Download CALCULUS 2 LECTURE NOTES 1 title page 2 some

CALCULUS 2
LECTURE NOTES
1 ... title page
2 ... some precalculus
4 ... review of the derivative
13... what is integration
13... antiderivatives
15... Euler approximation
17... Fundamental Theorem of Calculus
22... integral as area
28... applications- position/velocity/acceleration, economics, electricity
31... application- work in physics
32... applications (set up, do not solve)
37... techniques of integration- algebra
38... techniques of integration- substitution
42... area between curves
46... volume
56... arclength
58... surface area
60... application- consumer surplus in economics
61... application- fluid pressure & force
62... application- work applied to a gas
63... average value
64... application- center of mass (centroid)
69... integration by parts
72... integration of basic trig functions
73... integration of combined trig functions
78... integration by partial fractions
82... integration by substitution with trig functions ("trig substitution")
90... other techniques of integration
92... techniques of integration, overview
93... improper integrals
96... the integal as a limit
First, an important topic you may or may not have seen before.
Four ways to represent a mathematical situation
ex) y = x(x-4)
find y for x=11
...now, before we solve this,
what kind of representation is this?
symbolic representation (also called algebraic representation)
what are the variables?
x,y
numerical representation? (also called a table of values)
x
y
grapical representation?
verbal representation?
a rectangle has width 4 less than the length. let x be the length and let y
be the area.
ex) when i turn on the hot water faucet, the temperature starts out cold. it
stays cold for a little while, then slowly gets a little hotter, then quickly gets
real hot. it stays real hot.
what kind of representation is this?
verbal representation
what are the variables?
temperature (y), time (x)
graph representation?
ex)
what kind of representation is this?
graph
what are the variables?
x,y
symbolic representation?
y= -(x+2)(x-4)
... y = -(x2 - 2x - 8) ... y= -x2 +2x +8
numerical representation?
(-2,0) (4,0) ...or from the equation, (1,9) (0,8)...
verbal representation?
path of a thrown ball (because, thanks to gravity, it will travel in a parabola)
ex) value of a CD earning 5% annual interest
years
0
1
2
5
10
20
value
100010501102.50 1276.28 1628.89 2653.29
what type of representation is this?
numerical
what are the variables?
time, measured in years (t), value (A)
graph representation?
symbolic representation?
A = P(1+r)t or A = 1000(1.05)t
what are the 4 representations:
symbolic, graphical, numerical, verbal
why do we use different representations?
when is numerical better? in which problem would you prefer numerical? why?
if the information you want is on the list
often best to describe a real world situation
" graphical?
if the information you want can be found visually
e.g. is the highest point at positive x-value or negative x-value?
" symbolic?
for precise calculations
" verbal?
often best to describe a real world situation
Some things you should know about the derivative
all these things...
the slope at a point
the slope of the tangent line
(instantaneous) rate of change
e.g. miles/hour, $/unit, etc
...are equivalent, and they are called the derivative
notation:
"take the derivative" same as "differentiate" same as "D" same as d/dx
the derivative of
is
how do we find the derivative of f(x) ? (symbolically)
here are some rules - derivatives of basic functions
f(x) = 3
f '(x)=0
rule: f(x) = k ... f '(x) = 0
(for a constant k)
f(x) = 3x+2
f '(x) = 3
rule: f(x) = mx+b ... f '(x) = m
f(x) = x2
f '(x) = 2x
rule: f(x) = xn .. f '(x) = nxn-1
ex) f(x) = x1.7
f '(x) =
f(x) = sin(x)
f '(x) = cos(x)
f(x) = cos(x)
f '(x) = -sin(x)
also: (tan x)'=sec2x ... (sec x)'=sec(x)tan(x) ... (cot x)'=-csc2x .. (csc x)'=-csc(x)cot(x)
f(x) = ex
f '(x) = ex
f(x) = 2x
f '(x) = (ln2)2x
f(x) = ln x
f '(x) = 1/x
f(x) = log3x
use log rules: f(x) = lnx
f '(x) =
Derivatives of unusual functions
the derivative of inverse trig functions
ex) y = arctan(x) ... find y'
x = tan(y)
want: y'
take the derivative
1 = sec(y)2 y'
y=y(x) use the chain rule because
y' = 1
y is your inside function
sec2y
y' = [cos(y)]2
y' =
y' =
also:
y = arcsin(x)
y' =
(all six are in the text)
derivative rules for complicated functions
the product rule for derivatives
rule: (f·g)' = f '·g+ f·g'
extended product rule:
(f·g·h)' = f '·g·h + f·g'·h + f·g·h'
ex) find (x·sinx)'
ex) find the derivative of y=x·cos(x)
take f=x ... g=cos(x)
y' = (x cos(x) )'
y' = (x)'cos(x) + x( cos(x) )'
= 1cos(x) + x(-sin(x) )
y' = cos(x) - x sin(x)
ex) find the derivative of y = x2exsin(x)
take f=x2 ... g=ex ... h=sin(x)
y' = (x2)'exsin(x) + x2(ex)'sin(x) + x2ex(sin(x) )'
y' = 2xexsin(x) + x2exsin(x) + x2excos(x)
the quotient rule for derivatives
y= f
g
y' = (f)'g - f(g)'
(g)2
ex) y =
ex
x2
y' =
ex) y = cos x
x2 + 1
y' =
ex) y =
y' =
sinx
x2ex
the Chain Rule for derivatives
y = f(g(x)
y' = f'(g(x))·g'(x)
ex) y = sin(x2)
y' = [sin(x2)]'
= cos(x2) · (x2)'
= cos(x2)·2x
ex) y = (cosx)3
y' =
ex) f(x) = (ex+4x)5
f '(x) =
the extended chain rule:
y = f(g(h(x)))
y' = f '(g(h(x))·g'(h(x))·h'(x)
ex) y = sin(ln(x2+3))
y' = cos(ln(x2+3)) · [ln(x2+3)]'
= cos(ln(x2+3) · 1 · (x2+3)'
x2+3
= cos(ln(x2+3))· 1 ·(2x)
x2+3
Understanding the derivative using graphs
for a linear function, we know how to calculate the derivative
ex) rise/run OR y2-y1 / x2-x1
for a linear graph, we know how to "see" the derivative
ex) count boxes
what about for a non-linear graph?
in Calc 1, you learned how to calculate the derivative for f(x)
do you know how to "see" the derivative?
the derivative at a point
[use technology tool]
the derivative as a function
[use technology tool]
what is the derivative at x=1 ?
what is the graph of f '(x) ?
hw questions
Review of Linear Approximations / Tangent Lines
f(x) = ln(x2 + .19) ... at x=1,
find f(1) (with a calculator)
find the equation of the linear approximation, L(x)
approximate f(1.5)
f(1) = ln(12+.19) = ln(1.19) ≈ .1740 -> point (1,0.1740)
f '(x)= 1
(2x)
2
x +.19
f '(1) = 2(1) = 2
≈ 1.6807 -> slope 1.6807
2
1.19
1 +.19
f(x) ≈ L(x) = .1740 + 1.6807(x-1)
f(1.5) ≈ L(1.5) = .1740 + 1.6807(1.5-1) = 1.0144
in fact, f(1.5) = ln(1.52+.19) = .8920
what if we wanted to approximate f(2) ?
f(2) ≈ L(2) = f(1) + f '(1)·(2-1) = .1740 + 1.6807(2-1) = 1.8547
in fact, f(2) = ln(22+.19) = 1.4327
you know what? ... we can make a better approximation ... how?
f '(1.5) =
2(1.5) = 1.2295
1.52+.19
so we can estimate f(2) starting at the approximation for f(1.5)
so f(2) ≈ L(1.5) + f '(1.5) (x - 1.5) = 1.0144 + 1.2295(2-1.5) = 1.6292
what does this look like on the graph?
thats great, but...
...why are we doing this, since we already know f(x) ?
well,
what if we dont know f(x) ...??
Integration:
Differential Equations, Slope Fields, Euler Approximations, Antiderivatives
ex) suppose f '(x) = 2x
what can we figure out ?
symbolically:
what is f(x) ?
numerically:
what is f(3) ?
graphically:
draw f(x)
verbally:
if the velocity is v = 2x
how far have you gone
after x sec?
Symbolic:
we know f(x) = x2
right?
well, sort of
it could be x2... what else could it be?
it could be x2 +1
the full answer is f(x) = x2 + C
where C is some unknown constant
now, what if i told you that f(1)=2
can you find the constant?
2
2=1 +C
1=C
so...f(x) = x2 + 1
notice that we need to include "C"
we can find C only with that extra piece of information, which is called the
initial condition
when you are given f '(x) as your function, f(x) is called the antiderivative
ex) what is the antiderivative of f '(x) = 3x2 ?
f(x) = x3 + C
Antiderivative formulas
given the derivative, find the function
ex) f '(x) = 2x
f(x) = x2 + C
ex) g'(x) = x3 ... find g(x)
formula: f '(x) = xn ... f(x) = xn+1 +C
n+1
ok, lets make some more!
f '(x)
f(x)
cos(x)
sin(x)
sin(x)
-cos(x)
x
e
ex
1/ or x-1 ln(x)
x
sec2x
tan(x)
two more useful ones:
f(x) = arcsin(x) ... f '(x)= 1
1-x2
f(x) = arctan(x) ... f '(x)= 1
1+x2
[the list of all six inverse trig functions is in the text]
ex) f '(x) = 2x3 + cos(x)
the antiderivative f(x) =
Another Notation:
if you have f(x), the antiderivative is written F(x)
ex) f(x) = x2+3 ... find F(x)
note: when you see f(x) and F(x) written together, we usually assume that
F' = f
Numerical:
f '(x) = 2x ... if f(1) = 2, what is f(3) ?
f(3) ≈ L(3) = f(1) + f '(1)·(3-1)
= 2 + 2(1)·(2)
=2+4
=6
graphical:
that is a bad approximation ... what would give us a better one?
use "stopping points" x = 1, 1.5, 2, 2.5 ...and arrive at 3
each step is .5
f(3) ≈ f(1) + f'(1) (1.5-1) + ?
f '(1.5) (2-1.5) + ?
f'(2) (2.5-2) + ?
= 2 + 2(1) (.5) + 2(1.5) (.5) +
=2 +
1
+
1.5
+
=9
2(2) (.5) +
2
+
f'(2.5) (3-2.5)
2(2.5) (.5)
2.5
Euler approximation
given f '(x) and f(a), find a numerical approximation for f(b)
ex) f(2) = 5, f '(x) = ex/x ... approximate f(3.2) with 3 steps
with 3 steps, what is ∆x ?
we have to go from x=2 to x=3.2 in 3 steps...each step = .4 = ∆x
f(3.2) ≈
graph:
this sort of equation involving derivatives is called a differential equation.
we will represent a differential equation graphically
...how??
lets start with an easy differential equation: dy/dx = 2x
what does this mean?
at any point (x,y) the slope is 2x
ex) at (1,1) the slope is
at (4,7) the slope is
at (-2,3) the slope is
at (5,-2) the slope is
this is called a slope field
we can draw the function only if
we know the initial condition
ex) f(1)=2
we had: f(x) ≈ f(a) + f '(a)·∆x + ...
lets rewrite it as: f(x) - f(a) ≈ f '(a)·∆x + f '(a1)·∆x + f '(a2)·∆x +
where each ai is an x-value, they are evenly spaced by ∆x
the fancy notation for this is:
f(x) - f(a) ≈ Σ f '(ai)·∆x
as we know, this is an approximation. how do we get an exact answer?
take ∆x smaller and smaller
we have done this before....when?
with the derivative
so when you get "all the way small"
the approximation becomes exact
and we get new fancy notation:
f(x) - f(a) = ∫a f '(x) dx
where ∆x "becomes" dx and Σ "becomes" ∫ ("integral")
oooh, almost
theres a problem with the notation: x serves two roles: as the
variable in the f ', and as the partner value for a
heres how we fix the problem: we say f ' is a function of t
f(x) - f(a) = ∫a f '(t) dt
this is the Fundamental Theorem of Calculus #1
now let x be a particular value, say x=b. this gives
f(b) - f(a) = ∫a f '(t) dt
this is the Total Change Theorem
note that your start and endpoint are written next to the "∫". they are called the
boundaries of integration. (they are also called "limits of integration" - not to be
confused with "the limit"!)
note: if you are not precise and ignore the x/t variable conflict, its not a major
issue
ex) f '(x) = x3+3 ... f(2)=4 ... find f(x) ... find f(5)
Antiderivatives and the FTC - math examples
ex) find ∫cos(x) dx
=sin(x)+C
find ∫osin(x) dx
=[-cos(x) +C ]o
find ∫o cos(x) dx
= (-cos(pi)+C) - (-cos(0)+C)
= sin(x) |o
=1+1=2
= sin(t) - sin(0) = sin(t)
note that here, +C cancels itself
if g'(x)=cos(x) and g(0) = 1, find g(x)
g(x) - g(0) = ∫ cos(x) dx
[here, i am using the sloppy notation where x serves two roles
g(x) - 1 = sin(x)|0
its not that bad]
g(x) - 1 = sin(x) - sin(0)
g(x) = sin(x) + 1
ex) f '(x) = x3 ... find f(x)
f(x) = ∫x3 dx = x4 + C
4
ex) ∫1 1/x dx
also, if f(1)=1, find f(x)
f(x) = x4 /4 + C
1=f(1) = 14 /4 + C
1 = 1/4 + C
C = 3/4
f(x) = x4 /4 + 3/4
"the integral from 1 to 2 of 1/x dx"
" ln(x) from 1 to 2" OR "ln(x) evaluated from 1 to 2"
compare:
another notation for the FTC:
recall that ∫ f(x) dx = F(x) ... where F'(x) = f(x)
when you see F(x) and f(x) together, usually it is understood that F' = f
so you will often see the Fundamental Theorem as
∫a f(t) dt = F(x) - F(a) ... where F'(x) = f(x)
ex) find ∫2 x2 dx
∫2 x2 dx
f '(t) = t2 ... f(2) = 7 ... find f(t)
what does the Fundamental Theorem of Calculus #1 say, in words?
FTC1: ∫a f '(t)dt = f(x) - f(a)
"if you have a function, then you take the derivative, then you take the integral,
you get the function back"
(note that the notation is a little tricky...why?
because of the constant of integration, and "x" playing two roles)
writing the Theorem this way implies something else might also be
true...what?
discuss...
"if you have a function, then you take the integral, then you take the derivative,
you get the function back"
this is true, and its easy to show:
we will take the integral of f(x) ... say that F(x) is any antiderivative of f(x),
which means that F' = f
d/ (∫ f(t) dt)
= d/dx (F(x) - F(a) )
dx a
= F'(x) - 0
= f(x)
ex) d/dx (∫1 t2 dt) =
ex) d/dt [∫1 x2 dx ]
slightly more complicated:
ex) d/dx (∫sinx t3 dt) =
notice how the answer is not (sinx)3
you also have "cosx" ... from the chain rule
hw questions
Integral as an Area
there is another way to interpret a sum/integral
what is the area under f(x)=3 between x=1 and x=6 ?
thats easy...because its a rectangle
area = (3)(5) = 15
what is the area under f(x) = 3x between x=0 and x=4 ?
now,
what is the area under f(x) = x2 between x=1 and x=3 ?
oh, thats not easy
the way we will approach this is to approximate the area
how can we make an approximation of the area? ...discuss
there are several methods which will work
we will use the following method: approximation by rectangles
split up the interval from x=1 to x=3
make each section the same length (...this will be our ∆x )
ex) approximate the area under f(x) = x2 between x=1 and x=3 with 4
rectangles
rectangle bases are [1,1.5] [1.5,2] [2,2.5] [2.5,3]
we will approximate the function over each section with the same yvalue..which y-value?
we choose the y-value it starts with, on the left end
[note: there are other choices for which y-value to use]
Area ≈ sum of rectangle areas
= f(1) (.5) + f(1.5) (.5) + f(2) (.5) + f(2.5) (.5)
= 1(.5) + 2.25(.5) + 4(.5) + 6.25(.5) = 6.75
= f(1)·∆x + f(1.5)·∆x + f(2)·∆x + f(2.5)·∆x
[historical note: these methods originate with Archimedes
and the ancient Greeks]
you do:
ex) approximate the area under f(x) = .5x2 + 1 between x=2 and x=4, use
4 rectangles
width = ∆x = .5
approx area = f(2) ∆x + f(2.5) ∆x + f(3) ∆x + f(3.5) ∆x
= 3(.5) + 4.125(.5) + 5.5(.5) + 7.125(.5)
= 9.875
using rectangles with the y-value from the left-endpoint,
then adding up the areas
... this is called the Left Sum
notation: Ln ...where n is the number of rectangles
if we want a better approximation, what do we do?
take more rectangles, smaller ∆x
if we want the exact answer, what do we do
take ∆x "all the way small"
so we have Area ≈ Σ f(x)·∆x
and we are taking ∆x "all the way small"
well we know how this goes...
Area under f(x) between x=a and x=b is: ∫a f(x) dx
ex) the Area under f(x) = 2x between x=3 and x=5 is
Area = ∫3 2x dx
= x2 |3
= 52 - 3 2
= 16
the procedure is easy, but understanding why it works is hard
hard, but important...you have to know why so you know when to use it
ex) the Area under f(x) = ex between x=0 and x=4 is
Area = ∫o ex dx
= ex |o
= e4 - e o
≈ 53.6
ex) what is the area under f(x) = sin(x) between 0 and 2π
Area = ∫o sin(x) dx
= [-cos(x)]o
= -cos(2π) - -cos(0)
= -1 + 1
=0
how can that be? there certainly is area under the curve...
it depends how you define "area" and "under"
remember the approximation: we are adding up f(x)·∆x
well, sometimes f(x) is positive, and sometimes it is negative
so area above the x-axis is positive ... and area under the x-axis is negative
can you see why the integral = 0 ?
tricky case:
ex) ∫-3 x-2 dx
whats the problem?
moral: you must make sure that f(x) is defined
before you integrate
(more on this later in the semester)
remember we learned two different integral types: ∫b and ∫x
in terms of area:
∫b f(x)dx is the area under f(x) from x=a to x=b
what is ∫x ?
∫x f(x)dx is the area under f(x) from x=a to x=x
(ok, lets fix the notation)
x
∫ f(t)dt is the area under f(t) from t=a to t=x
notice that the value depends on x ... this is a function
we call it the Area Function
Af(x) = ∫x f(t)dt
note that the Area Function is a way of seeing the antiderivative if you graph f(x)
ex) f(x) = 2x ... graph F(x)
ok, we know the formula for F(x) ... but what if we didnt? how could we graph it?
graph of f
Af =
x || area under f from 0 to x
tie it all together: what is the derivative of the Area Function?
[Af(x)]' = d/dx [ ∫x f(t)dt ]
= d/dx [ F(x) - F(a) ]
where F(x) is an antiderivative of f(x)
= F'(x) - 0
= f(x)
conclusion?
the Area Function is a way to visualize the antiderivative
ex) f(x) = 1/x what is the Area function Af (if we start counting area at x=1)
ex) f(x) = x3 + 2x find Af (if we start counting area at x=3)
lets rewrite the FTC formula, starting with f(x) [instead of f '(x) ]
∫a f(x) dx = F(b) - F(a) ...where F'(x) = f(x)
in other words, when we start with f(x), we assume there is some function
F(x) which is the antiderivative ... we just have to figure out what it is
hw notes:
if you are asked to find f, then find it
ex) f '=2x, f(0)=4
finding f from f '' and handling constants
ex) f '' = et
antiderivative with constant in front
ex) ∫ 2/x dx
trig/inverse trig and using radians
ex) ∫πcos x dx
using the power rule (too much)
ex) ∫ (1-x2)1/2dx
Euler approximation results are *approximations*, use ≈ (or "approximately")
writing down integral with other variables in it... (a) understand different
variables (b) important for word problems (c) compare with related rates
hw questions
real world application
first lets recall...
the derivative of distance is:
the derivative of velocity is:
note: we also refer to distance as position or displacement
note: we call the position function p(x). why?
we dont want to write d(x), because when we take the derivative you
would see dd, and thats confusing
ex) if at time t, velocity is v=2t, what is the position function?
v(t)=p'(t)
p(t) = ∫p'(t) dt = ∫ 2t dt
=t2 +C
if the position at time t=1 is p=2, then:
2=12+C
1=C
p(t) = t2+1
ex) the acceleration of a free falling object is 9.8
[9.8 m/sec2 is the acceleration due to gravity ... a = 9.8]
what is the velocity at time t ?
v' = a
v(t) - v(0) = ∫o v'(x) dx
note: we want t as our variable, so we
use x as the "dummy variable"
v(t) - v(0) = ∫o a(x) dx
v(t) - v(0) = ∫o 9.8 dx
what function has a derivative of 9.8 ?
thats easy, its "9.8x"
what do we do with 0 and t ? [those are called "endpoints (or limits) of
integration"]
v(t) - v(0) = 9.8x ]0
v(t) - v(0) = 9.8t - 9.8(0)
= 9.8t
OR v(t) = v(0) + 9.8t
[note: this is just like 9.8t + C
in this problem, we know exactly what the constant C represents: the
velocity at t=0 ... call it v0
next
what is the distance (position) at time t ?
p' = v
p(t) - p(0) = ∫o p'(x) dx
p(t) - p(0) = ∫o v(x) dx
p(t) - p(0) = ∫o 9.8x + vo dx
what function has a derivative of 9.8x + vo ?
"9.8x2 /2 + vox"
p(t) - p(0) = 4.9x2 + vox|o
p(t) - p(0) = (4.9t2 + vot) - ( 4.9(0)2 + vo(0) )
= 4.9t2 + vot
OR p(t) = p(0) + vot + 4.9t2
ex) if an object has initial position 3 meters, and initial velocity 4m/sec, what
position is it at when t=2sec?
differential equations are very important in: math, physics,
chemistry, biology, sociology...
why? a lot of times, we describe a situation using the rate of change
(like the velocity)
lets look at some examples we can solve by hand
Economics
ex) suppose the derivative of the cost function C'(x) (known as the
marginal cost, MC) is 28 dollars/item. suppose making no items costs
$8000.
what is the total cost function C(x) ? (C is dollars, x is items)
what is the cost of 120 items?
C(x) = ∫ C'(x) dx
= ∫ 28 dx
= 28x + k dollars
C(0) = 28(0) + k
8000 = k
C(x) = 8000 + 28x
C(120) = 8000 + 28(120) = 11360 dollars
...or...
C(x) - C(0) = ∫0 C' dt
= ∫0 28 dt
= 28t |0
= 28(x) - 28(0)
C(x) - 8000 = 28x - 0
C(x) = 8000 + 28x
and
C(120) = 8000 + 28(120) = 8000 + 3360 = 11360 dollars
ex) electricity
current = charge
time
current is measured in amps (I)... charge in coulombs (Q) ... time in
seconds (t)
amps = coulombs
sec
∆Q
I
= /∆t
ex) suppose at t=0sec we have Q=4 coulombs, at t=4sec we have Q=12
coulombs
what can we say about the current?
current I =
note that this is the average current
ex) suppose the current is I = 3amps ... if we have 12 coulombs after 1
sec, how many do we have after 5 sec?
how is this calculus??
ex) suppose the current varies. at time t, the current is I = 1 + t
how much charge is collected between t=2sec and t=5sec (meaning,
what is the total change in charge) ?
∫2 I(t) dt = ∫2 Q'(t) dt
follow-up: if there are 4 coulombs at t=2 sec,
how many coulombs are there at t=5 sec?
Application in Physics: Work
Work is the amount of energy expended
Force is the amount of work per unit of distance (it is the same as weight)
note: in different countries, work and force are measured in different units. The
American units are confusing, so I will use the metric units:
Work is measured in Joules
Force is measured in Joules per meter (also called Newtons)
when Work and Force are written as functions of distance, W(x) and F(x),
then W'(x) = F(x)
therefore, W(x) = ∫ F(x) dx
the simple case: force is constant, so work is calculated by multiplication
ex) to push a desk, F=7 Joules/meter
how much work is done to push the desk from x=1 to x=6?
ex) if the amount of work done by a tow truck is W(x)=20x
a) how much work is done after 0 meters? after 5 meters?
b) what is the Force of the tow truck?
now, with calculus...
ex) suppose a magnet exerts a force of F(x)=x-2 joules/meter
a) how much work is done moving an object from 1 meter to 4 meters?
examples of Work and Force from some important laws in physics:
Hooke's law for springs: F = kx
ex) how much work is done by a spring with k=25 as it moves an object from 2
meters to 15 meters?
ex) suppose a spring exerts a force of 30 Joules/meter when extended 6
meters. how much work is done extending it from 6 to 12 meters?
Law of gravitation between two objects: F = G m1m2 OR k
x2
x2
where m1 and m2 are the masses of the objects, x is the distance
between
and G is a constant from physics ("universal gravity constant")
ex) two objects are currently 1000m apart and exerting a force of 5 Newtons.
how much work must be done to move them to 5000m apart?
here are some fancy examples (which we do not know how to solve by hand)
ex) Newton's law of Cooling
for a situation where you have an object of one temperature sitting in a room
of a different temperature
the rate of change of the temperature of the object is proportional to the
difference of temperature
why? in the same room, a very hot cup of tea will cool off faster than warm
pizza.
how can we write this as a differential equation?
dT = k(T - T0) ... T = temp of the object ... T0 = temp of the room ... t = time ...
dt
k = constant reflecting properties of the object
ex) Torricelli's Law of Draining
when a cylindrical tank filled with water is draining, the height of the water y
decreases at a rate proportional to √y ... in other words
dy = k√y
[k is a constant based on the size of the drain hole]
dt
ex) assume a certain drain has k = 2
when the water is 4 meters high, how fast is it draning? (meters/sec)
dy/ = 2√4 = 4 meters/sec
dt
when the water is 2 meters high, how fast is it draning? (meters/sec)
dy/ = 2√2 ≈ 2.828 meter/sec
dt
how much does the height of the water change from t=0 to t=3 sec ?
a difficult, interesting real world example - extra credit material
ex) population growth of the US
typically, you hear people say "the population grew 2% last year (and the same
every year, we assume)"
what does that mean? how can we write that?
dy = .02y
dx
thats very interesting...what do we do with that?
there are several approaches
first, the numerical approach:
ex) suppose the population is 300 million
what will it be next year?
.02(300) = 6
next year, 306 million
...write that as one equation: 306 = 300 + .02·300·1
hmmm
f(1) = f(0) + f '(0)·∆x
[∆x is 1 because we want
the increase for 1 year]
what will the population be in 2 years?
f(2) = f(0) + .02(300)·2 = 312 million people
well, that's not quite right
in the second year, the population increases by 2% of 306
so: f(2) = f(0) + f '(0)·1 + f '(1)·1
= 300 + .02(300)·1 + .02(306)·1
= 300 + 6 + 6.12
= 312.12 million people
ex) dy = .04y
dx
f(0) = 1000
find f(2) [break it up year-by-year...in other words, ∆x=1]
we will use f(x) = f(a) + f '(a) (x-a) repeatedly
f(1) = f(0) + f '(0)(1-0) = 1000 + .04(1000)(1) = 1000 + 40 = 1040
that gives the point (1,1040)
f(2) = f(1) + f '(1)(2-1) = 1040 + .04(1040)(1) = 1040 + 41.6 = 1081.6
note that we need the previous y-value to find the next y-value, so we must
calculuate y-values one by one
in the lab assignment, the technology did these calculations for us
hw questions
review
Techniques of integration - Algebra
we know how to do basic integrals if we recognize the function as the
derivative of another function
ex) ∫ 2x dx =
sometimes we need to do algebraic manipulation to turn the problem
into one we can do
ex) ∫ x(x+5) dx
rewrite this as:
= ∫ x2+5x dx
now it is in a form we are familiar with
=
x3
+
ex) ∫ 2 dx
x4
ex) ∫
5x2
+C
Techniques of Integration - Substitution
some examples need a little more help to get into a form we recognize
ex) ∫ 2x cos(x2) dx
what does it look like it might be the derivative of? ... discuss
(sin x2)' = cos(x2)·(x2)'
so, the key is x2 - it is the inside function from the chain rule. it appears
twice - once as the "inside function", and again, as its derivative. so any
expression that has that form, we know how to find the integral.
ex) ∫ (x4+1)7 4x3 dx
so what is that "inside function"?
x4+1
what is the "outside function"
well, its derivative is u7
so the original function is u8
∫ (x4+1)7 4x3 dx = (x4+1)8
check it: (x4+1)8 ' = 8(x4+1)7 · (x4+1)' = (x4+1)7 · 4x3
...yes
Lets make a nice notation to keep track of this process:
∫ (x4+1)7 4x3 dx
u = (x4+1) ... du = 4x3 we can rewrite this as du = 4x3 dx
dx
(more about that in one minute)
(substitution makes a new integral)
7
∫ u du
(which we know how to do)
8
=u +C
(then we substitute back)
4
= (x +1)8 + C
A comment about dx, dy, and du
Previously we have defined dy/dx as a derivative. we never carefully
defined "dy" or "dx" separately. they are called differentials.
lim ∆x = dx
lim ∆y = dy
∆x →0
∆x →0
We will often think of dx as a "very small ∆x", and this will help us do
problems.
We will often think of a differential as a term that we can manipulate with
algebra, as in the exercise above. The proof that you can do this is
beyond the scope of this class.
Another example:
ex) ∫ sin(tan(x)) sec2x dx
can we find u & du ?
u=
du=
ex) ∫ cos(t) esin(t) dt
Sometimes, the substitution does not work out perfectly, but we can
make it work.
ex) ∫ x (x2+7)11 dx
u = x2+7 du=2x dx
but we want to replace x dx ... what can we do?
...divide by 2 ... 1/2 du = x dx
= ∫ u11 1/2 du
= 1 u12 + C
2 12
=u12 + C
=(x2+7)12 + C
moral: we can always manipulate the constant, so it doesnt have to
match in the original problem, we can make it match
ex) ∫ sec2 (6x) dx
u=
ex) ∫ ln(x) dx
x
ex) ∫
will substitution work?
dx
x3-x
4
2
√x -2x -5
why should we check to see if substitution will work?
...because there is an "inside function". look for u & du
Techniques of integration - substitution/shifting the variable
There is another use of substitution which is sort of an algebra trick.
ex) ∫ x√x-3 dx
"x-3" under the square root is an obstacle. here, we can remove it with a
trick.
u=x-3 ... du=dx
u+3=x
= ∫ (u+3)√u du
= ∫ u3/2 + 3u1/2 du
= u5/2 + 3u3/2 + C or 2u5/2 + 2u3/2 + C
=2/5(x-3)5/2 + 2(x-3)3/2 +C
generally, we might use this trick when we have radicals of linear terms
combined with products or quotients
ex) ∫
t dt
√2t+1
sub: u=
what the indicators of possible substitution?
- inside function
- an expression and its derivative
(might have one OR the other OR both)
hw questions (integration by substitution)
Application of Integration - Area between curves
Previously, we saw that the area under a curve could be calculated using an
integral. The area could be approximated by rectangles, where each rectangle
had a height f(x) and a width ∆x, so each rectangle had an area of f(x)·∆x and
the total area approximation was Σf(x)∆x. This approximation becomes exact
by taking the limit, which we write as an integral.
ex) y=x2
area from 1 to 3 approximated with 4 rectangles = L4 = 6.75
area exactly = ∫13 x2dx = 8.66
Now what about the area between two curves?
how much area is between the two graphs,
f(x)=0.1x3 + 0.1x2 - 0.4x + 2.6 and g(x)=0.2x2 - 3
from x=-3 to x=2
we can approximate it with rectangles
what is the width?
what is the height? ...discuss
what is the formula for the
approximate area?
what is the formula for the
exact area?
the area between two functions f(x) and g(x) from x=a to x=b is: ∫ab f(x) - g(x)
dx
ex) find the area between f=sin(x)+2 and g=cos(x)-2 from x=0 to x=2π
ex) find the area bounded by the functions f(x)=x2 and g(x)=x+2
what are the boundary x-values? ... you have to find them - how?
at the intersection, f=g
x2 = x+2
x2 - x - 2 = 0
(x-2)(x+1)=0
x=2, -1
so the area is
∫-12 x+2-x2 dx
= x2 + 2x - x3 |-12
2
3
2
= (2 /2+2(2)-23/3) - ((-1)2/2+2(-1)-(-1)3/3)
= 20/6 - -7/6
= 4.5
note: here g is above f.
if you accidentally switch f & g and calculate ∫ f-g, you get the same
answer but negative. thats easy to fix.
What if the graphs create more than one region?
ex) find the area bounded by f=3x3 - x2 - 10x
and g= -x2 + 2x
f=g ... 3x3-x2-10x = -x2+2x
3x3-12x=0
3x(x2-4)=0
3x(x+2)(x-2)
x=-2,0,2
but notice that for -2<x<0, f>g
then 0<x<2, g>f
we want all area to count positively,
so we must do two integrals
examples
ex) find the area between f = 2x2 - 8x - 2 and g = x2 - 3x + 5
ex) find the area between f = x3 - x2 and g = 3x - 3x2
hw questions- enclosed area
Volume
method 1: known cross-section area
One of the earliest questions investigated, which helped lead to the creation
of calculus, was: What is the volume of a given object?
lets approach volume in the same spirit we used for area:
- for what basic object do we know the volume?
... a rectangluar solid. its volume is (L)(w)(h)
but what about a barrel? or a pyramid?
we will extend our method from finding 2-dimensional area
area = (h)(w)
when the height=f(x) is changing, we approximate with Σf(x)∆x
and the exact amount is ∫ab f(x) dx
now,
volume = (L)(w)(h) or (Area)(w)
when the area=A(x) is changing, we approximate with ΣA(x)∆x
and the exact amount is ∫ab A(x) dx
lets see how a single slice with width ∆x is a solid with volume A(x)∆x
http://socrates.bmcc.cuny.edu/jsamuels/math302/volume_known_crosssection.html
ex) what is the volume of a pyramid with a square base?
imagine the pyramid, instead of pointing up, is pointing to the side. the base
is at x=0 and the tip is at x=2. a cross-section (at each x-value) is a square,
and the side of the square = (2-x) [length measured in meters]
ex) what is the volume of a barrel which is 4m long, and the cross-section (at
each x-value) is a circle with radius r= 4 - 0.3(x-2)2 0<x<4?
Volume: known cross-section (y is the variable)
lets revisit the pyramid problem. what if we switched the variable from x to y?
the base is at y=0 and the top is at y=2. a cross-section at each y-value is a
square, and the side of the square = (2-y)
Volume =
note: we can interpret this in two ways.
one, we can use any variable we want - r, s, t, z, whatever.
two, when you draw the graph, the pyramid is pointed up, along the y-axis.
also, we are thinking of Area as a function of y, A(y).
ex) find the volume when the region is bounded by x=4-y2 in the first quadrant
and a cross-section at each y-value is an isoscales right triangle (the right
angle sits on the y-axis)
Volume - solid of revolution
using disks
what if the solid is created by taking a region bounded by y=f(x) and revolving
it around the x-axis? what does that look like? what is the cross-section?
what is the formula to find the volume?
http://socrates.bmcc.cuny.edu/jsamuels/math302/volume_disk.html
Volume = ∫ab A(x) dx = ∫ab πr2 dx ...where r=r(x) is a function of x
ex) find the volume when the region bounded by y=x2, x=1, x=3, y=0 is
revolved around the x-axis
ex) find the volume when the region bounded by y=√x and x=0 and x=4 is
rotated around the x-axis
what if it is rotated around a line that is not the x-axis?
ex) find the volume when the region bounded by y=x2+x+2 and y=1 and x=1
and x=2 is rotated around the line y=1
...what is the radius?
Volume: disk-with-hole
what if there is a "hole" in the middle created by a second function?
...just subtract that volume.
ex) take the region enclosed by y=x and y=x2 and rotate it around the x-axis.
what is the volume?
first, make a sketch to get the idea:
Volume = V1 - V2
= ∫01 A1(x) dx - ∫01 A2(x) dx
= ∫01 A1(x) - A2(x) dx
= ∫01 πr12 - πr22 dx
= ∫01 π(x)2 - π(x2)2 dx
[nicer as one integral]
ex) take the region enclosed by y=x and y=x2 and rotate it around y=2. what is
the volume?
Volume- solid of rotation, using disks, y is the variable
ex) find the volume when the region bounded by x=0, y=1, x=y1/2 is revolved
around the y-axis.
ex) find the volume when the region bounded by y=ln(x) & y=1 & y=2 and x=0
is revolved around the y-axis
picture:
note: if you understand differentials really well, there is another way to do the
problem - with x as the variable
moral: i will show you guiding principles for how to do the problems, but there
are many possible techniques
Volume - solid of revolution
using shells
http://socrates.bmcc.cuny.edu/jsamuels/math302/volume_shells.html
if a region is rotated around the y-axis, one method is to set up the integral
with disks and dy. there is another alternative.
ex) region bounded by y=x2-2x+3, y=0, x=0, x=3
find the volume when:
- rotated around the x-axis
...we know how
-rotated around the y-axis
...why cant we use disks?
ex) find the volume when the region bounded by y=2x2-x3 and y=0 is rotated
around the y-axis
[hint: use shells. why cant we use disks?]
ex) the region bounded by y=x2 & y=0 and x=2 find the volume when
-rotated around the x-axis
-rotated around the y-axis (use dy)
-rotated around the y-axis (use dx)
applications and fancier problems
ex) what is the volume of a sphere with radius R ?
ex) imagine a hemisphere resting on its flat side. where do you make a
horizontal cut that divides it in equal amounts? [set up, do not solve]
hw questions: volume
ex) the region bounded by y=3x-x2 and y=0 is revolved around the x-axis
Arc Length
when is it easy to find the length of a graph?
...when it is a straight line.
length = s = √(x2-x1)2 + (y2-y1)2 OR √(∆x)2+(∆y)2
what if it is not straight? ... approximate!
recall we approximated area with rectangles,
now we approximate length with straight lines
what is the length in the approximation?
s ≈ s1 + s2 + ... + sn = √(∆x1)2+(∆y1)2 + √(∆x2)2+(∆y2)2 + ... + √(∆xn)2+(∆yn)2
= Σ √(∆x)2+(∆y)2
Usually at this point we have a Σ with a ∆x on the end, and they become ∫ and
dx
There is a Σ but no ∆x ... so now what?
...an algebra trick
Σ √(∆x)2+(∆y)2 = Σ √(∆x)2+(∆y)2
= Σ √(∆x)2+(∆y)2 ∆x
=Σ
so,
s ≈ Σ √ 1 + (∆y/∆x)2 ∆x
s = ∫ √ 1 + (dy/dx)2 dx
notation: sometimes we write ds = √ 1 + (dy/dx)2 dx
notation: in higher math, sometimes we write ds2 = dx2 + dy2
note: we introduced ∆x in the algebra trick. we could also have introduced ∆y,
lets try some examples
ex) find the length of the graph, y=(x-1)3/2 1<x<9
ex) find the length of the graph, y= 1/3 x3 +
1<x<2
ex) set up [but do not solve] the length of the graph y=xex for 2<x<4
Surface area of a solid of revolution
this will build from what we know about arc length
first, the approximation
when we rotate a straight line around the axis, what surface area is created?
for a cylinder, it is 2πrL (where r is the radius and L is the length of the line)
if the line is at an angle, you dont get a cylinder, you get something called a
frustum.
the surface area is:
...we will skip some of the details...
2πrL (where r is the average radius and L is the length of the line)
when we approximate the graph with many short straight lines
(just like for arc length), then the first radius and the second radius are
approximately the same, so its just r. also, L is the same as s.
so
SA ≈ Σ 2πr √ 1 + (∆y/∆x)2 ∆x
SA = ∫ 2πr √ 1 + (dy/dx)2 dx
and r = y when rotating around the x-axis, r=x when rotating around the yaxis (draw the picture!)
note: rotating around the x-axis or the y-axis creates the same shape, so the
formula is the same (we do not have the problem of disks & shells)
ex) find the surface area when y=x is rotated around the x-axis, 0<x<4
ex) find the surface area when y=x2 is rotated around the y-axis, 1<x<3
Applications, more advanced
Economics: consumer surplus
suppose i value an item at more than i pay for it. thats great for me.
ex) my value for a shirt is $35, but i can buy them for $27. if i buy 30 shirts,
whats my "benefit"?
(35-27)(30) = $240
this is called the consumer surplus
in a market, different people place a different value on an item. this creates
what is called the consumer demand curve.
p = p(x) where x is the number of items and p is the price it will sell for as
determined by how many people will buy it at that price.
ex) p=100-x
if the market price is $35, how many are sold?
...x=65
if there are 40 items sold, what is the market price?
...p=$60
but note that lots of people would pay more than $60. for example, if x=1,
p=99, so 1 person would pay $99. his surplus is (99-60)(1)=39 ...etc...
call the market price P
the total consumer surplus is Σ (p(x)-P) ∆x
Consumer Surplus = ∫ p(x)-P dx
ex) what is the consumer surplus if p=100-x and 40 items are sold ?
note: p(x)-P is the surplus per item, so it is a rate. integrate it to get the total
surplus.
physics: fluid pressure
some definitions from Physics:
the hydrostatic pressure on an object at a depth h in a liquid with density w is:
P = wgh
where g is the acceleration due to gravity, g=9.8m/sec2
since pressure = force/area, P=F/A, then
F = PA
so F = wghA
ex) a plate with 8 m2 of area is lowered to a depth of 12m in water, which has
density w=1000 kg/m3 What is the pressure on the plate? what is the force?
note that "depth=1" occurs below the x-axis. this means we work with positive
y-values being below the x-axis. we are consistent through the whole
problem, so that is ok.
now the hard case: what if the plate is vertical, so not at the same depth? and
not rectanglular, so varying size?
ex) suppose a vertical rock with a flat face sits in the water. at depth y meters
it has width 0.4y, for 0<y<2. what is the pressure on the rock?
the pressure is constant at the same depth - and a horizontal slice has the
same depth everywhere (a vertical slice does not). so we take a slice ∆y (or
dy). for that slice, the depth is y, the area is length·width = f(y)∆y. the liquid
density is w=1000
the total force is the sum of all the forces on the slices
so
F = wghA
F ≈ Σ 1000·9.8·(y)·(0.4y)∆y
F = ∫02 9800y(0.4y)dy
the general formula is: F = ∫ab w·g·h·f(y)dy
ex) find the total force on a right triangle plate, with the base of 4meters sitting
4meters below the surface and with tip at the water surface [hint: at depth y,
what is the width?]
Application in physics: Work by a gas
imagine storing a gas in a bag. we want to know how much work it the gas
does to make the bag bigger or smaller ...
now,
in the case of a gas, force/area is called pressure
units: Newtons/m2
therefore, force = (pressure)(area)
previously, we learned that W = (force)(distance), so
Work = (force)(distance)
W = (pressure)(area)(distance)
W = (pressure)(volume)
so for a gas, work can be a function of volume, W(V)=PV
take the derivative and you can write:
W'(V) = P or dW = P
dV
from physics, in many situations it is true that (P)(V) is some constant, so: P=k/V
so W'(V) = k/V
W= ∫ k/V dV
(to learn more about the physics, look up the ideal gas law)
ex) for a container of gas, PV=2. recall that dW/dV=P. Find the amount of work
done by the gas as the volume changes from 3 m3 to 7 m3 .
math application: average value of a function
suppose we know the velocity function and we want to find the average
velocity
how could we do this...
...as an approximation: take a velocity reading every so often, then add them
up and divide by n
ex) v(t) = 0.06t2 for 0<t<10
averge velocity = vavg ≈ v(0) + v(2) + v(4) + v(6) + v(8) = v(0) +... + v(8)
10-0/
5
2
where 2 is ∆t. flip it to the numerator and you get:
vavg ≈ 1/b-a Σ v(t) ∆t
vavg = 1/b-a ∫ab v(t) dt
this works for any function, so
average value of f(x) for a<x<b is
favg = 1/b-a ∫ab f(x) dx
OR favg (b-a) = ∫ab f(x) dx
ex) find the average value of v(t)=0.06t2 for 0<t<10 v is m/sec, t is sec
one graphical interpretation:
if you graph the given function, the
integral from a to b is the area.
the rectangle from a to b with the
same area has height favg
another graphical interpretation:
if you treat the given function as a
derivative f ' and graph the slope field,
then for f(x) mark the starting point
x=a,y=f(a) and the ending point
x=b,y=f(b), then the average value f 'avg
Physics: center of mass (or centroid)
the center of mass (or centroid) is the balancing point. (think of people sitting on
a seesaw.) to find this point, we must consider mass, but we also must consider
position. (think of a lighter person balancing a heavier person on the seesaw.) a
system of two weights is in balance when: m1d1 = m2d2 ...where m1 m2 are
masses of each object and d1 d2 are the distances from the balance point x'
if the object is sitting at x1, the distance can be written as x1-x'
since one distance is on the positive side and one is on the negative side, this is
the same as: m1(x1-x') = m2(x'-x2)
(if we assume the origin is the balance point, then we get m1x1 + m2x2 = 0 )
solve for x'...
m1x1 - m1x' = m2x' - m2x2
m1x1 + m2x2 = m1x' + m2x'
x' = m1x1 + m2x2
m1 + m2
what if you have many objects?
x' = Σ mixi
Σ mi
some names:
x' is called the center of mass
M = Σ mixi is called the moment of the system
m = Σ mi is called the mass of the system
question: where is the centroid of a standard meter stick?
...at the half-meter mark. this is just obvious, by symmetry.
now, on to calculus...
where is the centroid of a meter stick that gets heavier (or wider) the closer you
get to the 1m end?
ex) find the centroid for a 1m stick which has width 0.1x at the x-distance
to find x', we need find the moment and the mass
mass = (mass/area)(area) = (density)(area) ... and lets assume that the material
has constant density 2.4 kg/m2
we know how to find area - its an integral
mass = ∫2.4 f(x) dx
mass = m = ∫ w L(x) dx
1
= 2.4 ∫0 0.1x dx
w=density, L=length
= 0.12x2 |01
= 0.12 kg
moment = (mass)(position)
= (density)(area)(position)
= (density)(length)(width)(position)
moment ≈ Σ 2.4x(0.1x)∆x
moment = ∫ 2.4x(0.1x)dx
in general, moment = ∫ wx L(x)dx
w=density, L(x)=length
note: what if the density varies as a function of x?
no problem- just put that in the formula.
note: these calculations find the x-coordinate of the centroid. what if you want to
find the y-coordinate of the centroid?
no problem- just do the calculation as a function of y with dy
(the textbook gives a fancy formula, we will do it this straightforward way)
ex) find the centroid of the region bounded by y=x and y=x2 with density 12
mass= m = ∫01 12 (x-x2)dx = 12(x2 - x3)|01 = 12(1/2 - 1/3 ) = 2 kg
x-moment = ∫ 12x(x-x2)dx
= ∫ 12x2 - 12x3 dx
= ∫ 4x3 - 3x4 dx
= 4(1)3 - 3(1)4 - 0
=1
x-centroid = 1/2
now for y ... x=y and x=√y
y-moment = ∫ 12y(√y-y)dy
= ∫ 12y3/2 - 12y2 dy
= 24/5 y5/2 - 4y3
= 24/5 - 4 - 0
= 4/5
y-centroid = 4/5 /2 = 2/5
centroid = (.5, .4)
hw questions
hw questions
on the exam, you will get this formula sheet:
Techniques of integration
some techniques we know so far:
antiderivative for functions we know from taking the derivative:
∫ x2 dx =
∫ 1/x dx =
∫ ex dx =
∫ sin(x) dx =
∫ cos(x) dx =
∫ sec2x dx =
∫ 1 dx =
1+x2
∫ 1
dx =
2
√1-x
adding and subtracting:
ex) ∫ ex + x2 dx
= ∫ ex dx + ∫ x2 dx = ...
using algebra:
ex) ∫ x(x+1) dx
= ∫ x2 + x dx = ...
substitution:
ex) ∫ esin(x)cos(x) dx
u=sin(x) ... du=cos(x)
u
= ∫e du = ...
ex) ∫ cos(4x)dx
u = 4x ... du = 4dx
= 1/4 ∫ cos(u)du
recall that substitution came from undoing one of the rules of
derivatives- the chain rule
Now, here is our next technique. It also comes from undoing one of the
derivative rules...
Integration by parts
This "undoes" the product rule.
for two expressions, u and v,
(u·v)' = u·v' + v·u'
It is not obvious how to "undo" this, since both sides of the formula have
derivatives. But here's what we do.
On the left side there is a derivative, so we integrate and get the original
function:
∫ (u·v)' dx = ∫ u·v' dx + ∫ v·u' dx
u·v
= ∫ u·v' dx + ∫ v·u' dx
Now, solve for one integral:
∫ u·v' dx = u·v - ∫ v·u' dx
I like using differentials, so I will turn it into that form.
Recall that u' = du/dx so du=u'dx, and dv=v'dx. now the formula is written:
∫u dv = u·v - ∫v du
This tells you that one expression with an integral equals another expression
with an integral. That's only helpful if the new integral is easier to do. As it turns
out, that happens a lot.
ex) ∫ x·cos(x)dx
u=x
dv=cos(x)dx
du = dx v=sin(x)
= uv - ∫v du
= x·sin(x) - ∫ sin(x) dx
= x·sin(x) + cos(x)+ C
check: [x·sin(x) + cos(x)]' = 1·sin(x) +x·cos(x) -sin(x) = x·cos(x) ...yes
same problem, in the other notation:
u=x
v'=cos(x)
u'=1
v=sin(x)
= u·v - ∫v·u' dx
= x·sin(x) - ∫sin(x)·1·dx
= x·sin(x) + cos(x)+ C
(both notations work just fine, use the one you prefer)
ex) ∫ arctan(x) dx
u = arctan(x) dv=dx
du= 1 dx
v=x
1+x2
= u·v - ∫ v du
= x·arctan(x) - ∫x· 1 dx
1+x2
u = 1+x2 ... du=2xdx
- 1/2 ∫ du/u
- 1/2 ln(u)
= x·arctan(x) - 1/2 ln(1+x2) + C
Now, the main question with any technique of integration - when do I use it?
And for this technique, how do i know which part is u and which part is dv ?
It turns out, for integration by parts, both of those questions have the same
answer.
The indicators for when and how to use integration by parts are:
1) there is a piece of the integral we would like to "differentiate away" to make the
integral easier (as in the first example). then set u to be the part you want to
differentiate away
2) we dont know how to do the integral of a function, but we can take the
derivative (as in the second example). then set u to be the part you know how to
take the derivative of.
3) (if using parts) dv should be something you can take the integral of (to get v)
Cheat sheet - common integrals for parts:
∫xneaxdx & ∫xnsin(ax)dx & ∫xncos(ax)dx ... let u=xn
∫xnln(x)dx & ∫xnarcsin(ax)dx & ∫xnarctan(ax)dx ... let dv=xndx
∫eaxsin(bx)dx & ∫eaxcos(bx)dx ... let dv=eaxdx
Repeated integration by parts
ex) ∫ x2sin(x)dx
[why is this a candidate for parts? ...discuss]
dv=sin(x)dx
du=2xdx v=-cos(x)
2
= -x cos(x) - ∫ -cos(x)2xdx
now what? ...
...do it again
u = 2x
dv=-cos(x)dx
du = 2dx v=-sin(x)
2
= -x cos(x) - [-2xsin(x) - ∫ -2sin(x)dx]
= -x2cos(x) + 2xsin(x) + 2cos(x) + C
u=x2
Shortcut for repeating integration by parts - the table method
ex) ∫ x2sin(x)dx
sign | u & deriv | v' & antideriv
+
x2
sin(x)
2x
-cos(x)
+
2
-sin(x)
0
cos(x)
2
= -x cos(x) + 2xsin(x) + 2cos(x) + C
ex) ∫ x5cos(x)dx
[why is this a candidate for parts?]
Using repeating to make an equation
ex) ∫ e2xsin(x)dx
u=e2x
dv=sin(x)dx
2x
du=2e dx v= -cos(x)
2x
= -e cos(x) - ∫ -cos(x)2e2xdx
u=e2x
dv=cos(x)dx
2x
du=2e
v=sin(x)
2x
2x
=-e cos(x) + 2[e sin(x) - ∫ 2sin(x)e2xdx ]
as one equation... ∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x) - 4∫e2xsin(x)dx
and now...
5∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x)
so...
∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x)
5
note: always pick u as the same thing. if you make one thing u and then in
the next step dv, you will wind up with what you started with - no progress.
note: you cannot use the table. it doesnt give an equation, or an integral.
Integrating basic trig functions
There are 6 trig functions. We know the integral for 2 of them. What about the
other 4?
∫ tan(x) dx
= ∫ sin(x) dx
cos(x)
u=cos(x) ... du=-sin(x)dx
du
=-∫ /u
= -ln(cos x) OR ln(cos x)-1 OR ln(sec x) +C
∫ sec(x)dx
= ∫ sec(x) sec(x)+tan(x) dx
sec(x)+tan(x)
2
= ∫ sec (x) + sec(x)tan(x) dx
sec(x)+tan(x)
u = sec(x)+tan(x) ... du= sec(x)tan(x)+sec2x dx
= ∫ du/u
= ln(u)
= ln(sec(x)+tan(x)) +C
you do (hint: use the same methods)
ex) ∫ cot(x) dx
ex) ∫ csc(x) dx
Techniques of Integration - combined trig functions
How do we do an integral which combines several trig functions, like
ex) ∫ sin3x cos4x dx
We use algebra tricks to do many of these integrals. First, here are the formulas
we will need. Then we will look at how to use them in some exercises.
trig:
+ sin2x = 1
2
tan x + 1 = sec2x
1 + cot2x = csc2x
cos2x
cos2x = 1 + cos(2x)
2
sin2x = 1 - cos(2x)
2
sin(A)cos(B) = 1/2 [sin(A-B) + sin(A+B)]
sin(A) sin(B) = 1/2 [cos(A-B) - cos(A+B)]
cos(A)cos(B) = 1/2 [cos(A-B) + cos(A+B)]
derivatives:
(sin x)' = cos x
(cos x)' = sin x
(tan x)' = sec2x
(sec x)' = sec(x)tan(x)
(cot x)' = -csc2x
(csc x)' = -csc(x)cot(x)
ex) ∫ sin3x cos4x dx
To solve this, we will use substitution. But first, some manipulations.
= ∫ sin2x cos4x sin(x)dx
= ∫ (1-cos2x)cos4x sin(x)dx
= ∫ (cos4x - cos6x) sin(x)dx
...and now...
Replacing sin2(x) with 1-cos2(x) gave us u, and putting sin(x) with dx gave us du
u = cos(x) ... du = -sin(x)dx
4
= -∫u - u6 du
= -u5 + u7
5
7
5
= -cos x + cos7x +C
5
7
notation: you can also write (cos x)5
ex) ∫ sin2x cos5x dx
What do you think will be the u? du?
= ∫ sin2x cos4x cos(x)dx
= ∫ sin2x (1-sin2x)2 cos(x)dx
= ∫ sin2x (1 - 2sin2x + sin4x) cos(x)dx
= ∫ (sin2x - 2sin4x + sin6x) cos(x)dx
u=sin(x) ... du=cos(x)dx
= ∫ u2 - 2u4 + u6 du
= u3 - 2u5 + u7
3 5
7
3
5
=sin x - sin x + sin7x +C
3
5
7
...and now...
Notice that we need an odd power so that when we pull out a sin(u) or cos(u) there
is a cos2u or sin2u, or a power of it, to replace.
So, what if both powers of sin and cos are even? We use a different formula.
ex) ∫ sin2x cos2x dx
= ∫ 1 - cos(2x) 1 + cos(2x) dx
2
2
= 1/4 ∫ 1 - cos2(2x) dx
= 1/4 ∫ 1 - 1+cos(4x) dx
2
1
1
1
= /4 ∫ /2 - /2 cos(4x) dx
= 1/8 ∫ dx - 1/8 ∫ cos(4x) dx
u = 4x ... du = 4dx
1
1
1
= /8 x - /8 ∫ /4 cos(u) du
= 1/8 x - 1/32 sin(4x) +C
Replace using a cos(2u) formula until you get something you can integrate.
We can do tricks like these with other trig functions as well.
ex) ∫ sec4x tan2x dx
We need to figure out - what will be u? du? Which function gets replaced?
= ∫ sec2x tan2x sec2x dx
= ∫ (tan2x + 1)tan2x sec2x dx
= ∫ (tan4x + tan2x) sec2x dx ...and now...
u=tan(x) ... du=sec2x dx
= ∫ u4 + u2 du
= u5 + u3
5
3
5
=tan x + tan3x +C
5
3
Notice that we need an even power of sec(u) to pull out a sec2(u) then have an
even power of sec(u) left over to replace with a power of tan2(u)+1
ex) ∫ sec3x tan3x dx
= ∫ sec2x tan2x sec(x)tan(x)dx
= ∫ sec2x (sec2x - 1) sec(x)tan(x)dx
= ∫ (sec4x - sec2x) sec(x)tan(x)dx ...and now...
u=sec(x) ... du=sec(x)tan(x)
= ∫ u4 - u2 du
= u5 - u3
5 3
= sec5x - sec3x +C
5
3
Notice that we need an odd power of tan(x) to pull out a sec(x)tan(x) then have
an even power of tan(x) left over to replace with a power of sec2x-1
In other cases, we need to use various techniques.
ex) ∫ sec3x dx
u = sec(x)
dv = sec2x dx
du = sec(x)tan(x)dx v = tan(x)
= sec(x)tan(x) - ∫ tan2(x)sec(x)dx
=
- ∫ (sec2x - 1)sec(x)dx
=
- ∫ sec3x - sec(x) dx
so ∫ sec3x dx = sec(x)tan(x) - ∫ sec3x dx + ∫ sec(x) dx
2∫ sec3x dx = sec(x)tan(x) + ln|sec(x)+tan(x)|
∫ sec3x dx = 1/2 (sec(x)tan(x) + ln|sec(x)+tan(x)|) +C
ex) ∫ tan2x dx
= ∫ sec2x - 1 dx
= tan(x) - x +C
The formulas relating csc and cot are the same as the formulas relating sec and tan
(but maybe with a negative sign), so the tricks are the same.
ex) ∫ csc4x cot4x dx
= ∫ csc2x cot4x csc2x dx
= ∫ (1+cot2x) cot4x csc2x dx
= ∫ (cot4x + cot6x) csc2x dx
u=cot(x) ... du = -csc2x dx
= -∫ u4 + u6 du
= -u5 - u7
5
7
5
= -cot x - cot7x +C
5
7
One more type of trig integral - mixed angles with sin and cos
ex) ∫ sin(5x)cos(2x)dx
= ∫ 1/2 [sin(3x) + sin(7x)] dx
= 1/2 [-cos(3x) - cos(7x)]
3
7
= -cos(3x) - cos(7x) +C
6
14
Here is one more technique: convert to sin and cos and hope for the best
ex) ∫ sec(x) dx
tan2x
=∫ 1
cos2x dx
cos(x) sin2x
= ∫ cos(x) dx
sin2x
u=sin(x) ... du = cos(x)dx
= ∫ u-2 du
= -u-1
= -[sin(x)]-1 OR -1
OR -csc(x)
+C
sin(x)
Integrals with combined trig functions
The Rules:
product of sin and cos &
power of sin is odd
factor out sin(x), replace every sin2x with 1-cos2x, u=cos(x)
power of cos is odd
factor out cos(x), replace every cos2(x) with 1-sin2(x), u=sin(x)
power of sin, cos both even
replace sin2u and cos2u using "cos(2u)" formulas
product of sec and tan &
power of sec is even
factor out sec2x, replace every sec2x with tan2x +1, u=tan(x)
power of tan is odd
factor sec(x)tan(x), replace every tan2x with sec2x -1, u=sec(x)
tan by itself, even power
turn only one tan2x into sec2x -1, multiply, solve one integral, repeat as needed
sec by itself, odd power
by parts, dv=sec2x dx
product of csc and cot - same as rules for sec and tan
otherwise, try replacing with sin and cos
Techniques of integration - partial fractions
How do we calculate ∫ x+5 dx ?
x2+x-2
It would help a lot to know that x+5 = 2 - 1
x2+x-2 x-1 x+2
= ∫ 2 - 1 dx
x-1 x+2
= 2 ln(x-1) - ln(x+2) +C
because now we get
So the main trick here involves algebra. How do we convert the complicated
rational function into several simple fractions?
ex)
x+5
2
x +x-2
= x+5 = ? + ?
(x+2)(x-1) x+2
x-1
We can assume that the numerator has a lower degree than the
denominator. (We will see why in the next problem.) Since denominator is
degree 1, we make the numerator with degree 0 (a constant).
So x+5
= x+5 = A + B
2
x +x-2
(x+2)(x-1) x+2
x-1
Now, multiply both sides by the original denominator. After cancellation,
x+5 = A(x-1) + B(x+2)
x+5 = Ax - A + Bx + 2B
x+5 = (A+B)x + (-A+2B)
Therefore, by matching coefficients (for x and for the constant),
x:
1 = A+B
const: 5 = -A+2B
Solve it using algebra...
Add equations to get 6=3B ... 2=B ... from equation#1, 1=A+2 ... A=-1
So x+5 = -1 + 2
...and now we can calculate the integral
x2+x-2
x+2
x-1
Here is a shortcut for solving...
x+5 = A(x-1) + B(x+2)
notice that the roots are x=1, -2
plug in x=1 to get 6=0+3B ... B=2
plug in x=-2 to get 3=-3A+0 ... A=-1
Note: this shortcut will work completely only with distinct linear factors, not
with the harder problems we will see later.
If the polynomial in the numerator had degree equal to or greater than the
denominator, you can divide to get (polynomial) + (fraction). This allows us to
assume that in every fraction we make, the denominator has a higher degree
than the numerator.
ex) ∫ x3 + x dx
x-1
x-1 | x3
+x
3
2
x -x
x2 + x
x2 - x
2x
2x-2
2
= ∫ x2 + x + 2 + 2 dx
x-1
= x3 + x2 + 2x + 2 ln(x-1) +C
3
2
Mental shortcut: to integrate the last term, you have to do a substitution. But
since u=x-1 and du=dx, we skipped writing it down. If there is a constant to keep
track of, make sure you keep track of it!
Slightly harder cases
What if, when the denominator factors, one factor is raised to a power?
ex) ∫ 4x
dx
2
(x-1) (x+1)
4x
= A + B + C
2
(x-1) (x+1) x-1 (x-1)2 x+1
...For that factor, each power ()1, ()2,... gets its own fraction, and the numerator
is a constant. You can prove that this is true using algebra. We will skip that.
We will simply see that it always works.
4x = A(x-1)(x+1) + B(x+1) + C(x-1)2
4x = Ax2 - A + Bx + B + Cx2 -2Cx + C
x2: 0 = A+C
x1: 4 = B-2C
x0: 0 = -A+B+C
Solve with algebra...
#1 + #3 gives 0=B+2C ... add that to #2 to get 4=2B ... 2 =B ... 4=2-2C, so C=-1
... 0 = A-1, so A=1
= ∫ 1 + 2 - 1 dx
x-1 (x-1)2 x+1
= ln(x-1) - 2(x-1)-1 - ln(x+1) +C
Note: what about the shortcut? You could plug in x=1 and get 4=2B. you could
plug in x=-1 and get -4=4C. You still need coefficient equation to find A.
Another tricky case - what if a factor is quadratic?
ex) ∫ 2x2 - x + 1 dx
x3 + x
= ∫ 2x2 - x + 1 dx
x(x2+1)
Now,
2x2 - x + 1 = A + Bx+C
x(x2+1)
x x2+1
So: if the denominator has degree 2, then the numerator can have degree 1
2x2 - x + 1 = A(x2+1) + (Bx+C)x
2x2 - x + 1 = Ax2 + A + Bx2 + Cx
x2: 2 = A + B
x1: -1 = C
x0: 1 = A
so C=-1, A=1, and then B=1
= ∫ 1 + x-1 dx
x x2+1
= ∫ 1 + x - 1 dx
x x2+1 x2+1
= ln(x) + 1/2 ln(x2+1) - arctan(x) +C
Note: what about the shortcut? There is only one root. You can plug in x=0
to get 1=A. You need coefficient equations for the rest.
What about repeated non-linear factors?
ex) ∫ x2 +x-3 dx
x(x2+6)2
Set up:
x2+x-3 = Ax+B + Cx+D + E
x(x2+6)2 x2+6
(x2+6)2
x
Multiply both sides by the original denominator, and cancel, to get:
x2+x-3 = (Ax+B)(x2+6)x + (Cx+D)x + E(x2+6)2
...then finish by more of the same calculations as before...
Integration by partial fractions (summary)
Indicator: the integral of a rational function (polynomial over polynomial)
Steps:
1 if the numerator has degree equal or higher to the denominator, divide
2 factor the denominator
3 set up fractions
- distinct linear factors OR repeated factors OR non-linear factors
OR repeated non-linear factors
4 multiply both sides by the original denominator and solve for the unknowns
- with distinct linear denominators, use shortcut of plugging in each root for x
- otherwise, match coefficients on each side and solve system of equations
5 rewrite original integral as sum of simpler fractions and integrate each
A comment: do you always integrate a polynomial over a polynomial this way?
ex) ∫ 3x2+3 dx
x3+3x-1
No! We are learning many techniques. To calculate integrals by hand, one big
challenge - probably the biggest - is to decide which technique to try first.
After we learn all of our techniques, we will talk more about these strategies.
Integration by substitution with trig functions ("trig sub")
There are integrals where we make a substitution where u="trig function"
ex) ∫ √9-x2 dx
x2
We can turn the expression under the radical into something we can take
the square root of - if we make a clever substitution:
x = 3sin(t) ... dx = 3cos(t) dt
= ∫ √9-(3sin(t))2 3cos(t) dt
(3sin(t))2
= ∫ √9-9sin2t 3cos(t) dt
9sin2t
= ∫ √9(1-sin2t) 3cos(t) dt
9sin2t
= ∫ 3√cos2t 3cos(t) dt
9sin2t
= ∫ 3cos(t) 3cos(t) dt
9sin2t
= ∫ 9cos2t dt
9sin2t
= ∫ cot2t dt
= ∫ csc2t - 1 dt
= -cot(t) - t +C
We must convert back to an expression in x ... how do we replace t ?
x = sin(t), so arcsin x = t
3
3
What about cot(t) ? Here we need the trig triangle
So cot(t) = √9-x2
x
= √9-x2 - arcsin x
+C
x
3
The substitution of sin(t) worked because √1-sin2t = cos(t)
note: we needed to properly choose the constant so it would factor out
note: the radical from the original integral will often reappear in the answer
We can apply a similar trick to other integrals.
ex) ∫
dx
√16+x2
What trig identity does this look like? What substitution should we use?
x=4tan(t) ... dx = 4sec2t dt
= ∫ 4sec2t
dt
√16+(4tan(t))2
= ∫ 4sec2t
dt
√16+16tan2(t)
= ∫ 4sec2t
dt
2
√16(1+tan (t))
= ∫ 4sec2t dt
4√sec2t
= ∫ sec(t) dt
= ln(sec(t)+tan(t)) +C
We must rewrite in terms of x ... so we need the trig triangle
So tan(t)=x and sec(t)= √16+x2
4
4
= ln( √16+x2 + x ) +C
4
4
note: the substitution of tan(t) worked because √1+tan2t = sec(t)
ex) ∫ √x2 - 1 dx
x
What trig identity does this look like? What substitution should we use?
x = sec(t) ... dx = sec(t) tan(t) dt
= ∫ √sec2t - 1 sec(t) tan(t) dt
sec(t)
= ∫ √tan2t tan(t) dt
= ∫ tan2(t) dt
= ∫ sec2t - 1 dt
= tan(t) - t
We must rewrite in terms of x ... so we need the trig triangle
so t = arcsec(x) and tan(t) = √x2 - 1
(Don't like to use arcsec? Note that t =arccos( 1/x )
= √x2 - 1 - arcsec(x) +C
note: the substitution of sec(t) worked because √sec2t - 1=tan(t)
Trig Sub - Integral of an expression with a rational exponent
ex) ∫
dx
(x2+1)3/2
Why is this a candidate for trig sub? And which one?
The rational exponent gives us a square root, and inside it there is an x2.
x = tan(t) ... dx = sec2t dt
= ∫ sec2t
dt
2
3/2
(tan t + 1)
= ∫ sec2t
dt
2
3/2
(sec t)
= ∫ sec2t dt
sec3t
= ∫ 1 dt
sec(t)
= ∫ cos(t) dt
= sin(t) +C
to rewrite in terms of x, use the trig triangle
=
x
+C
2
√x +1
Trig sub - completing the square
ex) calculate ∫
dx
2
√x +10x+26
look at the denominator ... x2+10x+26 = x2+10x+25 + 1
so one algebra step gives:
= ∫ dx
√(x+5)2 + 1
u=x+5 ... du=dx
= ∫ du
√u2+1
...and we know how to handle this with a trig sub, u=tan(t)
ex) ∫
x
dx
√3-2x-x2
It is a candidate for trig sub because there is a square root with x2 inside. But
it requires some algebra to put it in the proper form - complete the square.
=∫ x
dx
2
√3-(x +2x)
=∫ x
dx
2
√3-(x +2x+1)+1
=∫ x
dx
2
√4-(x+1)
u=x+1 ... du=dx ... also, x=u-1
= ∫ u-1 du
√4-u2
u=2sin(t) ... du=2cos(t) dt
= ∫ 2sin(t)-1
2cos(t) dt
2
√4 - (2sin t)
= ∫ 2sin(t) - 1 2cos(t) dt
√4(1-sin2t)
= ∫ 2sin(t) - 1 2cos(t) dt
2cos(t)
= ∫ 2sin(t) - 1 dt
= -2cos(t) - t +C
to rewrite in u, use the trig triangle
= -√4-u2 - arcsin(u/2) +C
and rewite in x
= -√4-(x+1)2 - arcsin(x+1/2) OR -√3-2x-x2 - arcsin(x+1/2) +C
Trig Substitution
Indicator: a square root with a quadratic inside it
Rules:
if necessary, first complete the square and do a substitution
when you have:
√a2 - x2
√a2 + x2
√x2 - a2
substitute:
x=a sin(t)
x=a tan(t)
x=a sec(t)
use the identity:
1-sin2t = cos2t
1+tan2t = sec2t
sec2t - 1 = tan2t
(note: also applies to rational exponents with a 2 in the denominator)
near the end, to undo the substitution, you probably will use the trig triangle
basic inverse trig integrals (which you can prove using trig substitution):
What about this old problem:
ex) ∫ √9-x2 dx
We know how to do this integral if the boudaries are -3 or 0 or 3, since then it
represents the area of one or two quarters of a circle. But what if the boundaries
are something else? or there are no boundaries?
We make the integral into something we can integrate with the substitution:
x=3sin(t) ... dx=3cos(t) dt
= ∫ √9-(3sin(t))2 3cos(t) dt
= ∫ √9-9sin2t 3cos(t) dt
= ∫ √9(1-sin2t) 3cos(t) dt
= ∫ 3√1-sin2t 3cos(t) dt
= ∫ 3√cos2t 3cos(t) dt
= ∫ 3cos(t) 3cos(t) dt
= 9 ∫ cos2t dt
= 9 ∫ 1/2 (1 + cos(2t)) dt
= 4.5 (t + .5sin(2t) )
We need to replace t with an expression in x ... but what?
x = sin(t), so arcsin x = t
3
3
We know how to handle trig functions of t, but what do we do with sin(2t)?
...we use the trig identity sin(2t)=2sin(t)cos(t)
from the triangle, sin(t)= x and cos(t) = √9-x2
3
3
= 4.5arcsin(x/3) + 2.25 · 2 · x · √9-x2
3
3
If the boundaries were 0<x<3, its a quarter-circle, and geometry tells us we
should get .25π(3)2=2.25π
= 4.5arcsin(3/3) + 2.25 · 2 · 3 · √9-32 - 4.5arcsin(0/3) + 2.25 · 2 · 0 · √9-02
3
3
3
3
= 4.5 (.5π) + 0 - (0+0)
= 2.25π
A problem that looks harder, but isn't
ex) ∫
x3
dx
2
3/2
(4x +9)
x = 3/2 tan(t) ... dx = 3/2 sec2t dt
= ∫ (3/2 tan t)3 · 3/2 sec2t dt
(4(3/2 tan(t))2 + 9)3/2
= 81/16 ∫ tan3t sec2t
dt
2
3/2
(9tan (t) + 9)
81
= /16 ∫ tan3t sec2t
dt
3/2
2
3/2
9 (tan (t)+1)
= 81/16 ∫ tan3t sec2t dt
27sec3t
= 3/16 ∫ tan3t dt
sec(t)
= 3/16 ∫ sin3t dt
cos2t
= 3/16 ∫ cos-2t sin2t sin(t) dt
= 3/16 ∫ cos-2t (1-cos2t) sin(t) dt
= 3/16 ∫ (cos-2t - 1) sin(t) dt
u = cos(t) ... du = -sin(t)dt
3
-2
= - /16 ∫ u - 1 du
= -3/16 (-u-1 - u)
= -3/16 (-cos-1t - cos(t) )
to rewrite in terms of x, we need the trig triangle
)
= 3/16 ( √4x2+9 - 3
2
3
√4x +9
2
OR √4x +9 - 9
+C
2
16
16√4x +9
General very useful tip for integration:
ex) ∫ 1 dx ... this looks like ____ so substitute u=_____
4-3x
you can always get rid of a linear expression in x with a substitution
ex) ∫ 2 dx ... this looks like _____ so substitute u=_____
1+4x2
Techniques of integration using algebra
- expand
ex) ∫ x(x+1)dx
= ∫ x2+x dx = ...
-complete the square
ex) ∫ 1
x2 + 2x + 2
=∫ 1
dx
2
(x+1) +1
= arctan(x+1) +C
ex) ∫ 1
√2x - x2
=∫ 1
dx = ∫ 1
dx = arcsin(x-1)
√1-1+2x-x2
√1-(x-1)2
- break up a fraction
ex) ∫ 1 + 2x dx
1+x2
= ∫ 1 dx + ∫ 2x dx
1+x2
1+x2
= arctan(x) + ln(1+x2)
- add and subtract in a fraction numerator (then split up the fraction)
ex) ∫ 2x
dx
2
x +2x+1
= ∫ 2x + 2 - 2 dx = ∫ 2x+2 dx - ∫ 2 dx
x2+2x+1
x2+2x+1
(x+1)2
= ln(x2+2x+1) - -2(x+1)-1
An adventure in integration - an example to show the tricky side
ex) Find ∫ sec3x dx
Techniques of integration - an overview of strategy
the number one rule - look for the simplest technique first
1 its a known formula
2 substitution
3 standard form for one of the techniques - trig sub, partial fractions, parts,
trig powers
4 something harder - an algebra trick, more than one technique, etc
Assume that every integral is easy, until easy methods dont work.
Then assume that the intergal is medium, until that doesnt work.
Then its hard. you can presume that, in the context of a class, you will not be
asked to do too many hard integrals in one assignment.
On the review sheet are integrals that (1) require many different techniques
(2) show how certain integrals can be solved in more than one way (3) show
how similar looking integrals require different techniques.
In addition to the review problems, let's try this exercise. take a
straightforward integral, then make it harder through a substitution.
ex) we know that ∫ cos(x) dx = sin(x) +C
replace x with something... try et ... then dx=etdt
now, ∫ cos(et) et dt = sin(et) +C
ex) we can solve ∫ ln(x) 1/x dx
u=ln(x) ... du= 1/x dx
= ∫ u du
= u2 +C
2
=(ln x)2 +C
2
now, replace x...
step 1: pick an integral we know how to do
step 2: replace x with an expression in t
This is a great way to study techniques of integration. It trains you to see,
when there is a complicated integral, the simpler integral hiding inside.
Improper Integrals
or, what to do when infinity appears somewhere in the integral
first, infinite x-value
∞
ex) ∫ dx
x2
After you do the integral, what does it mean to plug in ∞ ?
Or, to intepret this graphically,
· if f '(x)=x2, and assume f(1)=0, what is the value of f(x) at ∞ ?
· if f(x)=x2, what does it mean to measure the area for 1<x<∞ in other words,
under an infinitely long curve?
We want to know what happens "eventually" - the target value as that
endpoint approaches ∞. How do we say that mathematically?
...with a limit
ex) ∫∞ dx
x2
= limb→∞ ∫b x-2 dx
= limb→∞ [-x-1]b
= limb→∞ -b-1 - -(1)-1
= limb→∞ 1 - 1/b
=1
So, anytime an integral has ±∞ as a boundary,
1 set it up as a limit
2 do the integral
3 evaluate the limit
If the limit exists, the integral is defined, or converges. If the limit does not
exist, the integral in undefined, or diverges.
ex) ∫-∞ ex dx
= lima→-∞ ∫a ex dx
= lima→-∞ [ex]a
= lima→-∞ e0 - ea
=1-0
=1
ex) ∫∞ cos(x)dx
= limb→∞ ∫b cos(x)dx
= limb→∞ [sin(x)]b
= limb→∞ sin(b) - sin(0)
the limit is undefined, so the integral diverges
If both boundaries are infinite, you must take a limit for each one.
ex) ∫ 1 dx
1+x2
= lima→-∞ limb→∞ ∫a 1 dx
1+x2
= lima→-∞ limb→∞ [arctan(x)]a
= [limb→∞ arctan(b)] - [lima→-∞ arctan(a)]
= π/2 - -π/2
=π
note: this notation is different from standard textbook notation,
but it is equivalent
Improper integral - infinite function value
What if you want to integrate and, at one boundary, the function is infinite?
...take a limit
ex) ∫0 1 dx
√x
the function is undefined, and infinite, at x=0. So take a limit there.
= lima→0+ ∫a 1 dx
√x
= lima→0+ ∫a x-1/2dx
= lima→0+ [2x1/2]a
= lima→0+ 2(1)1/2 - 2a1/2
=2
note: to be precise, since the undefined x-value is a boundary, this is a onesided limit (thats what the + means). However, that detail will not affect the
outcome of problems we look at in this course.
What if the undefined x-value is in the middle of your interval?
...you must split the integral into two integrals so that the first integral ends at
the problem value, and the second integral begins there. Then each integral
must be evaluated with a limit.
ex) ∫-1 x-1/3 dx
= ∫-1 x-1/3 dx + ∫8 x-1/3 dx
= limb→0 ∫-1 x-1/3 dx + lima→0 ∫8 x-1/3 dx
= limb→0 [3/2 x2/3]-1 + lima→0 [3/2 x2/3]8
= limb→0- [3/2 (b)2/3 - 3/2 (-1)2/3] + lima→0+ [3/2 (8)2/3 - 3/2 (a)2/3]
= (0 - 3/2 ) + (6 - 0)
= 4.5
Lets revisit an integral we saw previously...
ex) ∫-3 x-2 dx
= ∫-3 x-2 dx + ∫2 x-2 dx
= limb→0 ∫-3 x-2 dx + lima→0 ∫2 x-2 dx
= limb→0 [-x-1]-3 + lima→0 [-x-1]2
= limb→0- [-(b)-1 - -(3)-1] + lima→0+ [-(2)-1 - -(a)-1]
both limits are undefined, so the integral diverges
the Integral as a Limit
recall: we discussed how to describe the derivative as a limit
ex) f '(a) = lim∆x→0 f(a+∆x) - f(a)
OR
f '(a) = limx→a f(x) - f(a)
∆x
x-a
now, we will do the same thing with the integral. First, the basic idea.
F(x) - F(a) ≈ Σ f(xi)·∆x
where F'(x) = f(x)
to make this an exact equality, and do it mathematically, we use a limit:
F(x) - F(a) = lim∆x→0 Σ f(xi)·∆x
instead of writing "lim..." every time, we call that the integral ∫
lim∆x→0 Σ f(xi)·∆x = ∫ f(x) dx
ex) lim∆x→0 Σ (xi)2 ·∆x ...rewrite as an integral
=
note: the xi are x-values, x1 x2 x3 ... and they are spaced ∆x apart
ex) lim∆x→0 Σ sin(xi) ·∆x ...rewrite as an integral
ex) ∫ ex dx ...rewrite as a limit
But what is ∆x ? What is the "i" in xi ?
Now we will be a little more precise. recall:
ex) for f(x)=x2+x on 1<x<4, find L6 (the area approximation with 6 rectangles)
Area ≈ L6 = f(1)·.5 + f(1.5)·.5+f(2)·.5+f(2.5)·.5+f(3)·.5+f(3.5)·.5 = ...
where did each of those numbers come from?
.5 ... 4-1 / 6
1 ... initial x-value, call it x0
1.5 ... (initial x-value) + ∆x, call it x1
2 ... x2 = (previous x-value) + ∆x
2.5 ... x3 = (previous x-value) + ∆x OR (initial x-value) + ____
3 ... x4 = (initial x-value) + 4·∆x
3.5 ... x5 = (initial x-value) + 5·∆x
Lets turn this into a general formula for Ln of f(x) when a<x<b
∆x =
xi =
also, in the sum, i starts at ___ and ends at ___
so Ln = Σi=0 f(xi)·∆x
Ln = Σi=0 f(a+i· )·( )
note: you could instead use Rn - then you start with x1 so i goes from 1 to n.
ex) rewrite ∫2 x3 dx as:
a) L5 approximation
b) Ln approximation
c) the exact value using a limit
a- ∫2 x3 dx ≈ L5 = Σi=0 (2+i(1/5))3 (1/5)
b- ∫2 x3 dx ≈ Ln = Σi=0 (2+i(1/n))3 (1/n)
c- ∫2 x3 dx = limn→∞Σi=0 (2+i(1/n))3 (1/n)
We know how to find (a). It turns out, we can also calculate (b) and (c).
Ln = Σi=0 (2+i(1/n))3 (1/n)
= Σi=0
Ln
= Σi=0
separate into four sums
pull out common factors, which is anything without an i
Ln
=
To evaluate this, we need some special sum formulas:
Ln = 8 . (n-1) + 12 . n(n-1) + 6 . n(n-1)(2n-1) + 1 . n2(n-1)2
n
n2
2
n3
6
n4 4
Ln = 8n-8 + 6n2-6n + 2n3-3n2+n + n4-2n3+n2
n
n2
n3
4n4
to find the integral exactly, take the limit
∫2 x3 dx = lim 8n-8 + 6n2-6n + 2n3-3n2+n + n4-2n3+n2
n→∞
n
n2
n3
4n4
since n→∞, find each limit by looking at the lead coefficients
= 8 + 6 + 2 + 1/4
= 16.25
Is that what we think it should be?
∫2 x3 dx =
= 81/4 - 16/4
= 16.25
ex) calculate ∫4 2x-1 dx (a) as the limit of a sum (b) using integration
(a) Ln = [2(4+i(3/n)) - 1] (3/n)
Ln
=
(7 + 6i/n)(3/n)
=
21 + 18i
n
n2
= 21(n-1) + 18 n(n-1)
n
n2 2
= 21n - 21 + 18n2 - 18n
n
2n2
∫4 2x-1 dx = limn→∞Ln
= lim 21n - 21 + 18n2 - 18n
n→∞
n
2n2
= 21 + 9
= 30
(b) ∫4 2x-1 dx = [x2 - x]4
= (72-7) - (42-4)
= 30
ex) find lim
n→∞
long way: calculate the limit
short way: convert to an integral
a=
b=
f(x)=
so the limit = ∫1 x5 dx
= x6 4
6 1
= 46 - 16
6 6
= 4095/6 OR 682.5
The tools of calculus are very powerful, and make a problem that would
otherwise be very long and tedious into something short and easy.
Things you will not be asked on the final:
y-coordinate for centroid
calculate an algebraically difficult integral for arc length
"integral as a limit" problem where you calculate the sum (so, no sum formulas)