integration made easy - university of texas at austin · 2015-10-25 · outline 1 - length,...

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

Outline

1 - Length, Geometric Series, and Zeno’s Paradox2 - Archimedes’ method for getting area under a parabolic arch3 - Area under a general curve4 - Application of definite integral–volume of a cone

Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 2 / 47

Basic notions of length

Suppose we have a strip of length 1For instance, a ruler that’s 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.



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?


Zeno’s paradoxEnter Zeno’s paradox.

Ancient Greek philosopher Zeno of EleaInvented several different paradoxes to support the doctrine of Parmenidesthat “belief in plurality and change is mistaken, and that motion is illusory."We’ll examine the “dichotomy" paradox


Zeno’s 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!


Zeno’s 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 =⇒ paradox!


Zeno’s paradox

So what gives?

Archimedes, the ancient Greek mathematician, offered a resolution based onideas from modern calculus.


Geometric seriesLet’s introduce some notation: ∑

= sum

5∑i=1

i = 1 + 2 + 3 + 4 + 5

i is a “dummy" variable. The lower bound i = 1 tells you where to the begin thesum. The upper bound, 5, tells you when to stop summing.

6∑k=3

2k = 6 + 8 + 10 + 12


Geometric series

Recall Zeno maintained Homer must complete an infinite amount of tasks tomake it to the bus stop.

After completing the first task, he’s 1/2 of the way thereAfter the second task, he’s 3/4 of the way thereAfter the third, 7/8 . . .

Where will Homer be after completing the Nth task?


Geometric series

Using a telescoping sum, we can calculate Homer’s position after completingthe Nth task.

Homer’s position =N∑

k=1

(1/2)k = 1− 2−N


Geometric series1st task:

1∑k=1

(1/2)k = 1/2 ⇐⇒ 1− (1/2)1 = 1/2

3rd task:

3∑k=1

(1/2)k = 1/2 + (1/2)2 + (1/2)3 = 7/8 ⇐⇒ 1− (1/2)3 = 1− 1/8 = 7/8

34th task:

34∑k=1

(1/2)k = 1− (1/2)34 =1717986918317179869184

≈ 0.9999999999417923391


Geometric seriesEven after completing “only" 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!


Geometric series

After a finite amount of tasks completed, Homer has traveled

N∑k=1

(1/2)k = 1− (1/2)N .

Let N tend to infinity:

∞∑k=1

(1/2)k = limN→∞

(1− (1/2)N)


Geometric seriesHow do we know what

∞∑k=1

(1/2)k = limN→∞

(1− (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.


Geometric seriesThe key statement is the following:

By choosing N “big enough", the difference between the finite sum

N∑k=1

(1/2)k

and 1 can be made arbitrarily small.

Textbook definition:The limit of a quantity SN as N →∞ equals L if ∀ε > 0, ∃N such that|L− SN| < ε.


Geometric series

So,∞∑

k=1

(1/2)k = 1

and Homer makes it to the bus after all! In general, for |r| < 1, the sum

∞∑k=0

rk =1

1− r

is called an infinite geometric series.



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.


Area under parabolic archLet’s consider a slightly different problem: how to calculate the area of aparabolic segment?More specifically, let’s consider the following: let

f (x) = 1− x2.

How do we calculate the area “under the arch"? That is, the area between thegraph and the x-axis?


Area under parabolic arch

Archimedes’ idea was to use the area of a triangle to approximate the areaunder the arch.We call this approximation by simple geometric shapes–a very powerful idea.

Area of triangle =12

bh

We “break up" the parabola into a bunch of different triangles as follows.



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.



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.



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 triangles–called triangle α and triangle β. 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



total area of triangles after Step 1 = A0 + (area of α) + (area of β)

=⇒ A1 = 1 + 1/8 + 1/8 = 5/4

What we really want is the total area of all triangles drawn at Step N.


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 =N∑

k=0

2k(

18

)k

A0



This sum can be simplified to

AN =N∑

k=0

(14

)k

A0.

Now, it should be clear that at each step, we don’t 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 Zeno’s paradox, we need to take an infinitenumber of steps and draw an infinite number of triangles.


Area under parabolic archLet

A∞ =∞∑

k=0

(14

)k

A0.

This is an infinite geometric series! We saw how to compute this earlier.

A∞ = A0

∞∑k=0

(14

)k

= A01

1− 1/4=

43

A0.

Since A0 = 1, we conclude the area under f (x) = 1− x2 is equal to 4/3.



The intuition is essentially identical to the one behind the resolution to Zeno’sparadox.

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.



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 “large enough" number of trianglesDifference between the exact area under the arch, and sum of the areas ofthe triangles will be arbitrarily small.


Area under a general curve

What if we want to calculate the area under a more general curve?


Some motivation

Why would we even want to compute the area under a general curve?

Let’s begin with a few simple examples.Suppose you’re in a car traveling along the highway. The road is not crowdedand you’re not in a rush, and the car is set to run on cruise control–let’ssuppose it’s set at 70mph.

You’re bored in the backseat, but instead of asking mom or dad “are we thereyet" for 5th time in the last 20 minutes, you decide to calculate yourself how faryou have traveled.


Some motivation

For simplicity, let’s 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 car’s speed versus time look like?


Some motivationOf course the graph is a constant function, with constant value 70 miles perhour.

Now what if you wanted to compute how far you’ve traveled in two and a halfhours?


Some motivationOf course, since we know our speed is a constant 70mph, and we want toknow how far we’ve traveled in 2.5 hours time, we simply multiply:

distance = (70 mph) · (2.5 hours) = 175 miles


Some motivation

What if you aren’t 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 . . .


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!).


Area under a general curveIn general your car’s velocity as a function of time might not look so “nice."Traffic and weather conditions, for instance, might cause nonlinearaccelerations. For example, what if your car’s velocity as a function of timelooked like this:


Area under a general curve

How would you go about finding the total distance traveled, i.e. area under thiscurve?


Riemann sums

Recall Archimedes’ 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 curveWe’ll approximate our areas with rectangles


Riemann sums

Our intuition for Riemann sums is: the more rectangles we use (equivalently,as we decrease ∆x), 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.


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) ≈N∑

i=1

∆x · f (xi).

It should be stressed that ∆x is simply the base of our rectangles, while f (xi) isthe height of each rectangle.


Riemann sums

Can anyone guess how we are going to define the definite integral?


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 = limN→∞

N∑i=1

∆x · f (xi)

where a = x0 and b = xN. We have a special notation for the definite integral:∫ b

af (x)dx.


Application of the definite integral

Now let’s investigate how we can apply this thing.Recall the volume of a coneVcone = (1/3)πr2hHow is this formula obtained?Recall also volume of a cylinderVcyl = πr2h


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 object–a coneTo get the volume of this cone, we approximate it by volumes of cylinders


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 ∆x?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 sum–the sum of the volumes of all thecylinders.Simplify this sum–the identity (∗) on the board will be useful.What is the limit as N goes to infinity? This will be your answer :-)


Thanks for listening! Questions?


integration made easy - university of texas at austin · 2015-10-25 · outline 1 - length,...

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