫ Let 21 9 ln 3xu = + . Then 2 21 9 ln 3xu = + whence

2 32 9 (ln9)(ln 3) 2(ln 3)9x xudu dx dx= = and, because 2

2

19ln 3

x u −= , we

have 3 2(ln 3)9 (ln3)( 1)xdu dudx

u= =

−. Therefore

2 2

2 2

1 9ln 3 1 9ln 32

21 ln 3 1 ln 3

2 22 2

2 2

1 1 1 1 1 11ln3 1 ln3 2 1 2 1

1 1 1 9ln 3 1 1 1 ln 3 11 9ln 3 1 ln 3 ln ln 2.25.ln 3 2 21 9ln 3 1 1 ln 3 1

u duL duu u u

+ +

+ +

⎛ ⎞= = + − =⎜ ⎟− − +⎝ ⎠

⎡ ⎤+ − + −= + − + + − ≈⎢ ⎥+ + + +⎢ ⎥⎣ ⎦

∫ ∫

8. A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm? Solution. According to the Hooke’s law the force required to hold the spring stretched x meters beyond its natural length is f (x) = kx . When the spring is stretched from 12 cm to 15 cm, the amount stretched is 3 cm = 0.03 m. This means that f (0.03) = 30 , so

0.03k = 30,k = 300.03

= 1000 . Thus f (x) = 1000x and the work done in stretching the

spring from 12 cm to 20 cm is

W = 1000xdx0

0.08

∫ = 500x2 |00.08= 3.2J 9. A tank full of water has the shape of a paraboloid of revolution; that is, its shape is obtained by rotating a parabola about a vertical axis. If its height is 4 ft and the radius at the top is 4 ft, find the work required to pump the water out of the tank. Solution. The equation of the revolved parabola is y = ax2 . Because f (4) = 4 we

have y = 14x2 and x = 2 y . Therefore the area of cross-section at height y is

A(y) = 4π y . Considering a thin slice of water from the height of y to y + Δy we see that the work required to raise the water in this slice to the top of the tank is approximately ΔW = 62.5 × 4π y(4 − y)Δy . Finally,

W = 250π y(4 − y)dy0

4

∫ =8000π3

lb-ft.

172 Calculus 2 Review 3

1. The curve ln , 1xy xx

= ≥ is revolved about the x-axis. Is the volume of the

resulting solid finite or infinite? Solution. We apply the disks method and see that the question is about the

convergence or divergence of the following improper integral. 2

21

ln x dxx

π∞

∫ . We will prove that the integral converges and find its value using integration by parts. Let 2 2

1ln ,u x dv dxx

= = . Then 1 12ln , du x dx vx x

= = − , and

2 2

2 21 11

ln ln ln2x x xdx dxx x x

π π π∞∞ ∞

= − +∫ ∫ . The first term in this expression is equal to 0. Indeed, at the lower limit we have ln1 0= and at the upper limit we have to look at

2lnlimx

xx→∞

. The last limit is an indeterminate form ∞∞

and we can apply the

L’Hospital’s rule. 2

12lnln 2ln 2 /lim lim lim lim 01 1x x x x

xx x xxx x→∞ →∞ →∞ →∞

⋅= = = = . Therefore

2

2 21 1

ln ln2x xdx dxx x

π π∞ ∞

=∫ ∫ . We apply integration by parts once again; this time we take

2

1ln ,u x dv dxx

= = . Then 1 1,du dx vx x

= = − , and 2 211 1

ln ln 12 2 2x xdx dxx x x

π π π∞∞ ∞

= − +∫ ∫ . Like in the previous expression we see that the first term is 0 and it remains to evaluate

211

1 12 2 2dxx x

π π π∞∞

= − =∫ . So, the volume is finite and its value is 2π . 2. The same question about the area of the surface of revolution. Solution. The area of the surface of revolution can be expressed as the following

integral. 2

1

2 ( ) 1 dyy x dxdx

π∞ ⎛ ⎞+ ⎜ ⎟⎝ ⎠∫ . We will prove, using the comparison test that the

integral diverges to ∞ . To this end notice the simple inequality 2

1 1dydx

⎛ ⎞+ ≥⎜ ⎟⎝ ⎠ whence

2

1 1 1dydx

⎛ ⎞+ ≥ =⎜ ⎟⎝ ⎠and

2

1 1 1

ln2 ( ) 1 2 ( ) 2dy xy x dx y x dx dxdx x

π π π∞ ∞ ∞⎛ ⎞+ ≥ =⎜ ⎟⎝ ⎠∫ ∫ ∫ . Let lnu x= then

1du dxx

= and 20

1 0

ln2 2xdx udu ux

π π π∞ ∞

∞= = = ∞∫ ∫ . Therefore

2

1

2 ( ) 1 dyy x dxdx

π∞ ⎛ ⎞+ =∞⎜ ⎟⎝ ⎠∫ and

the area of the surface of revolution is infinite.

3. Prove that the integral 212 2

dxx x

∞

−∞ + +∫ converges and find its value.

Solution.

2 2 2

1 1 1 (where 1)2 2 ( 1) 1 1

arctan .2 2

dx dx du u xx x x u

u π π π

∞ ∞ ∞

−∞ −∞ −∞

∞

−∞

= = = + =+ + + + +

⎛ ⎞= = − − =⎜ ⎟⎝ ⎠

∫ ∫ ∫

(Recall that lim arctan2u

u π→±∞

= ± )

4. Use the Ratio test to find out whether the series 0

!(2 )!

n

n

n en

∞

=∑ converges or

diverges.

Solution. Let !(2 )!

n

nn ean

= be the nth term of our series. Then the ratio 1nn

aa+ can be

simplified as 1 1

1 ( 1)! ! ( 1)! (2 )![2( 1)]! (2 )! ! [2( 1)]!

1( 1) 0.(2 1)(2 2) 2(2 1)

n n nn

nn

n

a n e n e n n ea n n n n e

en en n n

+ ++

→∞

+ += ÷ = ⋅ ⋅ =+ +

= + = →+ + +

By the ratio test the series converges.

5. Find the radius of convergence of the power series2 ln

n

n

xn n

∞

=∑ .

Solution. We will apply the ratio test to find the radius of convergence. Let

ln

n

nxan n

= . Then 1| | ln | || | ( 1) ln( 1)n

n

a n n xa n n+ =

+ +. Next notice that lnlim 1

( 1) ln( 1)nn n

n n→∞=

+ +.

Indeed, lim 11nnn→∞

=+

. To find the limit lnlimln( 1)n

nn→∞ +

we will apply the L’Hospital’s

rule,1

ln( 1) ( 1)lim lim lim 11ln 1x x xu uuu uu

→∞ →∞ →∞

+ += = =+

. Therefore 1lim | |nn

n

a xa+

→∞= . By the ratio test

the series converges absolutely if | | 1x < and diverges if | | 1x > . 6. Find the value of arcsin(0.499) with accuracy 710− using an appropriate Taylor series. Solution. A “brute force” solution would be to use the Maclaurin series for arcsin at the point 0.499. The solution suggested below is much more effective.

As often in such problems we will try to state the problem in such a way that we can use Maclaurin series of known pattern. To this end notice that arcsin 0.5

6π= .

Next we will use the formula ( )2 2arcsin arcsin arcsin 1 1a b a b b a− = − − − . We can easily prove this formula if we notice that sin(arcsin arcsin ) sin(arcsin )cos(arcsin ) cos(arcsin )sin(arcsin )a b a b a b− = − and recall that 2sin(arcsin ) and cos(arcsin ) 1x x x x= = − . In our case we get

221 1arcsin(0.499) arcsin 1 0.499 0.499 1 arcsin 0.0011543159

6 2 2π ⎛ ⎞⎛ ⎞⎜ ⎟− = − − − ≈⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

. Now we

will use the Maclaurin series 2 11

1 3 5 (2 1)arcsin2 !(2 1)

nn

n

nx x xn n

∞+

=

⋅ ⋅ ⋅ ⋅ −= ++∑

K .

If we compute the sum of the first two terms of the series above for 0.0011543159x = we will get

31arcsin(.499) 0.0011543159 (.0011543159) .52244445946 2 3π≈ − − ≈

⋅. All the digits are

correct. Remark. You will not be required to do it on the test but it is not difficult to estimate the error of our approximation. Indeed,

| Error |≤ 1⋅3⋅…⋅(2n−1)2n n!(2n+1)n=2

∞

∑ (0.0011543159)2n+1 ≤ (0.0011543159)2n+1n=2

∞

∑ =

= (0.0011543159)5 [(0.0011543159)2]nn=0

∞

∑ = (0.0011543159)5

1− (0.0011543159)2≈ .2049386957 ×10-14

7. Estimate the value of / 4

0

sin xdxx

π

∫ with accuracy 0.000001 using an appropriate Taylor series.

Solution. 2 1

0sin ( 1)

(2 1)!

nn

n

xxn

+∞

=

= −+∑ . Therefore

2

0

sin ( 1)(2 1)!

nn

n

x xx n

∞

=

= −+∑ and

/ 4/ 4 / 4 2 1 2 12

2 10 0 00 0 0

sin ( 1) ( 1) ( 1)(2 1)! (2 1)! 2 1 4 (2 1)(2 1)!

n n n n nn

nn n n

x xdx x dxx n n n n n

ππ π π+ +∞ ∞ ∞+

= = =

− − −= = =+ + + + +∑ ∑ ∑∫ ∫ .

Next we have to decide how many terms of the last series we want to use for approximation. The series is an alternating one so when we stop the error we make is not greater than the absolute value of the next term. We can compute that for

5n = the absolute value of the fifth term is 6.6543351347 10−× which is enough for the accuracy we need. Therefore we use the approximation / 4 3 5 7

0

sin .7589764 1152 614400 578027520

xdxx

π π π π π≈ − + − ≈∫ . 8. Find the solution of differential equation dy

dx= sinh x

x, y(0) = 1, in the form of a

Taylor series.

Solution. sinh x = x2n+1

(2n +1)!n=0

∞

∑ . Thus sinh xx =x2n

(2n +1)!n=0

∞

∑ and

y(x) = 1+ t2n

(2n +1)!n=0

∞

∑0

x

∫ = 1+t 2n

(2n +1)!= 1+ x

2n+1

(2n +1)(2n +1)!.

n=0

∞

∑0

x

∫n=0

∞

∑ 9. Find the Maclaurin polynomial of degree 4 for the function sec x using the techniques of division of Taylor series. Use this polynomial to approximate sec5! . Solution. Because sec0 = 1 and sec x is an even function its Maclaurin polynomial of degree 4 is of the form 1+ a2x2 + a4x4 . As sec xcos x = 1 and Maclaurin

polynomial of degree 4 for cos x is 1− x2

2!+ x

4

4! we have

(1+ a2x

2 + a4x4 )(1− x

2

2!+ x

4

4!) ∼1.

From it we obtain, a2 −12= 0, a2 =

12

and a4 −a22+ 14!

= 0, a4 =14− 124

= 524

.

Next, sec5! = sec π

36≈1+ 1

2π 2

362+ 524

π 4

364≈1.003819800. For comparison, a direct

approximation with a calculator provides 1.003819838. 10. Use binomial Maclaurin series to write and exact expression for the arc length of the curve y = x3,0 ≤ x ≤1/ 2 in the form of a Maclaurin series. Use the obtained series to estimate the arc length with the accuracy of 10-6. Solution.

L = 1+ 9x4 dx0

0.5

∫ = 1+1/ 2(1 / 2 −1)...(1 / 2 − n +1)

n!9n x4n

n=1

∞

∑⎛⎝⎜⎞⎠⎟0

0.5

∫ =

= 0.5 + 1/ 2(1 / 2 −1)...(1 / 2 − n +1)n!

9n 0.54n+1

4n +1.

n=1

∞

∑

This is an alternating series and we can check that

1/ 2(1 / 2 −1)(1 / 2 − 2)(1 / 2 − 3)(1 / 2 − 4)(1 / 2 − 5)6!⋅25

96(1 / 2)25 < 2.2 ⋅10−7

Therefore for the accuracy we need it is enough to take

L ≈ 0.5 + 1/ 2(1 / 2 −1)...(1 / 2 − n +1)n!

9n 0.54n+1

4n +1≈ 0.526268

n=1

5

∑

Math 172

Review for the final exam.

1. Find the integrals.(8 points each)

(a) ∫x arctanx dx;

Solution. We perform integration by parts taking u = arctanxand dv = xdx. Then

du =1

1 + x2dx and v =

x2

2.

Therefore∫x arctanx dx =

x2

2arctanx−

∫x2

2(1 + x2)dx =

=x2

2arctanx−1

2

∫ (1− 1

1 + x2

)=x2

2arctanx−x

2+

1

2arctanx+C =

=x2 + 1

2arctanx− x

2+ C.

(b) ∫1

t2 + 2t+ 2dt;

Solution. By completing the square we get∫1

t2 + 2t+ 2dt =

∫1

(t+ 1)2 + 1dt.

Let u = t+ 1 then du = dt and∫1

(t+ 1)2 + 1dt =

∫1

u2 + 1du = arctanu+C = arctan(t+1)+C.

1

(c) ∫cos3 x sin

12 x dx;

Solution. First we write this integral in the form∫cos2 x sin

12 x cosx dx =

∫(1− sin2 x) sin

12 x cosx dx

and then make the substitution u = sinx, whence du = cosxdx.Thus our integral becomes∫

(1− sin2 x) sin12 x cosx dx =

∫(1− u2)u

12 du =

=

∫u

12 − u

52 du =

2

3u

32 − 2

7u

72 + C =

=2

3sin

32 x− 2

7sin

72 x+ C.

(d) ∫e2x − e−2x

e2x + e−2xdx.

Solution. Probably the simplest way to solve the problem is tonotice that

d

dx(e2x − e−2x) = 2e2x − (−2)e−2x = 2(e2x + e−2x).

Thus, the numerator of the fraction we integrate equals to onehalf of the derivative of the denominator whence∫

e2x − e−2x

e2x + e−2xdx =

1

2ln(e2x + e−2x) + C.

Remark The answer also can be written in the form∫e2x − e−2x

e2x + e−2xdx =

1

2cosh 2x+ C.

2

2. Find the value of the improper integral (10 points)

∞∫2

1

x√x2 − 1

dx.

Solution. We perform the substitution x = sec t. Thendx = sec t tan tdt and

√x2 − 1 =

√sec2 t− 1 =

√tan2 t = tan t. We

have to find out how the limits of integration will change under thesubstitution. If x = 2 then sec t = 2 whence cos t = 1

2and

t = arccos 12

= π3. If x→∞ then cos t = 1

sec t= 1

xwhence

t = arccos 1x→ arccos 0 = π

2. Therefore after the substitution we get a

proper integral

∞∫2

1

x√x2 − 1

dx =

π2∫

π3

sec t tan t

sec t tan tdt =

π2∫

π3

dt =π

2− π

3=π

6.

3. Find the volume of the solid generated when the region bounded byy = 2− x, y =

√x and x = 0 is revolved about x-axis.(10 points)

Solution. The graph of the region is shown below. (On the test youdo not have to provide the graph) To find the point of intersection ofthe curves y = 2− x and y =

√x we have to solve the equation

2− x =√x. After we square both parts we get 4− 4x+ x2 = x or

x2 − 5x+ 4 = (x− 1)(x− 4) = 0. Only x = 1 is the solution of ouroriginal equation.

Next we apply the disks’ method to find the volume.

Vx = π

1∫0

[(2− x)2 − (√x)2] dx = π

1∫0

(x2 − 5x+ 4) dx =

= π(x3

3− 5x

2

2+ 4x

) 1∣∣∣0

= π(1

3− 5

2+ 4)

=11

6π.

4. Find the volume of the solid generated when the region bounded byy = 2− x, y =

√x and x = 0 is revolved about y-axis.(10 points)

3

Solution. We apply the cylindrical shells’ method.

Vy = 2π

1∫0

x[(2− x)−√x] dx = 2π

1∫0

(2x− x2 − x32 ) dx =

= 2π(x2 − x

3

3− 2

5x

52

) 1∣∣∣0

=8π

15.

4

y

-0.4-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.20.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

Centroid

Graphmatica 2.0e © 2005 kSoft, Inc. - region.gr

Equations on screen: 1. y=sqrt(x) 2. y=2-x 3. x=0

Data Plots: Data plot 1

5

5. Find the coordinates of the centroid of the region described in theprevious problem. (10 points) Solution. First we have to find thearea of the region.

A =

1∫0

(2− x−√x) dx =

(2x− x

2

2− 2

3x

32

) 1∣∣∣0

= 2− 12− 2

3=

5

6.

Now we can compute the coordinates of centroid.

xc =Vy

2πA=

8π

15÷ 5π

3=

8

25= 0.32,

yc =Vx

2πA=

11π

6÷ 5π

3=

11

10= 1.1.

6. Determine whether the series absolutely converges, conditionallyconverges, or diverges.(5 points each)

(a)∞∑k=2

13√k6 − 3k

;

Solution.We apply the limit comparison test. By keeping onlythe leading terms in the numerator and the denominator we getthe following series

∞∑k=2

13√k6

=∞∑k=2

1

k2.

The last series converges because the exponent of k in thedenominator is greater than 1. By limit comparison test ouroriginal series also converges.

(b)∞∑k=1

(−1)k kk2 + 2

;

Solution. This is an alternating series. Indeed,

• the sign of the terms alternates;

6

•limk→∞

k

k2 + 2= 0;

• The absolute values of the terms of the series are decreasing.To prove it it is enough to prove that the functiong(x) = x

x2+2is decreasing. Indeed, by the quotient rule

dg

dx=x2 + 2− x(2x)

(x2 + 2)2=

2− x2

(x2 + 2)2.

The last expression is negative if x >√

2 and therefore theabsolute values of the terms of the series are decreasingstarting from k = 2.

So the series converges as an alternating series. To see whether itconverges absolutely or conditionally let us look at the series ofabsolute values,

∞∑k=1

k

k2 + 2.

The limit comparison test tells us that this series converges ordiverges at the same time as the series

∞∑k=1

k

k2=∞∑k=1

1

k.

But the last series diverges and therefore the series of absolutevalues diverges and our original series converges conditionally.

(c)∞∑k=1

(n!)2

(2n)!;

Solution. We will apply the ratio test.

an+1an

=(n+ 1)!2

[2(n+ 1)]!÷ (n!)

2

(2n)!=((n+ 1)!

n!

)2 (2n)!(2n+ 2)!

=

=(n+ 1)2

(2n+ 1)(2n+ 2)=

n+ 1

2(2n+ 1)→n→∞

1

4< 1.

By ratio test the series converges.

7

7. Find the function to which the series∞∑k=1

(k − 1)xk+1

converges.(10 points)

Solution. The solution is based on two formulas. The first one is theinfinite geometric progression.

∞∑k=0

xk =1

1− x, −1 < x < 1. (?)

The second we obtain if we differentiate both parts of (?).

∞∑k=1

kxk−1 =1

(1− x)2, −1 < x < 1. (??)

Now we can compute the sum in our problem.

∞∑k=1

(k − 1)xk+1 =∞∑k=1

kxk+1 −∞∑k=1

xk+1 =

= x2∞∑k=1

kxk−1 − x2∞∑k=1

xk−1 =

(where i = k − 1))

= x2∞∑k=1

kxk−1 − x2∞∑i=0

xi =

(by (?) and (??))

=x2

(1− x)2− x

2

1− x=x2 − x2(1− x)

(1− x)2=

x3

(1− x)2.

Of course, the series converges and the formula is correct only on theinterval −1 < x < 1.

8. Find the Taylor series about x = a for the given function; expressyour answer in sigma notation (Σ); then find its radius of convergenceand the interval of convergence.(10 points each)

8

(a)

f(x) =1

2 + x,

at 0.

Solution. Plugging in −x instead of x into formula (?) we get

1

1 + x=∞∑k=0

(−1)kxk, −1 < x < 1. (? ? ?)

Therefore

1

2 + x=

1

2(x+ 12

=1

2

1

1 + x2

=1

2

∞∑k=0

(−1)k(x

2

)k=

=∞∑k=0

(−1)k xk

2k+1, −1 < x

2< 1, or − 2 < x < 2.

The radius of convergence of the Maclaurin series above is 2 andthe interval of convergence is (−2, 2).

(b)f(x) = lnx,

at 2.

Solution. It would be not very difficult to use the formula forTaylor series of function f at point a

∞∑k=0

f (k)(a)

k!(x− a)k, wheref (0) = f,

but it is easier to reduce the problem to one of our standardMaclaurin series. Recall that integrating both parts in (? ? ?) weget

ln (1 + x) =∞∑k=1

(−1)k+1xk

k, −1 < x < 1. (? ? ??)

Let u = x− 2 then x = u+ 2 and, using (? ? ??) we obtain

lnx = ln (u+ 2) = ln [2(1 +u

2)] = ln 2+ln (1 +

u

2) = ln 2+

∞∑k=1

(−1)k+1 (u/2)k

k=

9

=∞∑k=1

(−1)k+1 uk

2kk=∞∑k=1

(−1)k+1 (x− 2)k

2kk.

The series above converges if −1 < u/2 < 1, or −2 < u < 2, or−2 < x− 2 < 2, or 0 < x < 4. Therefore its radius ofconvergence is 2 and its interval of convergence is (0, 4).

10

SyllabusReview1Review2Review3Review-Final