APSC 173 Final Exam

Supplemental Learning
Review Packet

Matthias Jenne 2023

APSC 173 Final Exam SL Review Packet 2023

dx
1.  x2 − 4 x + 3

Solution:
dx dx dx
 x2 − 4x + 3
=
( x 2 − 4 x + 4) + 3 − 4
=
( x − 2) 2 − 1
, u = x − 2, du = dx

du
= → formula sheet →= cosh −1 (u ) + C = cosh −1 ( x − 2) + C
u −12

1
2.  ln( x)dx
0

Solution:
1
 ln( x)dx use integration by parts:  udv = uv −  vdu
0

1
[u = ln( x), dv = dx] → [du = dx, v = x ]
x
1 1 1
0 ln( x)dx = [ x ln( x)]0 − 0 x x dx
1

1 1
 ln( x)dx = [ x ln( x)] −  dx
1
0 0 0
1
 ln( x)dx = [ x ln( x)] −1
1
0 0

= 1ln(1) − lim+ p ln( p)  − 1

 p →0 
 
 
 
ln( p ) 
= 0 − lim+ −1
 p →0 1 
 p 
 
 L ' Hopital 
= −1
1
ln( p ) p 1
Note: lim+ = lim+ = lim+ − p 2 = 0
p →0 1 p →0 1 p →0 p
− 2
p p

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APSC 173 Final Exam SL Review Packet 2023

 1
3. Does 
1 3 + x4
dx converge?

Solution:
 dx
We know 1 xp
converges for p  1

dx
 converges
1 x4

1 1
 4 for all x
3+ x 4
x
 dx
 converges
1 3 + x4

 n−1
1
4. Find   
n=2  2 

Solution:
 n −1 1 2 3
1 1 1 1
  
n=2  2 
=   +   +   + ...
2 2 2
 1 0  1 0   1 1  1  2  1 3
=   −    +   +   +   + ...
 2   2    2   2   2 
0 1 2 3
1 1 1 1
= −1 +   +   +   +   + ...
2 2 2 2
 n

  2 
1
n =0

 n
1
= −1 +    (Geometric series)
n=0  2 

1
= −1 + =1
1
1−
2

Alternatively:
 n −1  n +1  n 1  n
1 1 1 1 1 1
  
n=2  2 
=  
n =0  2 
=     =  
n =0  2   2  2
  
n =0  2 
Geometric series

1 1
=  =1
 2 1− 1
2

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APSC 173 Final Exam SL Review Packet 2023


3 − 2n −1
5. Find 
n =1 5n −1

Solution:

3 − 2n −1  3 
2n −1

n =1 5n −1
= 
n =1 5
n −1
− 
n =1 5
n −1

 n −1 n −1
1 2
= 3   −   
n =1  5  n =1  5 

1 1 25
=3 − =
1 2 12
1− 1−
5 5

6. Does  nn converge? Use the integral test.
e n =1

Solution:
x
Let f ( x) = . f ( x) is posisitve, continuous and decreasing for the given bounds. Therefore apply the integral test.
ex
 x  1
1 e x dx → compare with 1 3 dx
x2
 x
  x dx converges
1 e


n
  n converges via the integral test
n =1 e

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APSC 173 Final Exam SL Review Packet 2023


(2n)!
7. Does  (n!)
n =1
2
converge?

Solution:
(2n)! (2(n + 1))!
an = , an +1 =
(n !) 2
((n + 1)!) 2
Apply ratio test
(2n + 2)! (2 n + 2)(2 n +1)(2 n )!

an +1 ((n + 1)!) 2
(2n + 2)! ( n !) 2 (2n + 2)(2n + 1)(2n)! n !n !
lim = lim = lim = lim
n → an n → (2n)! n → (( n + 1)!) 2
(2n)! n→ ( n + 1)( n + 1) n ! n ! (2n)!
2
(n !) ( n +1) n!

(2n + 2)(2n + 1) 2(2n + 2)(2n + 1) 2(2n + 2)(2n + 1) 4n + 2

= lim = lim = lim = lim = 4 1
n → (n + 1)(n + 1) n → 2(n + 1)(n + 1) n → (2n + 2)(n + 1) n → n + 1

 Diverges

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APSC 173 Final Exam SL Review Packet 2023


n!
8. Does n n =0
n
converge?

Solution:
n! ( n + 1)!
an = , an +1 =
n n
(n + 1) n +1
Apply Ratio Test:
(n + 1)!
an +1 (n + 1) n +1 (n + 1)! n n (n + 1)n ! n n
lim = lim = lim = lim
n → an n → n! n → ( n + 1) n +1 n ! n → ( n + 1) n ( n + 1) n !

nn
−n −n
 n +1
n
nn  n   1
= lim = lim   = lim   = lim 1 + 
n → ( n + 1)
 n +1  n   n
n n → n → n →

1 1 1
= lim n
= → Formula Sheet → 1
n →
 1
n
 1 e
1 +  lim 1 + 
 n n →
 n


n!
 n
converges
n =0 n


3( x + 1) n
9. (Multiple Choice) Select the correct interval of convergence for 
n=1 n 4n
.

a. −5  x  3
b. −5  x  3 (Correct)
c. −5  x  3
d. −5  x  3

Solution:

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APSC 173 Final Exam SL Review Packet 2023

3( x + 1) n 3( x + 1) n +1
an = , an +1 =
n 4n (n + 1)4n +1
Apply Ratio Test:
3( x + 1) n +1
(n + 1)4n +1 3( x + 1) n +1 n 4n 3( x + 1) n ( x + 1) n 4n
lim = lim = lim
n → 3( x + 1) n n → ( n + 1)4 n +1 3( x + 1) n n → (n + 1)4n 4 3( x + 1) n
n4n
n( x + 1) x +1
= lim =  1, x must be between -5 and 3. Plug in limits.
n → 4(n + 1) 4
For x = −5 :

3(−4) n 
3(−1) n 4n 
(−1) n

n =1 n 4
n
= 
n =1 n4n
= 3n =1 n
→ Converges via alternating series test

For x = 3 :
 
3(4) n 1

n =1 n 4
n
= 3n =1 n
→ Diverges


3( x + 1) n
 is convergent for − 5  x  3
n =1 n4n

( x − 3)n 
10. (Multiple Choice) Select the correct interval of convergence for  2 .
n=1 n
a. 2 x4
b. 2 x4
c. 2 x4
d. 2  x  4 (Correct)

Solution:

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APSC 173 Final Exam SL Review Packet 2023

( x − 3) n ( x − 3) n +1
an = , an +1 =
n2 (n + 1) 2
Apply Ratio Test:
( x − 3) n +1
an +1 (n + 1) 2 ( x − 3) n +1 n 2 ( x − 3) n ( x − 3) n 2
lim = lim = lim = lim
n → an n → ( x − 3) n n → ( n + 1) 2 ( x − 3) n n → (n + 1) 2 ( x − 3) n
n2
( x − 3)n 2
= lim = x − 3 , Converges when x − 3  1
n → (n + 1) 2
→ −1  x − 3  1
→2 x4
 Check x = 2 and x = 4
x = 2:

(−1) n

n =1 n
2
→ Alternating Series Test → Converges

x = 4:
 
1n 1

n =1 n
2
= 
n =1 n
2
→ Converges

( x − 3) n
 Converges for 2  x  4
n =1 n2

11.(Multiple Choice) Find the Taylor series for f(x) = ln(x) at a = 1.


(−1) n ( x − 1)( n+1)
a. 
n =0 (n + 1)!

(−1) n ( x − 1)( n+1)
b. 
n=1 (n + 1)!

(−1) n ( x − 1)( n+1)
c. 
n=1 (n + 1)

(−1) n ( x − 1)( n+1)
d. 
n =0 (n + 1)
(Correct)

Solution:

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APSC 173 Final Exam SL Review Packet 2023


f k (a)
f ( x) =  ( x − a ) k , f ( x) = ln( x), a = 1
k =0 k!
f (a ) = ln(a)
0

f 1 (a ) = a −1
f 2 (a) = (−1)a −2
f 3 (a) = (1)(2)a −3
f 4 (a) = (−1)(2)(3)a −4
(−1) k +1 (k − 1)!
For k = 0, f k (a) = 0. For k  1, f k (a) = → f k (1) = ( −1) k +1 ( k − 1)!
ak
k =1 k =2 k =3 k =k
k =0
 k
f (a) 0! 1! 2! (k − 1)!
f ( x) =  ( x − a ) k = 0 + ( x − 1)1 + (−1) ( x − 1) 2 + ( x − 1)3 + ... + (−1) k +1 ( x − 1) k
k =0 k! 1! 2! 3! k!
n=0 n =1 n=2

(−1) n ! n
(−1) n !
 n
(−1) ( x − 1) n +1
 n
f ( x) =  ( x − 1) = 
n +1
( x − 1) = 
n +1

n = 0 ( n + 1)! n = 0 ( n + 1) n ! n=0 ( n + 1)

1
12. Expand using partial fractions:
( x + 1)( x 2 − 4)

Solution:
1 1 A B C
= = + +
( x + 1)( x − 4) ( x + 1)( x − 2)( x + 2) ( x + 1) ( x − 2) ( x + 2)
2

1 = A( x − 2)( x + 2) + B ( x + 1)( x + 2) + C ( x + 1)( x − 2)

1
x = −1, A = −
3
1
x = 2, B =
12
1
x = −2, C =
4
1 1 1 1
 =− + +
( x + 1)( x − 4)
2
3( x + 1) 12( x − 2) 4( x + 2)

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APSC 173 Final Exam SL Review Packet 2023

13.Write the Taylor Series for f(x) = ex at a = 5 without having the constant ‘e’
in your solution.
Solution:
 
xn 5n
ex =  → e5 = 
n=0 n! n =0 n !

e5 ( x − 5) n
e x at a = 5 → 
n =0 n!

5k 
( x − 5) n
 e x at a = 5 →  
k =0 k ! n =0 n!
e5

14. Find the length of the line segment from (x,y) = (-1,-2) to (8,7) given
t3
x = t − 1, y = − 2
2

Solution:
t3
x = t 2 − 1, y = −2
3
dx dy
= 2t , = t
2

dt dt
When t = 0, x = −1, y = −2
When t = 3, x = 8, y = 7
 a = 0, b = 3
2 2
 dx   dy 
( 2t ) + ( t 2 ) dt = 
b b b
L= a
2
+ = 4t 2 + t 4 dt
2
    dt
a
 dt   dt  a

t 2 ( 4 + t 2 )dt =  t 4 + t 2 dt ,
b b
=
a a

du
let u = 4 + t 2 , du = 2tdt , = tdt
2
13
4+b2 13
1
2 3 
L= udu =  u du =  u 2   25.915
2
4+ a2
 3 4
4

15. Write the equation for a circle with radius 1 centered at the origin in polar
coordinates.
Solution: r = 1

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APSC 173 Final Exam SL Review Packet 2023

16. An antenna is mounted on top of a cell tower as shown in the figure below.
Its normalized radiation pattern can be described by r = cos( cos( )) . At
what distance, d, will a cell user not have coverage from the tower?

Solution:
Plotting the radiation pattern, we see:

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APSC 173 Final Exam SL Review Packet 2023

Notice when θ = -π/3 there is no antenna radiation. Alternatively, you could

find this by solving for θ in r = cos( cos( )) when r = 0. Knowing this angle, we
can solve for d via:
Letting  = − and recognizing the right-angle triangle formed by d , h and 
h h
tan( ) = d =
d tan( )
10
d = = 5.77 m

tan( )
3

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