AP Calculus D Final Exam Review Answers Chapter 7

1a. Use partial fractions.

€

9x − 32x −1( ) x +1( )

=A

2x −1+

Bx +1

€

⇒ 9x − 3 = A x +1( ) + B 2x −1( ) . Write two

equations, one for the x parts and one for the constant parts

€

9 = A + 2B−3 = A − B

. Solve two equations with

two unknowns by substitution or elimination to get A = 1 and B = 4. The integral becomes 1

2x −1+

4x +1∫ dx . Use “u” substitution and natural logarithms, to get 12 ln 2x −1 + 4 ln x +1 +C .

1b. Use integration by parts (u and dw). Let u = ln x giving du = 1x  dx , dw = x4  dx giving w = x55 .

Using the integration by parts formula, the original integral becomes ln x( ) x5

5( )− x5

5( ) 1x( )∫ dx

simplify the integral to ln x( ) x5

5( )− x4

5( )∫ dx = ln x( ) x55( )− x5

25 +C .

1c. Complete the square. 2

x2 − 6x + −3( )2( )+10− −3( )2 dx∫ = 2

x −3( )2 +1 dx∫ = 2 tan−1 x −3( )+C .

2a. Replace the limit value with b and write a limit equation as limb→∞

e− x

x1

b∫ dx . Solve using u

substitution with

€

u = − x giving du = − 12 x−12  dx . This gives lim

b→∞ -2 eu

1

b∫ du , solving the

integral gives limb→∞

-2eux=1

x=b, substitution u back in gives lim

b→∞ -2e− x

1

b. Inputting the limits

gives limb→∞

− 2e− b + 2e− 1 . Solving the limit gives

€

−2e− ∞ + 2e−1

€

= −2e−∞ + 2e−1 = − 2e∞+ 2e1

= − 2∞+ 2e = − 2∞ +

2e = 0+ 2e =

2e , therefore converges.

2b. Replace the limit value with b and write a limit equation as limb→1

x

1− x20

b∫ dx . To solve this

integral use u substitution with

€

u =1− x2 giving du = −2x  dx . This gives limb→1

-12

1u0

b∫ du ,

solving the integral gives limb→1

−12⋅u

12

12 x=0

x=b

, substitution u back in gives limb→1

− 1− x2

0

b.

Inputting the limits gives limb→1

− 1− b2 + 1− 02 . Solving the limit gives

€

− 1−12 + 1

€

= − 0 +1=1, therefore converges.

Final Exam Review Answers Chapter 8

3a. Area = x − x3( ) 0

1∫ dx

3b. Washer Method: Volume = π y3( )2− y2( )

2 

0

1∫ dy

3c. Volume = π2x−x32( )

2 

0

1∫ dx = π

8 x − x3( )2 

0

1∫ dx

4a. Arc length = 1+ 5− 2x( )2  0

5∫ dx

4b. Washer Method: Volume = π 5x − x2 + 2( )2− 2( )2  

0

5∫ dx

5a. The region to find the area of is the unshaded moon shape in the

diagram at the right. It is symmetrical to the x-axis, therefore can double and integral for the top portion of the region. Both graphs begin on the right side of the x-axis; therefore the lower limit is 0. To find the intersection (on top), set the equations equal to each other and solve for

€

θ .

€

2 1+ cosθ( ) = 6cosθ ⇒1+ cosθ = 3cosθ

⇒1= 2cosθ ⇒ 12 = cosθ and solve for

€

θ = cos−1 12( ) =π3

.

Therefore area = 2 ⋅ 12 6cosθ( )2 − 2 1+ cosθ( )( )2( )0

π3∫ dθ .

5b. Arc length = r2 + drdθ( )2

a

b∫ dθ , using

€

r = 2 1+ cosθ( ) = 2 + 2cosθ , then

€

" r = −2sinθ giving arc

length = 2+ 2cosθ( )2 + −2sinθ( )20

2π∫ dθ = 8+8cosθ

0

2π∫ dθ . (If you want to simplify)

5c. For C2,

€

r = 6cosθ and

€

" r = −6sinθ , giving the values

€

r = 6cos π6( ) = 6 3

2# $ % &

' ( = 3 3 and

€

" r = −6sin π6( ) = −6 12( ) = −3. Slope is

€

dydx

=" r sinθ + rcosθ" r cosθ − rsinθ

€

=−3sin π

6( ) + 3 3cos π6( )

−3cos π6( ) − 3 3sin π

6( )

€

=−3 12( ) + 3 3 3

2# $ % &

' (

−3 32

# $ % &

' ( − 3 3 1

2( )

€

=− 32 + 9

2

− 3 32 − 3 3

2

€

=3

−3 3

€

=−13

6. x t( ) = sin t , x ' t( ) = cost and y t( ) = t2 , y ' t( ) = 2t , then arc length = cos2 t + 2t( )2  dt−π

π

∫

7. The work to lift the bucket is constant, therefore work = 1400 (30) = 42,000 foot-lbs. The work to

lift the cable is a variable, therefore work = 40− y( )4 0

30∫ dy = 3,000 foot-lbs. This gives a total

work of 42,000 + 3,000 = 45,000 foot-lbs. 8a. Since the work occurs vertically, all values (and expressions)

must be done in terms of y. The integral for work requires the density (50 lb/ft3), the distance each cross section of the liquid is lifted (top of container minus cross section = 10 – y), and area of cross section (cross section is a circle with radius =

€

35 y ,

therefore area =

€

π 35 y( )2). The limits are the lowest and

highest y-values of the cone.

The work = 50 10− y( ) π 35 y( )2"

#$

%&'

0

10∫ dy .

8b. To find the force, think of a small band going around the tank. The area of the band will be the

circumference at that level times the band's height, area = 2πr  dy . The value for r will be the

same as part (a), therefore area = 2π 35 y( )dy . The force will be 50 10− y( ) 2π 3

5 y( )( )0

10∫ dy .

Final Exam Review Answers Chapter 9

9. limn→∞

  ln 5n2( )ln n3( )

=ln 5 ∞( )2( )ln ∞( )3( )

= ∞∞ , apply L'Hopital's rule, lim

n→∞ 

10n5n2

3n2

n3

= limn→∞

  10n4

15n4

= limn→∞

  23

=23≠∞ ,

therefore the sequence converges.

10a. Using nth-Term-Test: limn→∞

  n2n+3

, divide each term by n, limn→∞

  12+ 3

n

€

=1

2 + 3∞

=12

. Since the

limit ≠ 0, then by the Nth-term test n2n+3n=1

∞

∑ diverges. (This is not an alternating series so do not

have to check conditional convergence.)

10b. Using Limit Comparison Test: Compare to the series 1n32n=2

∞

∑ , this is a p-series with p = 32 >1 ,

therefore converges. Next, limn→∞

  n3n2 + n−1

⋅n

32

1 = lim

n→∞  n2

3n2 + n−1, divide all terms by n2 to get

limn→∞

  13+ 1

n−1n2

= 13+ 1∞− 1

∞2

€

= 13+ 0 − 0

=13

. Since 0 < limit < ∞ and 1n32n=2

∞

∑ converges,

then by the Limit Comparison Test n3n2 + n−1n=2

∞

∑ converges. Since the absolute series converges,

then by the Absolute Convergence Test −1( )n n3n2 + n−1n=2

∞

∑ converges absolutely.

10c. Check Absolute Convergence, −1( )n+11+ nn2n=1

∞

∑ = 1+ nn2n=1

∞

∑ . Using Limit Comparison Test with

1nn=1

∞

∑ , which is the harmonic series and therefore diverges. limn→∞

  1+ nn2 ⋅

n1

= limn→∞

  n+ n2

n2 , divide

all terms by n2 to get limn→∞

  1n+11

€

=1∞+11

€

=0 +11

=1. Since 0 < limit < ∞ and 1nn=1

∞

∑ diverges

then by the Limit Comaprison Test 1+ nn2n=1

∞

∑ diverges. This means that −1( )n+1 1+ n( )

n2n=1

∞

∑ does not

converge absolutely.

Check Conditional Convergence, (1) if

€

f x( ) =1+ xx2

then

€

" f x( ) =1 x2( ) − 1+ x( ) 2x( )

x2( )2

€

=−2 − 2xx3

which will always be negative for all values for x > 1. Since the derivative is negative, then f (x)

will always be decreasing and the terms of

€

1+ nn2

are decreasing, (2) limn→∞

  1+ nn2 , divide all the

terms by n2, limn→∞

  1n2 + 1

n

1

€

=1∞2

+ 1∞

1

€

= 0 + 01

= 0. Since both conditions are satisfied, then by

the Alternating Series Test −1( )n+1 1+ n( )

n2n=1

∞

∑ converges conditionally.

10d. Using Direct Comparison Test, start with

€

cos n( ) ≤1, meaning that

€

1+ cos n( ) ≤ 2 which leads to

€

1+ cos n( )n2

≤2n2

. The series 2n2n=1

∞

∑ is a p-series with p = 2 > 1, therefore 2n2n=1

∞

∑ converges. Since

2n2n=1

∞

∑ converges, then by the Direct Comparison Test, then 1+ cos n( )

n2n=1

∞

∑ converges (absolutely).

11. This is geometric series with

€

r = −35x . A geometric series will converge when

€

r <1, therefore

this series will converge when

€

−35x <1. Solving for x gives

€

−1<35x <1 therefore the values of x

that will make the series converge are

€

−53

< x <53

.

** Alternate version! ** Could start with ratio test. limn→∞

  35 x( )n+1

35 x( )n

= limn→∞

  35 x( )n 3

5 x( )35 x( )n

= limn→∞

  35 x =

35 x . Now

€

35 x <1 ⇒−1< 35 x <1 ⇒− 53 < x <

53 .

12. Start with the Ratio Test. limn→∞

  x − 2( )n+1

n+1⋅

nx − 2( )n

= x − 2 limn→∞

  nn+1

, divide all the terms by n

to get x − 2 limn→∞

  11+ 1

n

€

= x − 2 11+ 1

∞

€

= x − 2 . For the ratio to converge,

€

x − 2 <1

⇒−1< x − 2 <1 ⇒1< x < 3 . The radius of convergence will be

€

3−12

=1.

Test the endpoints, for x = 1,

€

−1( )n 1− 2( )n

nn=1

∞

∑

€

=−1( )n −1( )n

nn=1

∞

∑

€

=1nn=1

∞

∑ , this is the harmonic

series, which diverges. The other endpoint, x = 3,

€

−1( )n 3 − 2( )n

nn=1

∞

∑

€

=−1( )n 1( )n

nn=1

∞

∑

€

=−1( )n

nn=1

∞

∑ ,

this is the alternating harmonic series, therefore it converges conditionally. The interval of convergence is

€

1< x ≤ 3.

Final Exam Review Answers Chapter 10

13a. ex =1+ x + x2

2!+x3

3!+... = xn

n!n=0

∞

∑ giving e−x2=1+ −x2( )+

−x2( )2

2!+−x2( )

3

3!+...

=1− x2 + x4

2!−x6

3!+... =

−1( )n x2n

n!n=0

∞

∑

13b. e−x2

∫ dx = 1− x2 +x4

2−x6

3!+... dx∫

€

= x − x3

3+x5

5 ⋅ 2−x7

7 ⋅ 3!+ ... =

−1( )n x2n+1

2n+1( )n!n=0

∞

∑

14.

f x( ) = 1x

f 1( ) =1

f ' x( ) = −x−2 f ' 1( ) = −1

f " x( ) = 2x−3 f " 1( ) = 2

f '" x( ) = −6x−4 f '" 1( ) = −6

P3 =1− x −1( )+2 x −1( )2

2!−

6 x −1( )3

3!P3 =1− x −1( )+ x −1( )2 − x −1( )3

Pn = −1( )n x −1( )n

n=0

∞

∑

15. cos x =1− x2

2!+x4

4!−x6

6!+... =

−1( )n x2n

2n( )!n=0

∞

∑ giving cos 2x( ) =1−2x( )2

2!+2x( )4

4!−2x( )6

6!+...

=1− 4x2

2!+16x4

4!−64x6

6!+...  =1− 2x2 + 2x

4

3−4x6

45+...  . Then

cos 2x( )+ 2x2 −1= 1− 2x2 + 2x4

3−4x6

45+...

"

#$$

%

&'' +2x

2 −1 = 2x4

3−4x6

45+... . Then

cos 2x( )+ 2x2 −1x4

=

2x4

3−4x6

45+...

x4  =23−4x2

45+... . Finally lim

x→0 cos 2x( )+ 2x2 −1

x4  = lim

x→0 23−4x2

45+...

=23−4 ( )( )2

45+... = 2

3.

Final Exam Review Answers Chapter 11

16a. Move y to the left side and dx to the right side giving y dy∫ = e2x dx∫ which becomes

€

y2

2=12e2x + C

or

€

y2 = e2x + C , plug in point (0, 5) to get

€

5( )2 = e2 0( ) + C giving C = 24, therefore

€

y2 = e2x + 24

giving y = e2x + 24 .

16b. Move y to the left and dx to the right side giving 1y dy∫ = sin x( ) dx∫ which becomes

ln y = −cos x +C , rolling the log gives y = Ae−cos x . Substituting in the point gives 1= Ae−cosπ2( )

⇒1= Ae0 giving A = 1, therefore the final equation is y = e−cos x .

17.

18. dPdt

= 0.02P − 0.00004P2 = 0.02P 1− 0.002P( ) = 0.02P 1− P500

"

#$

%

&'

18a. k = 0.02, L = 500, A = 500− 5050

= 9 , P = 5001+ 9e−0.02t

18b. limt→∞

 P t( ) equals the limit value, therefore equals 500.

18c. Maximum rate of change =L2=5002

= 250

18d. Maximum rate of change occurs when t = 1klnA = 1

0.02ln9 = 50 ln9 ≈109.861

18e. Horizontal asymptotes P = 0 and P = 500.

19.

x y dydx

= xcos y( ) Δy = slope( )Δx

P0 2 0 2cos(0) = 2 2(0.5) = 1

P1 2.5 1 2.5cos(1) = 1.351 1.351(0.5) = 0.675

P2 3 1.675 -0.313 -0.157

P3 3.5 1.518 0.183 0.092

P4 4 1.61