Sem 2 Stewart Flashcards
method for 1st order differential equation
separating the variables
integrating factor if variables cannot be seperated
method for second order homogeneous differential equations
auxiliary equation and then general solution depending on roots of auxiliary equation
method for second order non-homogeneous differential equations
complementary function and particular integral in the same form as f(x) or multiply by x or x^2 if in same form as CF
form of general solution for real unequal roots
y=Ae^ax+Be^bx
form of general solution for real equal roots
y=(Ax+B)e^ax
form of general solution for complex roots
m=a+/-bi
y=e^ax(Acosbx+Bsinax)
steps for Riemann sums
- divide domain into n strips of width Δx=b-a/n
- find grid points
- evaluate riemann sum
(could be left, right, upper or lower)
midpoint rule
Δx[f(x1bar)+…+f(xnbar)]
where Δx=(b-a)/n and x bar is midpoint of interval
trapezium rule
the average of the L and R Riemann sums
1/2(Ln+Rn)
=Δx/2[f(x0)+2f(x1)+…+2f(xn-1)+f(xn)]
simpson’s rule
use parabolae instead of straight lines
n must be even
consider consecutive pairs of intervals and approximate the integral over each pair
Δx/3(y0+4y1+2y2+4y3+2y4+…+2yn-2+4yn-1+yn)
general steps for approximate integration
- find Δx
- find x0,x1,…,xn
- (midpoint) x1bar, x2bar…
- f(x1), f(x2),… etc
- apply formula
function is continuous on an interval if…
continuous at every number in interval
function is differentiable at a if..
f’(a) exists
differentiable on an open interval if differentiable at every point
F is an anti derivative of f if
F’(x)=f(x) for all x
what form are integrals in for integration by substitution?
∫f(g(x))g’(x)
u=g(x) is a differentiable function, continuous on I
∫f(g(x))g’(x) dx = ∫f(u) du
limits in integration by substitution
change limits using expression for u and sub in current limits
limits are g(a) and g(b)
trigonometric integrals
manipulate integrand using trig identities
can use substitution in reverse by setting x=g(f) and obtain inverse substitution
integral on closed interval [-a,a], if f is even (i.e. f(-x)=f(x))
∫a/-a f(x)dx = 2∫a/0 f(x) dx
integral on closed interval [-a,a], if f is odd (i.e. f(-x)=-f(x))
∫a/-a f(x)dx = 0
integration by parts formula
∫uv’=uv-∫vu’
what is needed to split into partial fractions?
proper function
degree numerator < degree denominator
if not, denominator into numerator
4 different forms of partial fraction
distinct linear factors
repeated linear factors
irreducible quadratic factors
repeated irreducible quadratic factors
distinct linear factors
A/ax+b + B/cx+d
repeated linear factors
A/ax+b + B/(ax+b)^2
irreducible quadratic factors
Ax+b/ax^2+bx+c
repeated irreducible quadratic factors
Ax+b/ax^2+bx+c + Cx+d/(ax^2+bx+c)^2
area between two curves y=f(x), y=g(x) and lines x=a, x=b
f(x)>/= g(x) for all x in [a,b]
A=∫a/b [f(x)-g(x)]dx
area between two curves y=f(x), y=g(x) and lines x=a, x=b
f(x) < g(x)
A=∫a/b |f(x)-g(x)| dx
volume: s is solid, lies between x=a and x=b.
if cross-sectional area of S in plane Px, through x and perpendicular to x=axis then V=
lim as n approaches infinity
∫ b/a A(x) dx
solids of revolution
obtained by revolving a region about a lines
v=∫ b/a A(x) dx
where A is area of cross-section
volume of revolution if cross -section is a disk
v=∫ b/a A(x) dx
A=pi(radius)^2
volume of revolution if cross-section is a washer
v=∫ b/a A(x) dx
A=pi(router)^2-pi(rinner)^2
volumes by cylindrical shells of revolution
rotating about y-axis, region under curve y=f(x), from a to b
v=∫b/a 2pix f(x) dx
=circumference.height.thickness
volumes by cylindrical shells of revolution
rotating about x-axis, region under curve x=f(y), from c to d
v=∫b/a 2piy f(y) dy
if f’ continuous on [a,b], length of curve is
L=v=∫b/a root(1+f’(x)^2 dx
(or root 1+ (dy/dx)^2 dx)
differential of arc length
ds=root 1+(dy/dx)^2 dx
arc length function
smooth curve equation y=f(x), s(x) is distance along curve from initial point P to Q
s(x) = ∫x/a root 1+ (f’t)^2 dt
surface area of revolution
f +ve, continuous derivative, surface area by rotating y=f(x) about x-axis
s= ∫b/a 2piy ds
where ds= root 1+ (dy/dx)^2 dx
what are parametric equations used for?
to describe curves not in the form y=f(x) i.e. fails vertical line test
parametric equations
dy/dx=
dy/dt / dx/dt
parametric area under y=F(x) from a to b, x=f(t), y=g(t)
t between alpha and beta
A=∫b/a F(x) = ∫beta/alpha g(t)f’(t) dt
parametric arc length
L=∫beta/alpha root (dx/dt)^2+ (dy/dt)^2 dt
parametric surface area of revolution
s=∫beta/alpha 2piy(t) root (dx/dt)^2+ (dy/dt)^2 dt
limits of vector functions
take each component separately
make sure to sub point into original equation to get in terms of t
d/dt [u(t)xv(t)] =
u’(t)xv(t) + u(t)xv’(t)
unit tangent vector to space curve r(t) is
T(t) = r’(t)/|r’(t)|
unit normal vector to space curve r(t) is
N(t)=T’(t)/|T’(t)|
improper integral
interval is infinite or f has an infinite discontinuty.
Type 1 improper integral
if ∫t/a f(x) dx exists for every t>/= a then
∫ infinity/a f(x) dx = lim as t appraoches infinity ∫t/a f(x) dx
same for negative infinity
convergent
limit exists
divergent
limit does not exist
type 2 improper integrals
if f continuous on [a,b) and discontinuous at b
∫b/a f(x)dx=lim t –> b- ∫t/a f(x) dx
if f continuous on (a,b] and discontinuous at a
∫b/a f(x)dx=lim t –> a+ ∫b/t f(x) dx
comparison theorem
if f(x) and g(x) continuous with f(x) >/= g(x)>/=0 for x>/=a
if ∫ infinity/a f(x) is convergent, ∫ infinity/a g(x) convergent
if ∫ infinity/a g(x) is divergent, ∫ infinity/a f(x) divergent
