AP Calc BC Chapter 6 Test

Definite integral

a number

evaluate using limits/bounds of integration

no need for +C

Indefinite integral

no numerical value

evaluate by finding the antiderivative

+C

Differential equation

dy/dx=

Slope fields show

the entire family for functions (accounting for +C)

Purpose of u-subs

to simplify the integration process by accounting for chain rule

2 things to look for when using u-subs

1. a function and its derivative

2. a composite function (function within a function)

Don't put u and x in the same

integral

When evaluating a definite integral after using a u-sub, remember to

calculate the new limits of integration

Separable differential equation standard form

dy/dx = f(x)g(y)

Steps to solve a separable differential equation

1. separate y's and x's on either side

2. integrate with a +C on the x side

3. find C using an initial condition

A=

±e^c

Initial value problem

the problem of finding a function y of x when we are given its derivative and its value at a particular point

Initial condition

the value of f for one value of x

When we have solved the differential equation, we have

found all the functions y that satisfy the differential equation

When we find the particular solution that fulfills the initial condition, we have

solved the initial value problem

Definition: slope field or direction field

a slope/direction field for the first order differential equation dy/dx = f(x,y) is a plot of short line segments with slopes f(x,y) for a lattice of points (x,y) in the plane

Slope fields enable us to

graph solution curves without solving the differential equation analytically

Slope fields are useful when

the formula given for dy/dx involves both x and y

Definition: indefinite integral

the set of all antiderivatives of a function f(x) is the indefinite integral of f with respect to c and is denoted by ∫f(x)dx

Once we have found one antiderivative F for a function f, all the other antiderivatives differ from F by

a constant

∫f(x)dx =

F(x) + C

C is

the constant of integration, an arbitrary constant

∫xⁿdx =

(xⁿ⁺¹)/(n+1) + C

∫dx/x =

ln|x| + C

∫(e^kx)dx =

(e^kx)/k + C

∫sinkxdx =

-coskx/k + C

∫coskxdx

sinkx/k + C

∫sec²xdx =

tanx + C

∫csc²xdx =

-cotx + C

∫secxtanxdx =

secx + C

∫cscxcotxdx =

-cscx + C

Constant multiple rule for integrals: ∫kf(x)dx =

k∫f(x)dx

Sum and difference rule for integrals: ∫[f(x)±g(x)]dx =

∫f(x)dx ± ∫g(x)dx

When we solve an initial value problem involving a first derivative, we have _____ arbitrary constant.

1

When we find a function from its 2nd derivative, we have to deal with _____ constants, one from each anti differentiation.

2

To find a function from its 3rd derivative, we'd need _____ constant values.

3

Each time we find an antiderivative, we need

an initial condition to tell us the value of C

Power rule for integration: if u is any differentiable function of x, then

∫uⁿdu = (uⁿ⁺¹)/(n+1) + C, n≠-1

Substitution method of integration: a change of variable can often turn an unfamiliar integral into

one that we can evaluate

∫cosudu =

sinu + C

∫sinudu =

-cosu + C

∫sec²udu =

tanu + C

∫csc²udu =

-cotu + C

∫secutanudu =

secu + C

∫cscucotudu =

-cscu + C

Steps to use substitution in indefinite integrals

1. substitute: u=g(x), du=g'(x)

2. evaluate by finding an antiderivative F(u) of f(u)

3. replace u by g(x)

Steps of u subs (in equation form)

∫f(g(x))*g'(x)dx = ∫f(u)du = F(u) + C = F(g(x)) + C

The substitution method works because

F(g(x)) is an antiderivative of f(g(x))*g'(x) whenever F is an antiderivative of f

To evaluate a definite integral by substitution, we can avoid

replacing u by g(x)

Substitution in definite integrals

same as for indefinite, but change the limits of integration from a to b to g(a) to g(b)

A differential equation y'=f(x,y) is separable if

f can be expressed as a product of a function of x and a function of y

When a separable differential equation is separated, the integrated equation provides the solution we seek by

expressing y as a function of x

dy/dx = g(x)*h(y) →

∫1/(h(y))dy = ∫g(x)dx

integration by parts formula

∫udv = uv - ∫vdu

when integrating by parts, u should be

a function that differentiates to 0

when integrating by parts, dv should be

easily integrable

tabular integration is used for

integrals of the form ∫f(x)g(x)dx, in which f can be differentiated repeatedly to become 0 and g can be integrated repeatedly without difficulty

tabular integration is

a way to organize the calculations of repetitions of integration by parts, saving much work

tabular integration setup

left column: f(x) and its derivatives (to 0)

right column: g(x) and its integrals

arrows starting with + and alternating signs

tabular integration: f(x) can be

any polynomial

law of exponential change: if y changes at a rate proportional to the amount present (dy/dt = ky) and y=y₀ when t=0, then y=

y₀e^kt, where k>0 represents growth and k<0 represents decay

k

the rate constant of the equation

radioactive decay

the process of radiation and change

radioactive element

an element whose atoms go spontaneously through the process of radioactive decay

if dy/dt = -ky, then y=

y₀e^(-kt)

half-life is

the time required for half of the radioactive nuclei present in a sample to decay

half-life=

(ln2)/k

compound interest: A(t) =

A₀(1+(r/k))^kt

k=# of times/yr interest is added

continuously compounded: A(t) =

A₀e^rt

population model: a differentiable function P growing at a rate proportional to the size of the population

dP/dt = kP

(dP/dt)/P=

k, a constant

P=

P₀e^kt

P₀

size of population at time t=0

logistic growth model

an exponential model for population growth that assumes unlimited growth

a logistic growth model is realistic only for a short period of time when P₀ is small because

relative growth rate is positive but decreases as population increases due to environment or economic factors

Carrying capacity, M

the maximum population that the environment is capable of sustaining in the long run

relative growth rate (RGR) is proportional to [1-(P/M)] with positive proportionality constant k: dP/dt/P=

k[1-(P/M)]

RGR: dP/dt =

(k/M)P(M-P)

the solution to the logistic differential equation (dP/dt = (k/M)P(M-P)) is called the

logistic growth model

rate of growth is proportional to both _____ and _____

P; (M-P)

if P>M, then the growth rate is negative and the population is

decreasing

solution to dP/dt = (k/M)P(M-P)

P=M/(1+Ae^-kt)

Euler's method: if we are given a differential equation dy/dx=f(x,y) and an initial condition y(x₀)=y₀, we can approximate the solution y=y(x) by its linearization

L(x) = y₀ + f(x₀,y₀)(x-x₀)

Euler's equation

yₙ₊₁=yₙ + f(xₙ,yₙ)∆x

-∫secθdθ=

-ln|secθ+tanθ| + c

-∫tanx=

ln|cosx| + c

for the formula dy/dt = ky, as y increases, the rate of growth

increases

k is specific to

temperature and humidity

for the formula dy/dt = ky, the change in population is dependent on

the population at that time

dy/dt = ky → y =

Ae^kt

logistic growth: dP/dt = kP →

P = Ae^kt

logistic growth equation 1

dP/dt = (kP/M)(M-P)

logistic growth equation 2

dP/dt = kP[1-(P/M)]

P

population

M

maximum carrying capacity

dy/dt = 30y(200-y)

lim(t→∞)y=

M, 200

logistic growth model: point of inflection occurs at

M/2

after M/2, the rate of growth

decreases

the rate is greatest at

M/2

if g(x)=∫(a to b)f(x)dx, then dg/dx=

db/dx*f(b) - da/dx*f(a)