Midterm1 Flashcards
Integration by parts formula
uv- integral v du
LIATE
Logarithmic,Inverse Trig, Algebraic, Trig, Exponential
if integral ( p(x)sin(ax))
let u = p(x) du=the rest
sqrt( a^2 - x^2)
sub with x=asin(theta), identity: cos^2(theta)= 1- sin^2(theta)
sqrt(a^2 +x^2)
sub with x=atan(theta) sec^2(theta) = 1+tan^2(theta)
sqrt(x^2 - a^2)
sub with x=asec(theta) tan^2(theta)=sec^2(theta)-1
Integrating rational functions
if degree P(x)>=Q(x) use long division
otherwise use partial fractions
when doing partial fractions, denominator has (x-2)^2
do a variable over each power up to 2
when doing partial fractions, denominator when factored has irreducible quadrativ
make it BX+C/irreducible.
ODE
ordinary diff eq- equation containing a fcn of one variable and one or more derivatives of that fcn
order of an ODE
order of the highest derivative
Eullers Method equation
yn = yn-1+ h * F(xn-1,Yn-1)
I(x) in diffeq
e^(integral(P(x))dx)
standard form of linear 1st order ODE
dy/dx + yP(x) = Q(x)
solve a linear 1st order ODE
y* I(x) = integral(Q(x) I(x) dx)
d/dx sinx
cosx
d/dx cosx
-sinx
d/dx tanx
sec^2x
d/dx cotx
-csc^2x
d/dx secx
secxtanx
d/dx cscx
-cscxcotx
d/dx sin-1(x)
1/sqrt(1-x^2)
d/dx tan-1(x)
1/(1+x^2)
d/dx cost-1(x)
-1/sqrt(1-x^2)
d/dx cot-1(x)
-1/1+x^2
integral 1/ax+b
(1/a)ln(ax+b) +C
integral tanx
-lnI(cosx)+c
integral secx
ln(secx+tanx)+c
range of arcsin
pi/2 to -pi/2
range of arccos
0 to pi
cycloid
x=r(theta - sin(theta))
y=r(1 - cos(theta))
equation for dy/dx parametrics
=(dy/dt) / (dx/dt) if dx/dt is not 0
area under parametric curve
A = integral a-b ( y(t) * x'(t) dt) wher where f(t) =x
if dx/dt =0and dy/dt=0 what is dy/dx
undefined
if dx/dt =0 and dy/dt != 0 what is dy/dx
vertical ( infinity)
Length of parametric curve a->b
integral a->b [(dx/dt)^2 + (dy/dt)^2] dt
converting x and y to polar
x= rcos(theta)
y =rsin(theta)
r^2 =
x^2 + y^2
tan(theta)
y/x
dy/dx of a polar curve
dy/dx = (dy/dtheta)/(dx/dtheta) = [(dr/dtheta)(sintheta) + rcos(theta)] / [ (dr/dtheta)cost(theta) - rsin(theta)]
Area of polar curve
A= integral a->b [ (1/2)r^2] dtheta
Arc length of polar curve
integral a->b [ sqrt( r^2 + (dr/dtheta)^2 ] dtheta
Improper integrals type 1
infinite integrals
Improper integrals type 2
discts integrals
integral a->infinity [ f(x)] dx
lim t->infinity[ integral a->t ( f(x) ) dx]
integral (- infinity -> b) [f(x)] dx
lim t-> - infinity ( integral t->b ( f(x) dx) )
integral -infinity -> infinity
integral -infinity -> a + integral a->infinity
p series, where convergent
integral 1->infinity 1/x^p
convergent for p> 1./
divergent p<=1
integral a->b DISCTS AT b
integral a->b = lim t->b- [ integral a->t ]
integral a->b DISCTS AT a
= limit t-> a+ [integral t->b]
linear electricity problem
L(dI/dt) + RI = E(t)
L is inductance
R is resistance
E(t) is voltage
if c is in [a,b], f(x) discts at c then integral a->b
= integral a->c + integral b-> c
comparison thm for improper integrals
supp f,g are cts with
f(x)>=g(x) >=0 for all x>=a
if integral a->infinity f(x) is convergent then so is a->infinity of g(x)
if integral a->infinity g(x) divergent so is a->infinity f(x)
{An} has limit L
lim n->infinity an =L or an->L as n->infinity
{An} converges to L
if {An} limit DNE or is infinite
diverges
Squeeze thm for sequences
if {an}{bn}{cn} aninfinity an = lim n->infinity cn =L then lim n->infinity bn=L
if lim abs[an] as n->infinity =0 theb lim n-> an=0
if {an} converges to L and fis a fcn that is cts at L then
lim n->infinity f(an) =f(L)
integral cotx
ln(sinx)
Second derivative for parametrics
dy/dx = (d/dx)(dy/dx) = [ d/dt(dy/dx) ] / [dx/dt]
concave up
second deriv >=0
concave down
second deriv <= 0
formula for a circle (h,k) and radius r
(x-h)^2 +(y-k)^2=r^2
arctan(infinity)
pi/2
Absolutely convergent
if the series of absolute values of an is convergent
Conditionally convergent
if its convergent but not absolutely
if a series is absolutely convergent
then it is convergent
ratio test
if lim abs[(an+1)/(an)]= L1 or infinite, then divergent
if =1 inconclusive
root test
if limn->infinity (root(abs(an))) = L 1 or infinity then divergent
if =1 then inconclusive
given a power series sigma(cn(x-a)^n) there are only three possibilities:
the series converges only when x=a
the series converges for all x
there isa positive number R such that the series converges if abs(x-a) < R and diverges if abs(x-a)>R
(1/(1-x)) = 1 +x +x^2 +x^3 + … =
sigma(x^n) for abs(x)<1
if f hasa power series representation at a then its represented by
sigma[ (f(^n)(a)/n!)(x-a)^n) ] for abs(x-a)<R
Maclaurian series
when taylor series is centered at 0. equation is:
sigma [ f(^n)(0)/n! * x^n ]
power series of sinx
sigma [ (-1)^n (x^2n+1)/(2n+1)! for all x
power series of cosx
sigma [ (-1)^n (x^2n)/(2n!) ]
geometric series
sigma [ar^n-1] = a/(1-r)
if a series is convergent then
lim n->infinity = 0
if limit n-> infinity an dne or is non zero
then series is divergent
integral test
if integral 1->infinity f(x) is convergent the series is convergent.
if integral divergent, series is divergent
limit comparison test
limn->infinity (an/bn) =c if c>0 then either both converge or both diverge
alternating series test
if decreasing and limn->infinity =0 then series converges`
lim n->infinity (1+(1/n))^n
e
taylor series 1/(1-x)
sigma x^n
taylor series e^x
sigma (x^n)/(n!)
taylor series sinx
sigma (-1)^n * (x^2n+1)/(2n+1)!
taylor series cosx
sigma (-1)^n * (x^2n)/(2n)!
estimating series using integral
Rn needs to be less than some decimal number. so Rn by thm >= integral n->infinity an. the answer to that < the error. solve for n
