5 Steps to a 5 AP Calculus BC 2019

438 Formulas and Theorems

(b) Givenr=f(θ)andα≤θ≤β, the area

of the region between the curve, the origin,

θ=αandθ=β:

A=

∫β

α

1

2

r^2 dθ or A=

1

2

∫β

α

[f(θ)]^2 dθ.

(c) Area between two Polar Curves:

Givenr 1 = f(θ)andr 2 =g(θ),

0 ≤r 1 ≤r 2 andα≤θ≤β, the area

betweenr 1 andr 2 :

A=

∫β

α

1

2

(

r 2

) 2

dθ−

∫β

α

1

2

(

r 1

) 2

dθ

=

∫β

α

1

2

((

r 2

) 2

−

(

r 1

) 2 )

dθ.

31. Series and Convergence:
    (a) Geometric Series:
    ∑∞

k= 0

ark=a+ar+ar^2 +ar^3 +···

+ark−^1 ··· (a=/0)

if|r|≥1, series diverges;

if|r|<1, series converges and the

sum=

a

1 −r

.

(Partial sum of the firstnterms:

Sn=

a−arn

1 −r

for all geometric series.)

(b) p- Series:

∑∞

k= 1

1

kp

= 1 +

1

2 p

+

1

3 p

+

1

4 p

···

+

1

kp

+···

ifp>1, series converges;

if 0<p≤1, series diverges.

(c) Alternating Series:

∑∞

k= 1

(− 1 )k+^1 ak=a 1 −

a 2 +a 3 −a 4 +···+(− 1 )k+^1 ak+···or

∑∞

k= 1

(− 1 )kak=−a 1 +a 2 −a 3 +a 4 −

···+(− 1 )kak+···, whereak>0 for all

ks.

Series converges if

(1) a 1 ≥a 2 ≥a 3 ··· ≥ak≥ ···and

(2) klim→∞ak=0.

(Note: Both conditions must be satisfied

before the series converges.)

Error Approximation:

IfS=sum of an alternating series,andSn=

partial sum of n terms,∣ then

∣error∣∣=|S−Sn|≤an+ 1.

(d) Harmonic Series:

∑∞

k= 1

1

k

= 1 +

1

2

+

1

3

+

1

4

+···diverges.

Alternating Harmonic Series:

∑∞

k= 1

(− 1 )k+^11

k

= 1 −

1

2

+

1

3

−

1

4

+···+

(− 1 )

k+ 11

k

+···converges.

(

∑∞

k= 1

(− 1 )k^1

k

=− 1 +

1

2

−

1

3

+

1

4

−

···+(− 1 )k

1

k

+···also converges.)

32. Convergence Tests for Series:
    (a) Divergence Test:
    Given a series

∑∞

k= 1

ak, if limk→∞ak=/0, then the

series diverges.

(b) Ratio Test for Absolute Convergence:

Given

∑∞

k= 1

akwhereak/=0 for allks and let

p=klim→∞

|ak+ 1 |

|ak|

, then the series

∑∞

k= 1

ak

(1)converges absolutely ifp<1;

(2)diverges ifp>1;

(3)needs more testing ifp=1.

(c) Comparison Test:

Given

∑∞

k= 1

akand

∑∞

k= 1

bkwith

ak>0,bk>0 for allks, and

a 1 ≤b 1 ,a 2 ≤b 2 ,...ak≤bkfor allks:

(1)If

∑∞

k= 1

bkconverges, then

∑∞

k= 1

ak

converges.

(Note that if the bigger series converges,

then the smaller series converges.)