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Integration Made Easy
Sean Carney
Department of MathematicsUniversity of Texas at Austin
October 25, 2015Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 1 / 47
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Outline
1 - Length, Geometric Series, and Zenoâs Paradox2 - Archimedesâ method for getting area under a parabolic arch3 - Area under a general curve4 - Application of definite integralâvolume of a cone
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Basic notions of length
Suppose we have a strip of length 1For instance, a ruler thatâs 1 ft longQuestion: how many ways can we chop up this ruler into smaller stripssuch that the sum of the lengths of those strips still equals 1?
The answer is easy if we restrict ourselves to a finite number of chops.
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Basic notions of length
What if we want to chop our ruler into an infinite amount of strips?
How is it possible to add an infinite amount of things and end up withsomething finite?
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Zenoâs paradoxEnter Zenoâs paradox.
Ancient Greek philosopher Zeno of EleaInvented several different paradoxes to support the doctrine of Parmenidesthat âbelief in plurality and change is mistaken, and that motion is illusory."Weâll examine the âdichotomy" paradox
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Zenoâs paradox
Homer wants to catch the bus that will take him to the Olympic games.He begins to walk towards the bus stop.To get to the stop, he must get halfway there.Before he gets halfway there, he must get a quarter of the way thereBefore he gets a quarter of the way there . . .This description requires Homer to complete an infinite amount of tasks!
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Zenoâs paradoxFurthermore, there is no first distance to run, so the trip to the bus stop cannoteven begin! Indeed:
Name a first distance to begin with. For example, 1/8 of the way to the stopBefore starting with 1/8, you have to travel 1/16 of the way thereThis argument works for any starting distance
trip cannot ever begin + trip can never be completed =â paradox!
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Zenoâs paradox
So what gives?
Archimedes, the ancient Greek mathematician, offered a resolution based onideas from modern calculus.
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Geometric seriesLetâs introduce some notation: â
= sum
5âi=1
i = 1 + 2 + 3 + 4 + 5
i is a âdummy" variable. The lower bound i = 1 tells you where to the begin thesum. The upper bound, 5, tells you when to stop summing.
6âk=3
2k = 6 + 8 + 10 + 12
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Geometric series
Recall Zeno maintained Homer must complete an infinite amount of tasks tomake it to the bus stop.
After completing the first task, heâs 1/2 of the way thereAfter the second task, heâs 3/4 of the way thereAfter the third, 7/8 . . .
Where will Homer be after completing the Nth task?
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Geometric series
Using a telescoping sum, we can calculate Homerâs position after completingthe Nth task.
Homerâs position =Nâ
k=1
(1/2)k = 1â 2âN
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Geometric series1st task:
1âk=1
(1/2)k = 1/2 ââ 1â (1/2)1 = 1/2
3rd task:
3âk=1
(1/2)k = 1/2 + (1/2)2 + (1/2)3 = 7/8 ââ 1â (1/2)3 = 1â 1/8 = 7/8
34th task:
34âk=1
(1/2)k = 1â (1/2)34 =1717986918317179869184
â 0.9999999999417923391
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Geometric seriesEven after completing âonly" 34 tasks, Homer is pretty darn close to the busstop!
Zeno said Homer had to complete an infinite amount of tasks to get to thebusZeno maintained this is impossibleMathematically speaking, he was mistakenResolution to the paradox occurs when we sum up an infinite amount ofterms!
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Geometric series
After a finite amount of tasks completed, Homer has traveled
Nâk=1
(1/2)k = 1â (1/2)N .
Let N tend to infinity:
ââk=1
(1/2)k = limNââ
(1â (1/2)N)
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Geometric seriesHow do we know what
ââk=1
(1/2)k = limNââ
(1â (1/2)N)
even is?
The intuition is simple.Consider finite sums for N larger and larger.We saw for N = 34, the sum was pretty close to 1.But it was not 1!However, if we take a larger N, we can get even closer to 1.
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Geometric seriesThe key statement is the following:
By choosing N âbig enough", the difference between the finite sum
Nâk=1
(1/2)k
and 1 can be made arbitrarily small.
Textbook definition:The limit of a quantity SN as N ââ equals L if âΔ > 0, âN such that|Lâ SN| < Δ.
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Geometric series
So,ââ
k=1
(1/2)k = 1
and Homer makes it to the bus after all! In general, for |r| < 1, the sum
ââk=0
rk =1
1â r
is called an infinite geometric series.
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Basic notions of length
To answer our original question: is it possible to divide a strip of length 1 intoan infinite amount of pieces such that the sum of the length of the pieces is stillequal to 1?
YES! Just take the first piece to be length 1/2. Then the next piece length 1/4,and so on and so forth.
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Area under parabolic archLetâs consider a slightly different problem: how to calculate the area of aparabolic segment?More specifically, letâs consider the following: let
f (x) = 1â x2.
How do we calculate the area âunder the arch"? That is, the area between thegraph and the x-axis?
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Area under parabolic arch
Archimedesâ idea was to use the area of a triangle to approximate the areaunder the arch.We call this approximation by simple geometric shapesâa very powerful idea.
Area of triangle =12
bh
We âbreak up" the parabola into a bunch of different triangles as follows.
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Area under parabolic arch
Algorithm for dividing area under arch into triangles:First: Identify base of parabolic segmentNext: Draw line that is parallel to that base, and also tangent to theparabolaLastly: Draw triangle that connects two endpoints of the base with point oftangency
We will use this algorithm repeatedly to draw more and more triangles. Thereis an important relationship between the triangles, expressed in the followingtheorem.
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Area under parabolic arch
Procedure for estimating area under the arch:Step 0: first draw one triangleStep 1: draw two more trianglesStep 2: draw four more trianglesStep 3: draw eight more triangles . . .
Can you guess the general pattern? Our objective is to now add up the areasof all the triangles at Step N, using the theorem stated.
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Area under parabolic arch
At Step 0, our triangle T0 has base (or width) of length 2, height of length 1.Hence area of T0 is
A0 = (1/2)(2)(1) = 1.
At Step 1, we draw two more trianglesâcalled triangle α and triangle ÎČ. Ourtheorem tells us that their areas are equal. It also tells us the width of α isequal to the 1/2 the width of T0, and that the height of α equals 1/4 the heightof T0. Hence
Area of α =12
(12· 2)(
14· 1)
=18
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Area under parabolic arch
total area of triangles after Step 1 = A0 + (area of α) + (area of ÎČ)
=â A1 = 1 + 1/8 + 1/8 = 5/4
What we really want is the total area of all triangles drawn at Step N.
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Area under parabolic archTo get the total area of all triangles drawn at Step N, we should observe:
At Step N, we draw 2N more trianglesAll of these new triangles have the same areaThe area of one of these triangles is 1/8 the area of the previous triangle
Exercise: Compute the total area AN at Step N.
AN = A0 + 21(
18
A0
)+ 22
(18
(18
A0
))+ . . .+ 2N
((18
)N
A0
)
AN =Nâ
k=0
2k(
18
)k
A0
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Area under parabolic archTo get the total area of all triangles drawn at Step N, we should observe:
At Step N, we draw 2N more trianglesAll of these new triangles have the same areaThe area of one of these triangles is 1/8 the area of the previous triangle
Exercise: Compute the total area AN at Step N.
AN = A0 + 21(
18
A0
)+ 22
(18
(18
A0
))+ . . .+ 2N
((18
)N
A0
)
AN =Nâ
k=0
2k(
18
)k
A0
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Area under parabolic arch
This sum can be simplified to
AN =Nâ
k=0
(14
)k
A0.
Now, it should be clear that at each step, we donât quite capture of ALL of thearea under the arch. There will always be some sliver of area left.
Similar to the idea used to resolve Zenoâs paradox, we need to take an infinitenumber of steps and draw an infinite number of triangles.
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Area under parabolic archLet
Aâ =ââ
k=0
(14
)k
A0.
This is an infinite geometric series! We saw how to compute this earlier.
Aâ = A0
ââk=0
(14
)k
= A01
1â 1/4=
43
A0.
Since A0 = 1, we conclude the area under f (x) = 1â x2 is equal to 4/3.
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Area under parabolic arch
The intuition is essentially identical to the one behind the resolution to Zenoâsparadox.
After a finite number of steps, Homer never made it to the bus stop.This was the source of the paradox.When you took the mathematical limit, Homer made it just fine.Here, we can never cover the entire area of the arch with a finite number oftriangles.Take the limit, however, and consider an infinite number of triangles.Then we get the exact area under the arch.
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Area under parabolic arch
One last emphasis of the intuition behind the mathematical limit.The area under the arch is always greater than the sum of the areas of afinite number of triangles.Consider, however, a âlarge enough" number of trianglesDifference between the exact area under the arch, and sum of the areas ofthe triangles will be arbitrarily small.
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Area under a general curve
What if we want to calculate the area under a more general curve?
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Some motivation
Why would we even want to compute the area under a general curve?
Letâs begin with a few simple examples.Suppose youâre in a car traveling along the highway. The road is not crowdedand youâre not in a rush, and the car is set to run on cruise controlâletâssuppose itâs set at 70mph.
Youâre bored in the backseat, but instead of asking mom or dad âare we thereyet" for 5th time in the last 20 minutes, you decide to calculate yourself how faryou have traveled.
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Some motivation
For simplicity, letâs say the cruise control was turned on at noon, and that thehighway is straight, with little to no twists and turns.
What would a graph of your carâs speed versus time look like?
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Some motivationOf course the graph is a constant function, with constant value 70 miles perhour.
Now what if you wanted to compute how far youâve traveled in two and a halfhours?
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Some motivationOf course, since we know our speed is a constant 70mph, and we want toknow how far weâve traveled in 2.5 hours time, we simply multiply:
distance = (70 mph) · (2.5 hours) = 175 miles
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Some motivation
What if you arenât traveling at a constant velocity for the entirety of the trip?For example, you might run into construction zones.Speed limits are changingYour car has to accelerate/decelerate to keep upHow would you calculate your distance traveled now?
For example . . .
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Some motivation
The answer is not quite so obvious as the previous example. The notion,however, that total distance traveled equals area under the curve, still holds.
Exercise: compute the total distance traveled for the velocity function shown onthe board. Hint: you will need to know how to compute the area of a trapezoid.
Express your answer as a fraction (and be careful with your units!).
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Area under a general curveIn general your carâs velocity as a function of time might not look so ânice."Traffic and weather conditions, for instance, might cause nonlinearaccelerations. For example, what if your carâs velocity as a function of timelooked like this:
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Area under a general curve
How would you go about finding the total distance traveled, i.e. area under thiscurve?
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Riemann sums
Recall Archimedesâ method for finding area under a parabolic archUsed simple geometric shapes, who area was known, to approximate theareaRiemann sums use this idea to approximate area under a curveWeâll approximate our areas with rectangles
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Riemann sums
Our intuition for Riemann sums is: the more rectangles we use (equivalently,as we decrease âx), the better the sums of the areas of those rectangles willapproximate the area under a curve.
Just like what Archimedes expected as he increased the number of triangleshe used to approximate the area under the parabolic arch.
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Riemann sums
A general Riemann sum, with N rectangles, gives the following approximationto the area under a curve y = f (x).
Area under f (x) âNâ
i=1
âx · f (xi).
It should be stressed that âx is simply the base of our rectangles, while f (xi) isthe height of each rectangle.
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Riemann sums
Can anyone guess how we are going to define the definite integral?
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The definite integral
Just like we did with the infinite geometric sum and the area under a parabolicarch, to define the definite integral, we take the limit as N tends to infinity of ourfinite Riemann sums.
definite integral of f (x) from a to b = limNââ
Nâi=1
âx · f (xi)
where a = x0 and b = xN. We have a special notation for the definite integral:â« b
af (x)dx.
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Application of the definite integral
Now letâs investigate how we can apply this thing.Recall the volume of a coneVcone = (1/3)Ïr2hHow is this formula obtained?Recall also volume of a cylinderVcyl = Ïr2h
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Estimate volume of a coneFor simplicity suppose that the radius of the cone equals its height (r = h). Andconsider the function y = x .
We will rotate this line about the x-axisThis should give us a three-dimensional objectâa coneTo get the volume of this cone, we approximate it by volumes of cylinders
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Volume of a coneExercise: calculate the volume of a cone (with radius = height) by revolvingy = x around the x-axis.
First break the interval 0 to R into N pieces uniform pieces. What is âx?What is xi?For N pieces, you should have N cylinders. What is the volume of eachcylinder?
I What is the the height of each cylinder? The radius?I It may help to do a simpler case first, like N = 4
Now write down the Riemann sumâthe sum of the volumes of all thecylinders.Simplify this sumâthe identity (â) on the board will be useful.What is the limit as N goes to infinity? This will be your answer :-)
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Thanks for listening! Questions?
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