Math 2260E Exam 2

calc 2

sin²(x) =

½ - ½ cos(2x)

cos²(x) =

½ + ½ cos(2x)

sin(ax)sin(bx) =

½ cos((a-b)x) - ½ cos((a+b)x)

sin(ax)cos(bx) =

½ sin((a-b)x) + ½ sin((a+b)x)

cos(ax)cos(bx) =

½ cos((a-b)x) + ½ cos((a+b)x)

integral of sec²(x)

tan(x) + C

integral of sec(x)tan(x)

sec(x) + C

integral of tan(x)

ln|sec(x)| + C

integral of sec(x)

ln|sec(x) + tan(x)| + C

tan²(x) =

sec²(x) - 1

when integrating odd powers of sec(x), must use ___

integration by parts, set u = sec(x)

trig subs: remember you can back sub with u-sub forumla !!

!!

partial fraction decomp: if power on top is greater or equal to power on bottom,

divide fractions (and use placeholders!!)

what must you do when you have a repeated linear factor in partial fraction decomp that is repeated ‘n’ times?

include a distinct fraction for every power of that fraction from 1 to ‘n’

for integrals involving sqrt(a² - x²)

let x = a*sin(theta) and sub; use reference triangle to put into terms of x after (or adjust bounds with substitution if definite)

for integrals involving sqrt(a² + x²)

let x = a*tan(theta) and sub; use reference triangle to put into terms of x after (or adjust bounds with substitution if definite)

for integrals involving sqrt(x² - a²)

let x = a*sec(theta) and sub; use reference triangle to put into terms of x after (or adjust bounds with substitution if definite)

improper integrals

rewrite as limit, if (both) exists, integral converges to limit

completing the square:

add and subtract (b/2)², can’t have coefficient in front of x²

limit of arctan(x) as x approaches infinity

pi/2

improper integrals can also be improper if

within their bounds is a point where the integral DNE

order of differential equation:

highest order of derivative that appears in equation (ex. y’’’)

equilibrium solution of direction field

when y = constant, or when slope (y’) = 0

euler’s method formula for x_n

x_n = x_o + nh, where n = number of steps and h = step size

euler’s method formula for y_n

y_{n =} y_n-1 + h*f(x_n-1, y_n-1), where f(x,y) is given by what y’ =

seperable differential equations form

y’ = f(x)*g(y)

seperable differential equations steps

1) check for constant solutions (when g(y) = 0)
2) isolate variables to each side
3) integrate
4) solve for y
5) if possible, use initial condition to solve for C

logistic equation (rate version)

dP/dt = r*P*(1 - (P/L)), where r = growth rate, P = population and L = the limiting/maximum value or carrying capacity

logistic equation (curve version)

P = L/(1+C*e^-r*t)

equation for change in population

P’ = birth rate - death rate + net immigration rate

squeeze theorem for sequences

a_n <= b_n <= c_n; if lim(n—>inf) a_n = L = lim(n—>inf) c_n, then b_n converges and lim(n—>inf) b_n = L

0/inf converges to ___

zero

extreme values of sin(x) and cos(x) [for squeeze theorem]

-1, 1

harmonic series

infinte sum of (1/n); diverges to infinity very slowly

geometric series

sum from one to infinty of ar^n-1, where a is the initial term and r is the common ratio

partial sum forumla for geometric sum

S_k = (a(1 - r^k)) / (1 - r) for r cannot = 1

infinite sum forumla for geometric sum

S_inf = a / (1 - r) for |r| < 1; S_inf diverges for |r| >= 1

telescoping series

most terms cancel except some at start and some at end; example: sum from one to infinity (b_n - b_n+1)

telescoping series formulas

S_k = b₁ - b_k-1; S_inf = b₁ - B where B is what b_k+1 converges to as k approaches infinity, if b_k+1 diverges, so does S_inf

divergence test

if a_n does not = 0 as n approaches inf, test applies and series is divergent; if test doesn’t apply, series can be convergent or divergent

integral test

if f is continuous, decreasing, and f(n) = a_n for n >= N, series converges if identical improper integral converges

p-series test

sum 1 / n^p converges if p > 1, diverges if p <= 1

ratio test

take limit as n approaches infinity of ratio of a_n+1 to a_n; if L > 1, diverges; if L < 1, converges; if L = 1, test is inconclusive

redo Quiz 5 problems ! and alternating series test problem !

!

limit comparison test

take limit of a_n/b_n as n approaches infinity; if lim = L does not = 0, both converges or diverge; if lim = 0 and b_n converges, a_n converges; if lim = inf and b_n diverges, a_n diverges

alternating series test

converges if a_n is decreasing and and lim a_n as n approaches infinity = 0

alternating series test absolute convergence

if |a_n| converges

alternating series test conditional convergence

if |a_n| diverges but a_n converges

root test of a_n

let p = lim as n approaches inf of (nth-root(|a_n|)); if 0 <= p <= 1, a_n converges absolutely; if p > 1 or = inf, a_n diverges; if p = 1, test is inconclusive

limit rule for any polynomial p(n)

lim as n approaches infinity of (p(n))^1/n = 1

definition of constant e

lim as n approaches infinity of (1 + 1/n)ⁿ

power series: radius of convergence

distance from domain endpoints to center