Calculus II Final Exam
derivative of sinu
(cosu)u'
derivative of cosu
-(sinu)u'
derivative of cotu
-csc^2u du/dx
derivative of cscu
-cscucotu du/dx
Root test
-everything raised to the kth power -if p is between 0 and 1, the limit converges if p is greater than 1 the limit diverges if p is one, the test is inconclusive -determine ak first
Trig Integrals
-find a way to simplify what's given -reduction formulas given on exam
integral of tanx
-ln|(cosx)| or ln|(secx)|
integral of cscx
-ln|cscx+cotx|+c
Ratio Test
-take ratio of ak+1 over ak as limit approaches infinity -if r is between 0 and 1, the limit converges if r is greater than 1 the limit diverges if r is one, the test is inconclusive -use with factorials, k raised to a power
derivative of inverse tanu
1/(1+u^2)du
derivative of inverse sec
1/[|u|sqrt(u^2-1)du
derivative of inverse sinu
1/sqrt(1-u^2)du
Derivative of loga(u)
1/ulna * u'
Telescoping Series
A series whose partial sums eventually only have a fixed number of terms after cancellation -find what cancels and find a pattern for what's left over, take limit of that
Comparison test
Find Bn>An if it converges then An converges. Find Bn<An if it diverges An diverges.
Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) for all x ̸= a and limx→a f(x) = limx→a h(x) = L, then limx→a g(x) = L
Integration by parts
Integral of udv = uv - integral of vdu L-log I-inverse trig A-algebra (x) T-trig E-exponential u—>dv
On what conditions on R does R^n converge for a geometric sequence
R=|R|<1 R=1 converges |R|>1 diverges
Explicit Formula for Sequence
Relates any term in a sequence to only the first term and the common ratio. a(n)=f(n)
sum of an infinite geometric series
S=a/1-r a=first term r=common ratio
Recurrence Relation
a(n+1)=f(an)
Derivative of a^u
a^u(lna)(u')
integral of a^u
a^u/lna + C
quotient rule
d/dx (g(x)/ h(x)) = [(h(x) g'(x) - g(x) h'(x))]/ h(x)^2
chain rule
d/dx f(g(x)) = f'(g(x)) g'(x)
Derivative of e^u
e^u du
Infinite series
if goes to a #-converges if goes to infinity-diverges
Integral test
if integral converges, series converges -same convergence behavior
limit comparison test
if lim as n approaches ∞ of ratio of comparison series/general term is positive and finite, then series behaves like comparison series if the limit is 0 and the comparison series converges, then the original series converges if the limit is infinity and the comparison series diverges, then the original series diverges
∫sin^m(x)cos^n(x)
if m is odd and positive, split off sinx and rewrite the resulting power of sinx in terms of cosx(1-cos^2x=sinx), then use u=cosx if n is odd and positive, split off cosx and rewrite the resulting even power of cosx in terms of sinx then use u=sinx if m and n are both even, use half-angle formulas to transform the integrand into a polynomial in cos2x and apply preceding strategies once again to powers of cos2x greater than one
Partial Fractions
integral(P(x)/Q(x))dx P,Q are polynomials P/Q is a rational function (degree of P less than Q) case 1-(x) is a product of n linear factors (no irreducible quadratics) integral(3x+5)/(x^2-4x-5) =A/(x-5)+B/(x+1) =find A and B case 2-Q(x) has repeated linear factors integral(3/x(x+1)^2)=A/x+B/(x+1)+C/(x+1)^2 solve for coefficients, plug in and integrate case 3-irreducible quadratic factor integral (x-10)/x(x^2+25) x-10=A/x+(Bx+C)/(x^2+25) find coefficients, plug in and integrate
Alternating Series Test
lim as n approaches zero of general term = 0 and terms decrease, series converges
Divergence test
lim_(k->infinity) [a_k] doesn't equal zero, then the series Σa_k must diverge -however if the limit does equal 0, the series could converge or diverge
Improper integrals
limit as b --> infinity; don't plug infinity into the integral if limit goes to infinity-function diverges if limit goes to a number-function converges integral 1/x^p p>1=limit converges 1/1-p p≤1=limit goes to infinity
integral of du/u
ln |u| + C
integral of secx
ln|secx+tanx|+C
integral of cotx
ln|sinx|+c
inverse tangent of infinity
pi/2
derivative of tanu
sec^2u du
derivative of secu
secutanu du/dx
integrals involving a^2-x^2
substitute x=asinθ a^2-x^2=a^2-a^2sin^2θ =a^2(1-sin^2θ) =a^2cos^2θ
derivative of lnu
u'/u
product rule
uv' + vu'
integrals involving x^2-a^2
x=asecθ x^2-a^2=a^2secθ-a^2 =a^2(secθ-1) =a^2tan^2θ
integrals involving x^2+a^2
x=atanθ x^2+a^2=a^2tanθ+a^2 =a^2(tan^2θ+1) =a^2sec^2θ
exponential growth/decay over time
y(t)= y°ekt y'(t)=(y°e^kt)k y'(t)/y(t)=k k>0-growth k<0-decay y°=y(0)
remained in alternating series
|Rn|≤an+1
