CALCULUS 2LECTURE NOTES

1 ... title page2 ... some precalculus4 ... review of the derivative13... what is integration13... antiderivatives 15... Euler approximation17... Fundamental Theorem of Calculus22... integral as area28... applications- position/velocity/acceleration, economics, electricity31... application- work in physics32... applications (set up, do not solve)37... techniques of integration- algebra38... techniques of integration- substitution42... area between curves46... volume56... arclength58... surface area60... application- consumer surplus in economics61... application- fluid pressure & force62... application- work applied to a gas63... average value64... application- center of mass (centroid)69... integration by parts72... integration of basic trig functions73... integration of combined trig functions78... integration by partial fractions82... integration by substitution with trig functions ("trig substitution")90... other techniques of integration92... techniques of integration, overview93... improper integrals96... the integal as a limit

First, an important topic you may or may not have seen before.

Four ways to represent a mathematical situationex) 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,ynumerical 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 representationwhat 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 interestyears 0 1 2 5 10 20value 100010501102.50 1276.28 1628.89 2653.29what type of representation is this?

numericalwhat 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 listoften best to describe a real world situation

" graphical?

if the information you want can be found visuallye.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 pointthe 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) = 3f '(x)=0

rule: f(x) = k ... f '(x) = 0 (for a constant k)

f(x) = 3x+2f '(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 xf '(x) = 1/x

f(x) = log3xuse 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' = 1

sec2yy' = [cos(y)]2y' =

y' =

also:y = arcsin(x)

y' =

(all six are in the text)

y=y(x) use the chain rule becausey is your inside function

derivative rules for complicated functions

the product rule for derivativesrule: (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 = fgy' = (f)'g - f(g)'

(g)2

ex) y = ex

x2

y' =

ex) y = cos xx2 + 1

y' =

ex) y = sinxx2ex

y' =

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 derivativeex) rise/run OR y2-y1 / x2-x1

for a linear graph, we know how to "see" the derivativeex) 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)

x2+.19f '(1) = 2(1) = 2 ≈ 1.6807 -> slope 1.6807

12+.19 1.19f(x) ≈ L(x) = .1740 + 1.6807(x-1)f(1.5) ≈ L(1.5) = .1740 + 1.6807(1.5-1) = 1.0144in 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) = 1.8547in fact, f(2) = ln(22+.19) = 1.4327

you know what? ... we can make a better approximation ... how?

f '(1.5) = 2(1.5) = 1.22951.52+.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 Approxi mations, Antiderivatives

ex) suppose f '(x) = 2xwhat can we figure out ?

symbolically: numerically: graphically: verbally:what is f(x) ? what is f(3) ? draw f(x) if the velocity is v = 2x

how far have you goneafter x sec?

Symbolic :we know f(x) = x2

right?well, sort ofit could be x2... what else could it be?it could be x2 +1

the full answer is f(x) = x2 + Cwhere C is some unknown constant

now, what if i told you that f(1)=2can you find the constant?

2 = 12 + C1 = Cso...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 conditionwhen 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 formulasgiven the derivative, find the function

ex) f '(x) = 2xf(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)ex ex

1/x or x-1 ln(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

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 3each step is .5

f(3) ≈ f(1) + f'(1) (1.5-1) + ?f '(1.5) (2-1.5) + ?

f'(2) (2.5-2) + ?f'(2.5) (3-2.5)

= 2 + 2(1) (.5) + 2(1.5) (.5) + 2(2) (.5) + 2(2.5) (.5)= 2 + 1 + 1.5 + 2 + 2.5= 9

graphical:

Euler approximationgiven 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 stepswith 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 = 2xwhat 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 fieldwe 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 ∆xthe 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 exactand we get new fancy notation:

f(x) - f(a) = ∫a f '(x) dxwhere ∆x "becomes" dx and Σ "becomes" ∫ ("integral")

oooh, almosttheres a problem with the notation: x serves two roles: as the variable in the f ', and as the partner value for aheres how we fix the problem: we say f ' is a function of tf(x) - f(a) = ∫a f '(t) dt

this is the Fundamental Theorem of Calculus #1now let x be a particular value, say x=b. this givesf(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 issueex) f '(x) = x3+3 ... f(2)=4 ... find f(x) ... find f(5)

Antiderivatives and the FTC - math examplesex) 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 rolesg(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) also, if f(1)=1, find f(x)f(x) = ∫x3 dx = x4 + C

4 f(x) = x4 /4 + C1=f(1) = 14 /4 + C 1 = 1/4 + CC = 3/4 f(x) = x4 /4 + 3/4

ex) ∫1 1/x dx "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' = fd/dx (∫a f(t) dt) = d/dx (F(x) - F(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 rectanglearea = (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 easythe 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 workwe will use the following method: approximation by rectangles

split up the interval from x=1 to x=3make 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 y-

value..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 rectangleswidth = ∆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 dotake ∆x "all the way small"

so we have Area ≈ Σ f(x)·∆xand 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 isArea = ∫3 2x dx

= x2 |3 = 52 - 32

= 16the procedure is easy, but understanding why it works is hardhard, 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 isArea = ∫o ex dx

= ex |o= e4 - eo

≈ 53.6

ex) what is the area under f(x) = sin(x) between 0 and 2πArea = ∫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)·∆xwell, sometimes f(x) is positive, and sometimes it is negativeso area above the x-axis is positive ... and area under the x-axis is negativecan 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 definedbefore you integrate(more on this later in the semester)

remember we learned two different integral types: ∫b and ∫xin terms of area:

∫b f(x)dx is the area under f(x) from x=a to x=bwhat 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 functionwe 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)=4finding f from f '' and handling constants

ex) f '' = et

antiderivative with constant in frontex) ∫ 2/x dx

trig/inverse trig and using radiansex) ∫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 applicationfirst lets recall...the derivative of distance is: the derivative of velocity is:

note: we also refer to distance as position or displacementnote: 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 +Cif the position at time t=1 is p=2, then:2=12+C1=Cp(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) dxv(t) - v(0) = ∫o 9.8 dxwhat 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 ]0v(t) - v(0) = 9.8t - 9.8(0)

= 9.8tOR 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

nextwhat is the distance (position) at time t ?

p' = vp(t) - p(0) = ∫o p'(x) dxp(t) - p(0) = ∫o v(x) dxp(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|op(t) - p(0) = (4.9t2 + vot) - ( 4.9(0)2 + vo(0) )

= 4.9t2 + votOR 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

Economicsex) 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) + k8000 = kC(x) = 8000 + 28xC(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 - 0C(x) = 8000 + 28xandC(120) = 8000 + 28(120) = 8000 + 3360 = 11360 dollars

ex) electricitycurrent = charge

timecurrent is measured in amps (I)... charge in coulombs (Q) ... time in seconds (t)amps = coulombs

secI = ∆Q/∆t

ex) suppose at t=0sec we have Q=4 coulombs, at t=4sec we have Q=12 coulombswhat 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 + thow 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 expendedForce 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 JoulesForce 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 multiplicationex) to push a desk, F=7 Joules/meterhow 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)=20xa) 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/metera) 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 kx2 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 Coolingfor a situation where you have an object of one temperature sitting in a room

of a different temperaturethe 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 Drainingwhen 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 = 2when the water is 4 meters high, how fast is it draning? (meters/sec)dy/dt = 2√4 = 4 meters/secwhen the water is 2 meters high, how fast is it draning? (meters/sec)dy/dt = 2√2 ≈ 2.828 meter/sec

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 UStypically, 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 = .02ydx

thats very interesting...what do we do with that?there are several approachesfirst, the numerical approach:ex) suppose the population is 300 million

what will it be next year?.02(300) = 6next year, 306 million...write that as one equation: 306 = 300 + .02·300·1

hmmmf(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 peoplewell, that's not quite right

in the second year, the population increases by 2% of 306so: 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 f(0) = 1000dx 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) repeatedlyf(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 onein 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 functionex) ∫ 2x dx =

sometimes we need to do algebraic manipulation to turn the problem into one we can do

ex) ∫ x(x+5) dxrewrite this as:

= ∫ x2+5x dxnow it is in a form we are familiar with

= x3 + 5x2 + C

ex) ∫ 2 dxx4

ex) ∫

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 dxso what is that "inside function"?

x4+1what 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 dxdx (more about that in one minute)

(substitution makes a new integral)∫ u7 du

(which we know how to do)= u8 + C

(then we substitute back)= (x4+1)8 + C

A comment about dx, dy, and duPreviously 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 →0We 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 dxbut 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) dxu=

ex) ∫ ln(x) dxx will substitution work?

ex) ∫ x3-x dx√x4-2x2-5

why should we check to see if substitution will work?...because there is an "inside function". look for u & du

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=dxu+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)

Techniques of integration - substitution/shifting the variable

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 = 6.75area 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 rectangleswhat is the width?

what is the height? ...discuss

what is the formula for the approximate area?

what is the formula for theexact 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+2what are the boundary x-values? ... you have to find them - how?

at the intersection, f=gx2 = x+2x2 - x - 2 = 0(x-2)(x+1)=0x=2, -1so the area is ∫-12 x+2-x2 dx = x2 + 2x - x3 |-12

2 3= (22/2+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 + 2xf=g ... 3x3-x2-10x = -x2+2x3x3-12x=03x(x2-4)=03x(x+2)(x-2)x=-2,0,2but notice that for -2<x<0, f>g

then 0<x<2, g>fwe 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 + 5ex) find the area between f = x3 - x2 and g = 3x - 3x2

hw questions- enclosed area

Volumemethod 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 areaarea = (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_cross-section.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 revolutionusing 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 [nicer as one integral]= ∫01 πr1

2 - πr22 dx

= ∫01 π(x)2 - π(x2)2 dx

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 revolutionusing 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 trickΣ √(∆x)2+(∆y)2 = Σ √(∆x)2+(∆y)2

= Σ √(∆x)2+(∆y)2 ∆x

= Σ

so,s ≈ Σ √ 1 + (∆y/∆x)2 ∆xs = ∫ √ 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 approximationwhen 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.soSA ≈ Σ 2πr √ 1 + (∆y/∆x)2 ∆xSA = ∫ 2πr √ 1 + (dy/dx)2 dx

and r = y when rotating around the x-axis, r=x when rotating around the y-axis (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) = $240this 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-xif the market price is $35, how many are sold?...x=65if 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 Pthe total consumer surplus is Σ (p(x)-P) ∆xConsumer 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 = wghwhere g is the acceleration due to gravity, g=9.8m/sec2

since pressure = force/area, P=F/A, thenF = PAso 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=1000the total force is the sum of all the forces on the slicesso F = wghAF ≈ 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), soWork = (force)(distance) W = (pressure)(area)(distance) W = (pressure)(volume)so for a gas, work can be a function of volume, W(V)=PVtake the derivative and you can write:W'(V) = P or dW = P

dVfrom physics, in many situations it is true that (P)(V) is some constant, so: P=k/Vso W'(V) = k/VW= ∫ 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 velocityhow could we do this......as an approximation: take a velocity reading every so often, then add them up and divide by nex) 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)5 10-0/2

where 2 is ∆t. flip it to the numerator and you get:vavg ≈ 1/b-a Σ v(t) ∆tvavg = 1/b-a ∫ab v(t) dtthis works for any function, soaverage value of f(x) for a<x<b is

favg = 1/b-a ∫ab f(x) dxOR 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 systemm = 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-distanceto find x', we need find the moment and the massmass = (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 integralmass = ∫2.4 f(x) dx mass = m = ∫ w L(x) dx

= 2.4 ∫01 0.1x dx w=density, L=length= 0.12x2 |01

= 0.12 kgmoment = (mass)(position)

= (density)(area)(position) = (density)(length)(width)(position)

moment ≈ Σ 2.4x(0.1x)∆xmoment = ∫ 2.4x(0.1x)dx

in general, moment = ∫ wx L(x)dxw=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 kgx-moment = ∫ 12x(x-x2)dx

= ∫ 12x2 - 12x3 dx= ∫ 4x3 - 3x4 dx= 4(1)3 - 3(1)4 - 0= 1

x-centroid = 1/2now for y ... x=y and x=√yy-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 = 2/5centroid = (.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 = √1-x2

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)= ∫eudu = ...

ex) ∫ cos(4x)dxu = 4x ... du = 4dx

= 1/4 ∫ cos(u)durecall 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' dxNow, solve for one integral:∫ u·v' dx = u·v - ∫ v·u' dxI 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)dxu = 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) dxu = arctan(x) dv=dxdu= 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 away2) 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]u=x2 dv=sin(x)dxdu=2xdx v=-cos(x)

= -x2cos(x) - ∫ -cos(x)2xdx now what? ... ...do it again

u = 2x dv=-cos(x)dxdu = 2dx v=-sin(x)

= -x2cos(x) - [-2xsin(x) - ∫ -2sin(x)dx]= -x2cos(x) + 2xsin(x) + 2cos(x) + C

Shortcut for repeating integration by parts - the table methodex) ∫ x2sin(x)dx sign | u & deriv | v' & antideriv

+ x2 sin(x)- 2x -cos(x)+ 2 -sin(x)- 0 cos(x)

= -x2cos(x) + 2xsin(x) + 2cos(x) + C

ex) ∫ x5cos(x)dx [why is this a candidate for parts?]

Using repeating to make an equationex) ∫ e2xsin(x)dx

u=e2x dv=sin(x)dxdu=2e2xdx v= -cos(x)

= -e2xcos(x) - ∫ -cos(x)2e2xdxu=e2x dv=cos(x)dxdu=2e2x v=sin(x)

=-e2xcos(x) + 2[e2xsin(x) - ∫ 2sin(x)e2xdx ]as one equation... ∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x) - 4∫e2xsin(x)dxand now... 5∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x) so... ∫e2xsin(x)dx = -e2xcos(x) + 2e2xsin(x)

5note: 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)= ∫ sec2(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, likeex) ∫ 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:cos2x + sin2x = 1tan2x + 1 = sec2x1 + cot2x = csc2x

cos2x = 1 + cos(2x) 2sin2x = 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 dxTo 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 duu = cos(x) ... du = -sin(x)dx

= -∫u4 - u6 du= -u5 + u7

5 7= -cos5x + cos7x +C

5 7 notation: you can also write (cos x)5

ex) ∫ sin2x cos5x dxWhat 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 ...and now...

u=sin(x) ... du=cos(x)dx= ∫ u2 - 2u4 + u6 du= u3 - 2u5 + u7

3 5 7=sin3x - sin5x + sin7x +C

3 5 7

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= 1/4 ∫ 1/2 - 1/2 cos(4x) dx= 1/8 ∫ dx - 1/8 ∫ cos(4x) dx

u = 4x ... du = 4dx= 1/8 x - 1/8 ∫ 1/4 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 dxWe 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=tan5x + 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 dxdu = 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) dx2∫ 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= -cot5x - 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 bestex) ∫ 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 oddfactor out cos(x), replace every cos2(x) with 1-sin2(x), u=sin(x)

power of sin, cos both evenreplace 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 powerturn only one tan2x into sec2x -1, multiply, solve one integral, repeat as needed

sec by itself, odd powerby 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-2It would help a lot to know that x+5 = 2 - 1 because now we get x2+x-2 x-1 x+2

= ∫ 2 - 1 dx x-1 x+2

= 2 ln(x-1) - ln(x+2) +C

So the main trick here involves algebra. How do we convert the complicated rational function into several simple fractions?

ex) x+5 = x+5 = ? + ?x2+x-2 (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 x2+x-2 (x+2)(x-1) x+2 x-1Now, multiply both sides by the original denominator. After cancellation,x+5 = A(x-1) + B(x+2)x+5 = Ax - A + Bx + 2Bx+5 = (A+B)x + (-A+2B)Therefore, by matching coefficients (for x and for the constant),x: 1 = A+Bconst: 5 = -A+2BSolve it using algebra...Add equations to get 6=3B ... 2=B ... from equation#1, 1=A+2 ... A=-1So 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, -2plug in x=1 to get 6=0+3B ... B=2plug 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 x3 - x2

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 (x-1)2(x+1) 4x = A + B + C(x-1)2(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 + Cx2: 0 = A+Cx1: 4 = B-2Cx0: 0 = -A+B+CSolve 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+1So: if the denominator has degree 2, then the numerator can have degree 12x2 - x + 1 = A(x2+1) + (Bx+C)x2x2 - x + 1 = Ax2 + A + Bx2 + Cxx2: 2 = A + Bx1: -1 = Cx0: 1 = Aso 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 + Ex(x2+6)2 x2+6 (x2+6)2 xMultiply 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, divide2 factor the denominator3 set up fractions- distinct linear factors OR repeated factors OR non-linear factors OR repeated non-linear factors4 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 equations5 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) dt9sin2t

= ∫ √9(1-sin2t) 3cos(t) dt9sin2t

= ∫ 3√cos2t 3cos(t) dt9sin2t

= ∫ 3cos(t) 3cos(t) dt9sin2t

= ∫ 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 = t3 3What 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 √16(1+tan2(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 xWhat 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 (tan2t + 1)3/2

= ∫ sec2t dt (sec2t)3/2

= ∫ 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 √x2+1

Trig sub - completing the square

ex) calculate ∫ dx √x2+10x+26look at the denominator ... x2+10x+26 = x2+10x+25 + 1so 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 √3-(x2+2x)= ∫ x dx √3-(x2+2x+1)+1= ∫ x dx √4-(x+1)2

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 √4 - (2sin t)2

= ∫ 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) +Cand 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: substitute: use the identity: √a2 - x2 x=a sin(t) 1-sin2t = cos2t √a2 + x2 x=a tan(t) 1+tan2t = sec2t √x2 - a2 x=a sec(t) 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 dxWe 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 = t3 3We 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 3If 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 (4x2+9)3/2

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 (9tan2(t) + 9)3/2

= 81/16 ∫ tan3t sec2t dt 93/2(tan2(t)+1)3/2

= 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= -3/16 ∫ u-2 - 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 ) 3 √4x2+9 OR √4x2+9 - 9 +C 16 16√4x2+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 substitutionex) ∫ 2 dx ... this looks like _____ so substitute u=_____

1+4x2

Techniques of integration using algebra

- expandex) ∫ x(x+1)dx

= ∫ x2+x dx = ...

-complete the squareex) ∫ 1 x2 + 2x + 2

=∫ 1 dx (x+1)2 +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 fractionex) ∫ 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

x2+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 sideex) Find ∫ sec3x dx

Techniques of integration - an overview of strategy

the number one rule - look for the simplest technique first1 its a known formula2 substitution3 standard form for one of the techniques - trig sub, partial fractions, parts, trig powers4 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) +Creplace x with something... try et ... then dx=etdt

now, ∫ cos(et) et dt = sin(et) +C

ex) we can solve ∫ ln(x) 1/x dxu=ln(x) ... du= 1/x dx

= ∫ u du= u2 +C 2=(ln x)2 +C

2now, replace x...

step 1: pick an integral we know how to dostep 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 Integralsor, what to do when infinity appears somewhere in the integral

first, infinite x-valueex) ∫∞ 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 limit2 do the integral3 evaluate the limitIf 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 limitex) ∫0 1 dx √xthe 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 one-sided 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 limitex) f '(a) = lim∆x→0 f(a+∆x) - f(a) OR f '(a) = limx→a f(x) - f(a)

∆x x-anow, 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)·∆xinstead 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 / 61 ... initial x-value, call it x0

1.5 ... (initial x-value) + ∆x, call it x1

2 ... x2 = (previous x-value) + ∆x2.5 ... x3 = (previous x-value) + ∆x OR (initial x-value) + ____3 ... x4 = (initial x-value) + 4·∆x3.5 ... 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 approximationb) Ln approximationc) the exact value using a limita- ∫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 sumspull 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 4Ln = 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 limitshort way: convert to an integrala=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 centroidcalculate an algebraically difficult integral for arc length"integral as a limit" problem where you calculate the sum (so, no sum formulas)