Mu Math Calc 2 Final Exam

Front 1

Kth term test

Back 1

Lim as k -> Infinity of a-subk != 0 then summation of a-subk diverges

Front 2

Integral Test

Back 2

integral from 1 to infinity of f(x)dx and sum from k=1 to infinity of a-subk either both converge or both diverge

Front 3

Basic comparison test (BCT)
(0<=a-subk<=b-subk for all values of k)

Back 3

if sum of b-subk converges then sum of a-subk converges

if sum of a-subk diverges then sum of b-subk diverges

Front 4

Limit Comparison Test (LCT)
a-subk and b-subk are always positive

Back 4

limit as k -> infinity of (a-subk)/(b-subk) = L
where L>0
the two series either both converge or both diverge

Front 5

Alternating Series Test (AST)

Back 5

a-subk > 0 for all values of k and a-subk is decreasing sequence and limit as k -> infinity of a-subk = 0 then sum of (-1)^(5+1) x a-subk converges

Front 6

AbSolute Convergence Test (ACT)

Back 6

sum of abs(a-subk) converges then the sum of a-subk converges absolutely

Front 7

Ratio Test

Back 7

Sum of a-subk has no zero terms
limit as k -> infinity of abs((a-subk+1)/(a-subk)) = L
L<1 converge absolutely
L>1 series diverges
L=1 no conclusion try another method

Front 8

Root Test

Back 8

Sum of a-subk is a series so that limit as k-> infinity of kth root(abs(a-subk)) = L
L<1 converge absolutely
L>1 series diverges
L=1 no conclusion try another method

Front 9

half angle formula

Back 9

(sinx)^2 = 1/2(1-cos2x)
(cosx)^2 = 1/2(1+cos2x)
(sin2x)^2 = 1/2(1-cos4x)
(cos2x)^2 = 1/2(1+cos4x)
(sin3x)^2 = 1/2(1-cos6x)
(cos3x)^2 = 1/2(1+cos6x)

Front 10

integral of sinx

Back 10

-cosx + C

Front 11

integral of cosx

Back 11

sinx + c

Front 12

integral of secx

Back 12

ln|secx + tanx| + c

Front 13

integral of tanx

Back 13

ln|secx| + c

Front 14

integral of secxtanx

Back 14

secx + c

Front 15

integral of (secx)^2

Back 15

tanx + c

Front 16

sin(theta)

Back 16

opp/hyp

Front 17

cos(theta)

Back 17

adj/hyp

Front 18

tan(theta)

Back 18

opp/adj

Front 19

Arc Length

Back 19

integral a to b of root(1+(f-prime(x))^2) is the length of x on the interval [a,b]

Front 20

Arc length steps

Back 20

1 find derivative of the function
2 apply to arc length formula
3 substitute trig values for x values
4 solve for the integral
5 substitute x values back in and solve

Front 21

integral cscx

Back 21

ln|cscx - cotx| + c

Front 22

integral cotx

Back 22

-ln|cscsx| + c

Front 23

integral cscxcotx

Back 23

-cscsx + c

Front 24

integral (cscx)^2

Back 24

-cotx + c

Front 25

integral of 2^x

Back 25

(2^x)/ln2 + c

Front 26

when there's addition in the numerator split it up. integral of (2x+5)/x^2+1

Back 26

becomes integral of (2x)/x^2 + 1 dx + integral of 5/x^2 + 1 dx

Front 27

integral of 1/sqrt(a^2 - x^2)

Back 27

(sinx)^-1

Front 28

integral of 1/(x^2 + a^2)

Back 28

(tanx)^-1

Front 29

integral of 1/sqrt(x^2 - a^2)

Back 29

(1/a)(sec(x/a))^-1

Front 30

integration by parts

Back 30

u v - integral of v du

Front 31

trig sub

Back 31

sqrt(a^2 - x^2) means x = asin(theta)
sqrt(x^2 + a^2) means x = atan(theta)
sqrt(x^2 - a^2) means x = asec(theta)

Front 32

area under the curve

Back 32

A = integral from a to b of f(x) - g(x) dx

Front 33

Geometric Series Test

Back 33

sum of (r)^k as k -> infinity = a/(1-r)
if r <1 converge else diverge

Front 34

harmonic series

Back 34

always diverges
1/k

Front 35

p series

Back 35

1/k^p
p<= 1 diverge
p > 1 converge

Front 36

MaClaurin Series

Back 36

lim k -> infinity of |(a-sub(k-1))/a-subk|
replace values of k with k + 1

lim n -> infinity of x^n/n! can be used to set up a MaClaurin Series

Front 37

integral of g-prime(x)/g(x)

Back 37

ln(g(x))