Unit 1 – Functions
Def – for every x there is only 1 y
Domain, Range, and Inverses
One-to-one: for every y, only 1 x
Properties of Inverses:
To find domain:
D f = R f −1
• If fraction(s), set
bottom(s) ≠ 0 and solve
R f = D f −1
• If radical, set inside ≥ 0,
and solve
• If radical is in bottom of So, to find range of f(x):
• find inverse of f
fraction, set inside > 0,
• find it’s domain
and 0
• change x’s to y’s (this is
To find inverse: (fcn must be
range of f)
1-1)
Use your grapher to tell whether
• Switch x’s and y’s
a function has an inverse (it has
• Solve for y, this is
to pass vertical and horizontal
inverse.
line tests)
•
Even fcn: f(–x) = f(x)
• Symmetric over y-axis
• Polynomial function is even
if all exponents are even
(constant ok)
• Even trig: cos, sec
Odd fcn: f(–x)= –f(x)
• Symmetric over origin
• Polynomial function is odd
if all exponents are odd (no
constant)
• Odd trig: sin, tan, csc, cot
Domains and Ranges for Trig:
1. y = sin x and y = cos x
D : all reals
R : [–1,1]
2. y = tan x
D: x ≠ odd multiples of π2
R: all reals
3. y = cot x
D: x ≠ multiples of π
R: all reals
4. y = sec x
D: x≠ odd multiples of π2
R: y ≤ –1 or y ≥ 1
5. y = csc x
D: x ≠ multiples of π
R: y ≤ –1 or y ≥ 1
Domains and Ranges for
Inverse Trig Functions:
−1
1. y = sin x
D: –1 ≤ x ≤ 1
R: − π2 ≤ y ≤ π2
2. y = cos −1 x
D: –1 ≤ x ≤ 1
R: 0 ≤ y ≤ π
3. y = tan −1 x
D: all reals
R: − π2 < y < π2
3. ( x a ) b = x ab
3. log x n = n log x
n
x=x
5.
b
xa = x
n
a
b
Reflection Formulas
Over x-axis:
y = f(x) becomes y = –f(x)
(change the sign of function)
(change the sign of all x’s)
log a x = n ⇔ a n = x
Rules of Logs:
1. log xy = log x + log y
2. log xy = log x − log y
4.
y = ( f o g )( x) = f ( g ( x))
• wherever there is an x in
f, put in g
o uses parentheses
• Rg = Df
Over y-axis
y = f(x) becomes y = f(–x)
Exponents and Log Functions:
Rules of Exponents:
1. x a ⋅ x b = x a +b
a
2. xxb = x a −b
1
Composite functions
4. log n x = 1n log x
5. log a b = lnln ba
Domain and Range:
y = log x (regardless of base)
D: x > 0
R: all reals
y = a x (regardless of base)
D: all reals
R: y > 0
Graphs of Functions:
Basic Trig Graphs (sin, cos, tan)
y = tan x
y = cos x
Exponential Graphs: y = a x
0<a<1
Log Graph
a>1
Absolute Value Function
y =| x |
y =| x − a |
a
y = sin x
UNIT 2 – LIMITS
If f (x) is a polynomial function, then
lim f ( x) = f (c)
x →c
If f (x) and g (x) are polynomial functions and c
is a real number, then
lim
x →c
f ( x ) f (c )
=
g ( x ) g (c )
provided g (c) ≠ 0
Provided that x → a:
1. lim (f + g) = lim f + lim g
2. lim (f – g) = lim f – lim g
3. lim fg = lim f · lim g
4. lim f = lim f provided lim g≠ 0
g lim g
5. lim k ·f = k lim f (k is a constant)
6. lim f n = (lim f )n
A left-handed limit approaches c from the left
side on the number line:
i.e. if c is 6, then the left-hand limit would
be values like 5.9, 5.99, 5.999, etc…
A right-handed limit approaches c from the
right side on the number line:
i.e. if c is 6, then the right-hand limit would
be values like 6.1, 6.01, 6.001, etc…
A limit DOES NOT EXIST only if the lefthand limit ≠ right-hand limit
Limits to Infinity
If f and g are polynomial functions, then
A function y = f(x) has a horizontal asymptote
if either:
⎧0, deg f < deg g
f ( x) ⎪
= ⎨∞, deg f > deg g
lim
x →∞ g ( x )
⎪a ,
⎩ b deg f = deg g
a and b are leading coeff.
if
lim f ≠ lim+ f
x →a −
, then
x →a
lim f = DNE
x →a
lim f ( x) = b
or
x →∞
and b is not ∞ (it has to be a number)
A function y = f(x) has a vertical asymptote at
x=a if either:
lim− f ( x) = ±∞ or lim+ f ( x) = ±∞
x→a
To find a vertical asymptote:
• Set the bottom of the fraction = 0 and
solve for x
• Test the left-hand and right-hand limits at
each solution
• If any turn out to be ±∞, then it is an
asymptote.
• The left/right do not have to be equal
Continuity:
• To prove a function is continuous at x = c:
1. Show f(c) exists
2.
Rateavg =
f ( x2 ) − f ( x1 ) y 2 − y1
=
x2 − x1
x2 − x1
Finding Speed: If f(t) is the position function at
time t, then
•
The speed of the object at time t
(instantaneous speed) is:
•
f (t + h) − f (t )
speed = lim
h →0
h
If f(t) is the position function of a freefalling object, then if the object if
dropped from a height h, then the
function representing the position of the
object would be:
p(t) = h – f(t)
• To find the instantaneous speed, use the
following limit:
speed = lim
h →0
p(t + h) − p (t )
h
x →c
a.
•
•
•
•
Determine f(a)
Plug into the above equation
• Expand the expression on top
• If fractions multiple top and bottom by
common denominator
• Everything in the top without h will
cancel out due to +/–
Factor out the h in the top and cancel the h
with the bottom
Plug the 0 into all the remaining h to get
the rate.
Slope of the Tangent to a Curve (or slope of the
curve) at x=a:
•
Use the formula:
slope = m = lim
h →0
f ( a + h) − f ( a )
h
• Note: f (a) is the y-coordinate at x=a.
• Follow same directions as
Instantaneous Rate of Change
• Use Point-Slope Equation to find
equation of the tangent (use point
(a, f(a)) and m)
If you get
0
0
or
∞
∞
, then use
L’Hopital’s Rule
b.
If you get
0
k
, where k is any
number, then answer is 0
c.
If you get
k
0
, then use left and right
limits to get the same value. Answer
will be ∞, –∞, or DNE
d.
• Instantaneous Rate of Change of y=f(x)
at x=a:
f ( a + h) − f ( a )
Rate = lim
h →0
h
lim f ( x) = f (c)
x→a
Basic Steps to Finding Limits:
1. Plug the number in first.
a. If it is a number, then you’re done
2. If the function is a fraction:
• If you are viewing a graph of a function, if
the curve is discontinuous at points that are:
•
Holes
•
Asymptotes
•
Gaps
•
Jumps
To find a horizontal asymptote:
• Take the limit as x → ∞ and as
x → –∞
o If it is NOT ∞ or –∞, then the
asymptote is y = k where k is the limit
o If it is ±∞, then no asymptote.
Rates of Change:
• Average Rate of Change of y=f(x):
Show
lim f ( x) = b
x → −∞
lim
x →0
If you get
sin x
=1
x
k
∞
lim
x →∞
, then it is 0
sin x
=0
x
a sin bx ab
=
x →0
cx
c
lim
To find an end behavior of a function
f ( x) , simply write a fraction of the
h( x ) =
g ( x)
leading coefficients
•
The slope of the Normal is the negative
reciprocal of the slope of the tangent
• To find the equation, use the same
point as the tangent and the neg.
reciprocal slope
UNIT III – Intro to the DERIVATIVE
Def: The derivative is the slope of the curve or
the slope of the tangent to the curve.
f ' ( x) = lim
h →0
ƒ
f ( x + h) − f ( x)
h
Notation
y'
dy
dx
If f’(x) exists, we say that f has a
derivative (is differentiable) at x.
df
dx
The derivative of the function f(x) at the point
x = a is the limit
f ( x) − f (a)
f ' (a ) = lim
x →a
x−a
provided the limit exists.
d
f (x)
dx
d2y
y
“y prime”
“dy dx”
“the derivative of y with
respect to x”
“df dx”
“the derivative of f with
respect to x”
“d dx of f at x”
“the derivative of f at x”
The second derivative
C
B
Slope = 2
d y
D
Slope = –1
y = f (x)
A
dx 2
n
Slope = 0
E
Slope = 4
F
Slope = –1 Slope = 0
th
The “n ” derivative
dx n
x
y = f ‘(x)
** Know all the different notations **
Suppose you have a function like this:
⎧x2 , x ≤ 0
f ( x) = ⎨
⎩ 2 x, x > 0
To find(or determine) if it has a derivative:
Take the derivative of both functions
•
⎧2 x , x ≤ 0
f ' ( x) = ⎨
⎩2, x > 0
•
For it to have a derivative at a critical
point (in this case, 0), then f’(x) would
have to be the same for each part.
o In this case, it isn’t.
o There is no derivative at 0 for
this example
Property: y = f (x) will have no derivative at x
= a where the graph has:
1.
Th: If f has a derivative at x = a, then f is
continuous at x = a.
Differentiability Æ Continuity
The relationship between Differentiability and
Continuity:
1. If a function is differentiable at x = a,
then it is continuous at x = a
2. If a function is continuous at x = a, then
it does not have to be differentiable at x =
a
3. If a function is NOT continuous at x = a,
then it does NOT have a derivative at x =
a.
3.
A corner: One-sided derivatives differ
a vertical tangent: where the slopes of
the secant lines approach either +∞ or –∞
from both sides (this example: +∞)
•
•
•
Rules:
d
1. dx
(c ) = 0 where c is a constant
2.
3.
4.
5.
2.
a cusp: where the slopes of the secant
lines approach ∞ from one side and –∞
from the other side (an extreme case of a
corner)
4.
A point of discontinuity:
When the derivative crosses the x-axis, then
it will be a relative maximum or minimum.
If the graph of f’ is above the x-axis, then f
is increasing. If f’ is below the x-axis, then f
is decreasing.
When f’ is a max or a min, then f is a point
of inflection.
6.
d
dx
d
dx
( x n ) = nx n −1
[cf ( x )] = cf ' ( x )
(u ± v) = u '±v'
d
Product Rule: dx
(uv) = uv'+vu '
d u
Quotient Rule: dx ( v ) = vu ' −2uv '
v
d
dx
7. Chain Rule:
d
dx
(u n ) = nu n −1 du
dx
or dy = dy ⋅ du or ( f o g )' ( x) = f ' ( g ( x)) ⋅ g ' ( x)
dx
du dx
Position, Velocity, Acceleration, & Speed:
s (t) = position of a particle at time t
v (t) = s’(t) = velocity of the particle at time t
a (t) = v’(t) = acceleration of particle at time t
ƒ
Marginal Cost, Revenue, and Profit:
ƒ
Economists use derivatives as well for
rates of change. This is referred to as
marginals.
ƒ
The marginal cost of production is the
rate of change of the cost, c(x), with
respect to the level of production, denoted
c’(x)
ƒ
Most functions in calculus will be
differentiable, which means no corners,
no cusps, no points of discontinuity, or
vertical tangent lines within their
domains. Their curves will be smooth
with a well-defined slope at each point.
Marginal Revenue is the rate of change
of the revenue, r(x), with respect to the
level of production, or r’(x)
.
|v(t)| = speed of the particle at time t
j (t) = a’(t) = jerk of the particle at time t
ƒ
ƒ
ƒ
Profit = revenue – cost
P(x) = r(x) – c(x)
Marginal Profit: P’(x)=r’(x) – c’(x)
the rate of change of the profit with respect
to level of production
Derivatives of Trigonometric Functions:
1.
3.
5.
(tan u ) = sec u ⋅
d
du
dx (csc u ) = − csc u cot u ⋅ dx
d
du
dx (sec u ) = sec u tan u ⋅ dx
6.
d
dx
2.
3.
4.
d
dx
2
d
dx
2.
du
dx
4.
5.
7.
Implicit Differentiation
ƒ
taking the derivative without solving
for y before taking the derivative.
ƒ
Use if you can not solve for y
ƒ
BE CAREFUL about Chain Rule
and Product Rule
Implicit Differentiation Process:
1. Differentiate both sides of the
equation with respect to x
2.
Collect the terms with
on one
side of the equation
3.
4.
Factor out
Solve for
sin x
tan x = cos
x
cos x
cot x = sin x
sec x = cos1 x
csc x = sin1 x
cot x = tan1 x
Angles to Know:
Pythagorean Identities
6.
sin 2 x + cos 2 x = 1
(cot u ) = − csc 2 u ⋅ du
dx
dy
dx
Double Angle Formulas
9. sin 2x = 2sin xcos x
10. cos 2x = sin2x – cos2x
11. cos 2x = 1 – 2sin2x
12. cos 2x = 2cos2x – 1
2 tan x
13. tan 2 x = 1− tan 2 x
Trigonometric Identities
(sin u ) = cos u ⋅ du
dx
d
du
dx (cos u ) = − sin u ⋅ dx
1.
dy
dx
dy
dx
Logarithmic Differentiation:
Sometimes you can use the properties of logs
to help simplify before find the derivative.
Ex. Find y’ if y = xx, x > 0
You can not use any of the rules/properties
given because of the x’s in both the base and
the exponent. In this case, we will use logs
(natural logs):
1 + tan 2 x = sec 2 x
cot 2 x + 1 = csc2 x
8.
Derivatives of Inverse Trig Functions:
d
du
1
sin −1 u =
dx
1 − u 2 dx
d
du
1
cos −1 u = −
dx
1 − u 2 dx
sin
cos
tan
0
π
π
0
1
0
½
2
3
2
6
3
π
4
2
3
2
2
1
3
3
2
½
3
π
2
π
3π
2
1
0
∞
0
–1
0
–1
0
∞
Derivatives of Exponentials and Logarithms:
h
Property: lim e − 1 = 1
h
h→0
u
Derivative of e :
d
dx
(e u ) ⋅ du
dx
d
1 du
tan −1 u =
dx
1 + u 2 dx
Derivative of ln u: d ln u = 1 du or
dx
u dx
1
d
du
csc −1 u = −
dx
u 1 − u 2 dx
For a > 0 and a ≠ 1:
1
d
du
sec −1 u =
dx
u 1 − u 2 dx
d
du
1
log a u =
⋅
dx
u ln a dx
1 du
d
cot −1 u = −
dx
1 + u 2 dx
ln y = ln x x
ln y = x ln x
d
d
ln y =
( x ln x)
dx
dx
1 dy
⎛1⎞
= x⎜ ⎟ + ln x (1)
y dx
⎝x⎠
d u
du
(a ) = a u ln a
dx
dx
1 dy
= 1 + ln x
y dx
dy
= y (1 + ln x)
dx
dy
= x x (1 + ln x)
dx
u'
u
UNIT IV – APPLICATIONS OF THE DERIVATIVE
Let f be a function with domain D. Then
f(c) is the
(a) absolute maximum value on D if
and only if f(x) ≤ f(c) for all x in D.
(b) absolute minimum value on D if
and only if f(x) ≥ f(c) for all x in D.
Let c be an interior point of the domain of
function f. Then f(c) is a:
(a) local maximum (relative max) at c
if and only if f(x) ≤ f(c) for all x in
some open interval containing c.
(b) local minimum (relative min) at c if
and only if f(x) ≥ f(c) for all x in some
open interval containing c
Steps to find local/absolute extrema:
1.) Find the critical points
a. Any points of discontinuity
b. Take the derivative
c. Set = 0 and solve for x
2.) Substitute the critical points and
endpoints into the original equation:
3.) The highest value is absolute max
and lowest is absolute minimum:
(see graph to the right
Mean Value Theorem
If y = f(x) is a continuous function at
every point of the closed interval [a,b]
and differentiable at every point of its
interior interval (a,b), then there is at
least one point c in (a, b) at which
f ' (c ) =
f (b) − f (a )
b−a
To show how Mean Value is satisfied:
1. Use the above formula, plugging in
the interval bounds on the right side
and c into the derivative
2. Solve for c
3. If c is in the interval, then it is
satisfied, otherwise it isn’t.
Rolle’s Theorem
Let f be differentiable on (a,b) and
continuous on [a,b]. If f(a) = f(b) = 0,
then there exists at least one point c
in (a,b) such that f’(c) = 0.
Let f be a function defined on an interval I
and let x1 and x2 be any two point in I
(a) f increases on I if:
x1 < x2 Î f(x1) < f(x2).
f ' (x) > 0
(b) f decreases on I if:
x1 < x2 Î f(x1) > f(x2).
f ' (x) < 0
A differentiable function y = f(x) is
(a) concave down on an open
interval (a,b) if y’ is decreasing
on (a,b)
(b) concave up on an open interval
(a,b) if y’ is increasing on (a,b)
Second Derivative Test for Local Extrema
(1) If f’(c) = 0 and f”(c) < 0, then f has a
local maximum at x = c.
(2) If f’(c) = 0 and f”(c) > 0, then f has a
local minimum at x = c.
First Derivative Test (Local Extrema)
ƒ Find the first derivative of the function
ƒ Set = 0 and solve for x (Find the critical
points)
ƒ If a there are values not in the domain
of f’ then they are critical points as well
ƒ Set up a table (or number line) show
intervals around each of the critical
points.
ƒ If a C.P. = 0 and f’ < 0 before it and f’ >
0 after it, then it is a relative min
ƒ If a C.P. = 0 and f’ > 0 before it and f’ <
0 after it, then it is a relative max
Second Derivative Test (Concavity &
Ponits of Inflection)
ƒ Find the Second Derivative f”
ƒ Set = 0 and solve (critical points
again)
ƒ Any points where the second
derivative does not exist are also
critical points
ƒ Set up a table of intervals around the
C.P.’s (just like the 1st Derivative Test)
ƒ The points where there is a concavity
change (a sign change on the second
derivative) is called a point of
inflection.
Solving Related Rate Problems:
(1) Draw a picture of the situation, label
quantities that vary (change)
(2) Find an equation (formula) relating the
quantity with the unknown rate of
change to quantities whose rates of
change are known.
(3) Differentiate both sides of the equation
with respect to time t and solve the
derivative that will give the unknown
rate of change.
(4) Evaluate the derivative at the
appropriate point.
Particles Moving Along a Line:
Suppose s(t) is the position of a moving
particle:
ƒ If s(t) is increasing (s’ > 0), then the
particle is moving to the right
ƒ If s(t) is decreasing (s’ < 0), then the
particle is moving to the left
ƒ If v(t) is increasing (v’ > 0), then the
particle is accelerating
ƒ If v(t) is decreasing (v’ < 0), then the
particle is decelerating
Linearization:
If f is differentiable at x = a, then the
approximation function
L( x ) = f (a ) + f ' (a )( x − a )
is the linearization of f at a
__________________________________
Economics:
r(x) = the revenue from selling x items
c(x) = the cost of producing the x items
p(x) = the profit from selling x items
p(x) = r(x) – c(x)
r’(x) = marginal revenue
c’(x) = marginal cost
p’(x) = marginal profit = r’(x) – c’(x)
Th: Maximum profit (if any) occurs at a
production level at which marginal
revenue equals marginal cost.
p(x) = r(x) – c(x)
will have a maximum if p’(x) = 0
Which means r’(x) - c’(x) = 0 or when
r’(x) = c’(x)
Differentials
Let y = f(x) represent a function that is
differentiable on interval (a,b). The
differential of x, denoted by dx is any
non-zero real number. The differential
of y, denoted by dy is
dy = f’(x)dx or ∆y = f’ (x)∆x
To find errors or approximations:
1. Convert value to a easy value to
work with and a ∆x
2. Use the derivative and the formula
above to find ∆y
3. To find the approximation, find the y
value from the x you picked and
add the ∆y to it.
Using the Graph of f’:
y = f’(x)
Rel. min
Absolute
Change
Relative
Change
Percent
Change
Estimated
∆f = f ( a + dx ) − f (a )
df = f’(a) dx
∆f
f (a)
df
f (a )
∆f
× 100
f (a)
df
× 100
f (a )
Derivative of Functions with Function
in Exponents:
1. Take the natural log of both sides
•
This will move exponent
down
2. Use implicit differentiation to take
the derivative of the equation
3. Solve the equation for y’
y = x sin x
ln y = sin x ln x
1
y
y ' = sin x ⋅ 1x + ln x ⋅ cos x
y ' = y ( sinx x + ln x ⋅ cos x)
The graph of f would look something like
the one added to the graph to the left:
Rel. max
Where f’ crosses the x-axis is
where f has a max/min (see arrows).
True
y=f (x)
y=f’(x)
As you can see where f’ crosses the xaxis, then f is at a relative max/min.
Wherever f’ is below the x-axis, then
f is decreasing.
Formulas to Know:
1.
Vsphere = 43 πr 3
2.
SAsphere = 4πr 2
3.
Vcone = 13 πr 2 h
4.
Vcube = e3
5.
6.
SAcube = 6e 2
V pyramid = 13 Bh
7.
V prism = Bh
8.
Atrapezoid = 12 ( B1 + B2 )h
9.
A∆ = 12 bh
(B=area of base)
Wherever f’ is above the x-axis,
then f is increasing.
Where f’ is a max/min, then f has a
point of inflection.
How to solve Max/Min Problems:
1. Understand the Problem – Read
the problem carefully. Identify
information you need to solve the
problem.
2. Develop a Mathematical Model of
the Problem – use formulas from
geometry to help set up equations.
3. Identify the Critical Points and
Endpoints – Use the derivative to
determine max or min of the
problem. Find where the derivative
is equal to 0 or fails to exist.
4. Determine whether the Critical
Point is a max or min – use the
First/Second Derivative Tests to
determine whether the C.P is max
or min.
Ex. A rectangle is to be inscribed under one
arch of the sine curve. What is the largest
area the rectangle can have, and what
dimensions gives that area?
y
Using the area formula:
A = lw = (π − 2 x) sin x
Maximize using the derivative:
A' = (π − 2 x) cos x + sin x(−2)
P
0 = (π − 2 x) cos x − 2 sin x
Q
x
π
Let the distance from the origin to the rectangle
along the x-axis be labeled x, then the distance
from π to the rectangle is also x. This makes
the coordinate π – x.
The length of the rectangle would then be
π – 2x. The height would be sin x.
0 = π − 2 x − 2 tan x
Using the graph of the expression, we
get the following values for the zeros:
x=0.71046 and x=2.43
You can reject 2.43.
Plugging into the area formula
A = (π − 2 x) sin x
The area is A=1.122
Length=1.7207
Width = 0.65218
UNIT V – INTEGRATION
Indefinite Integral
The set of all antiderivatives of a
function f(x) is the indefinite
integral of f with respect to x and
is denoted by:
Properties of Indefinite Integrals:
Let k be a real number:
1. Constant Multiple Rule:
∫ kf ( x)dx = k ∫ f ( x)dx
The Fundamental Theorem of Calculus
1.
2.
d x
f (t )dt = f ( x)
dx ∫a
If F’(x) = f(x), then
∫
If k = –1, then,
∫ f ( x)dx
∫ − f ( x)dx = − ∫ f ( x)dx
And if F(x) is an antiderivative of
f(x) as defined in the Fundamental
Theorem, then
2. Sum and Difference Rule:
∫ ( f ( x) ± g ( x))dx = ∫ f ( x)dx ± ∫ g ( x)dx
b
a
f ( x)dx = F (b) − F (a )
Examples:
d x 3
(t − t 2 ) dt = x 3 − x 2
dx ∫7
If the upper limit is not just “x” then you have
to multiply by the derivative of the bound:
∫ f ( x)dx = F ( x) + C
d x2
cos tdt = cos( x 2 ) ⋅ 2 x
dx ∫7
where F’(x) = f(x) and C is ANY
constant value (called the
constant of integration).
If the lower limit is not a constant, then you
use will subtract:
d x2
cos tdt = cos( x 2 ) ⋅ 2 x − cos(3 x) ⋅ 3
dx ∫3 x
Integrals to Know:
∫ sin xdx = − cos x + C
∫ cos xdx = sin x + C
∫ tan xdx = − ln cos x + C
∫ tan xdx = ln sec x + C
∫ cot xdx = ln sin x + C
∫ sec xdx = tan x + C
∫ csc xdx = − cot x + C
∫ dx = ln x + C
∫ x dx = + C , n ≠ −1
∫ e dx = e + C
2
2
∫
∫u
∫
1
1 − u2
1
−1
du = sin u + C
u2 − 1
1
a2 − u 2
1
du = sec −1 u + C
x n +1
n +1
kx
1
k
b
a
a
b
∫ f ( x)dx = − ∫ f ( x)dx
2.
Zero:
a
du = sin −1
u
+C
a
1 −1 u
∫ u u 2 − a 2 du = a sec a + C
1
1
−1 u
∫ a 2 + u 2 du = a tan a + C
∫ f ( x)dx = 0
a
3.
Constant Multiple:
b
b
a
a
∫ kf ( x)dx = k ∫ f ( x)dx
4.
for any value k
Sum and Difference:
b
b
b
a
a
a
∫ [ f ( x) ± g ( x)]dx = ∫ f ( x)dx ± ∫ g ( x)dx
5.
1
x
n
Properties of Definite Integrals:
1. Order of Integration:
Additivity:
b
c
c
a
b
a
∫ f ( x)dx + ∫ f ( x)dx = ∫ f ( x)dx
kx
Average Value of a Function
The average value of f(x) on the
closed interval [a,b] is:
b
1
f avg ( x) =
f ( x)dx
b − a ∫a
Riemann Sums:
Finding the area under the curve using
rectangles of equal width whose height
is based on which type of Riemann sum
it is:
1. Left: Use the left side of the
rectangle to determine the
height (use f(left side))
2. Right: Use the right side of the
rectangle to determine the
height (use f(right side))
3. Middle: Use the midpoint of the
two sides of the rectangle to
determine the height (use
f(midpoint)).
What does a Definite Integral
mean?
1. The area under the function
2. It is the net change in amount
s (t ) = ∫ v(t )dt
This gives the net change in
position of the particle given the
velocity function.
Trapezoidal Rule:
Used to estimate the area of the curve
using trapezoids.
u-Substitution:
If the integrand is not a simple integrand (like
the list below), you can use a substitution to
make it simpler.
Let f be continuous on [a, b]. The b
Trapezoidal Rule for approximating∫ f ( x ) dx Just make sure you choose a u equal to a
a
is given by:
function whose derivative is also multiplying
b−a
the u.
[ f ( x0 ) + 2 f ( x1 ) + 2 f ( x2 ) + ... + 2 f ( xn −1 ) + f ( xn )]
2n
where n is the number of equal sized
subintervals.
Steps:
1.
2.
3.
4.
Ex.
Let u = a function
Find du/dx
Solve for the derivative part time
dx
Integrate the new integral
∫ ( x + 7)
8
dx
let u = x + 7 Î du/dx = 1 Î du=dx
∫ u du =
1
8
∫ x sin( x
2
7
Ex.
u 8 = 18 ( x + 7)8 + C
Area between 2 Curves:
If you have 2 curves f(x) and g(x),
then the area between the curves
from x=a to x=b is given by:
b
∫ [ f ( x) − g ( x)]dx
a
Where f(x) is the curve ABOVE
g(x).
If the bounds are not given:
1. Find the bounds using your
calculator (if allowed) by
graphing and finding the points
of intersection.
Note: If your area is negative, put
absolute value signs around f(x)g(x) and make the result positive.
)dx
Let u = x2 Î du/dx = 2x Î ½ du = xdx
1
2
Volume of Solids of Revolution:
Given a function y = f(x). If you revolved
the function about the x-axis, then the
volume of the solid formed would be:
∫ sin udu = −
1
2
cos u = − 12 cos( x 2 ) + C
Washer Method:
If the region is not totally bordered by axis of
revolution (maybe using two functions to create
the region), then use the following:
b
V x = π ∫ f ( x) 2 dx
a
Provided the region from x=a to x=b, is
bordered by the x-axis (or line of
revolution) – Disc Method.
If the line is not the x-axis, use the
following:
b
V x = π ∫ [ f ( x) − k ] dx
2
a
where the axis of revolution is the line
y=k.
You can use similar formulas when
revolving the y-axis or the line x=k.
b
V x = π ∫ f ( x) 2 − g ( x) 2 dx
a
or
b
V x = π ∫ R 2 − r 2 dx
a
Where R is the outer radius and r is the inner
radius.
Volumes of Solids of Known
Cross Sections
If you asked for the volume of a
solid whose base is a region
bounded by one or two functions
and whose known cross sections
are some geometric figure, then do
the following:
1. Set base=f(x)–g(x)
2. Write the area formula for the
geometric figure
3. Use the base in the formula to
determine the area function
A(x).
4. Use the following integral:
b
V ( x) = ∫ A( x)dx
a
You may need to find the bounds
using intersection points on the
graph.
Formulas to Know for Cross
Sections:
Square: A=base2
Rectangle: A=base x height
(the height will be given somehow)
Equilateral Triangl: A = 43 base 2
Semi Circle:
A = 14 π ⋅ base 2