Integration Made Easy

Integration Made Easy Sean Carney Department of Mathematics University of Texas at Austin Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 1 / 47

Outline 1 - Length, Geometric Series, and Zeno s Paradox 2 - Archimedes method for getting area under a parabolic arch 3 - Area under a general curve 4 - 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 1 For instance, a ruler that s 1 ft long Question: how many ways can we chop up this ruler into smaller strips such 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. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 3 / 47

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 with something finite? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 4 / 47

Zeno s paradox Enter Zeno s paradox. Ancient Greek philosopher Zeno of Elea Invented several different paradoxes to support the doctrine of Parmenides that belief in plurality and change is mistaken, and that motion is illusory." We ll examine the dichotomy" paradox Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 5 / 47

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 there Before he gets a quarter of the way there... This description requires Homer to complete an infinite amount of tasks! Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 6 / 47

Zeno s paradox Furthermore, there is no first distance to run, so the trip to the bus stop cannot even begin! Indeed: Name a first distance to begin with. For example, 1/8 of the way to the stop Before starting with 1/8, you have to travel 1/16 of the way there This argument works for any starting distance trip cannot ever begin + trip can never be completed = paradox! Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 7 / 47

Zeno s paradox So what gives? Archimedes, the ancient Greek mathematician, offered a resolution based on ideas from modern calculus. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 8 / 47

Geometric series Let s introduce some notation: = sum 5 i = 1 + 2 + 3 + 4 + 5 i=1 i is a dummy" variable. The lower bound i = 1 tells you where to the begin the sum. The upper bound, 5, tells you when to stop summing. 6 2k = 6 + 8 + 10 + 12 k=3 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 9 / 47

Geometric series Recall Zeno maintained Homer must complete an infinite amount of tasks to make it to the bus stop. After completing the first task, he s 1/2 of the way there After the second task, he s 3/4 of the way there After the third, 7/8... Where will Homer be after completing the N th task? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 10 / 47

Geometric series Using a telescoping sum, we can calculate Homer s position after completing the Nth task. N Homer s position = (1/2) k = 1 2 N k=1 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 11 / 47

Geometric series 1st task: 3rd task: 1 (1/2) k = 1/2 1 (1/2) 1 = 1/2 k=1 3 (1/2) k = 1/2 + (1/2) 2 + (1/2) 3 = 7/8 1 (1/2) 3 = 1 1/8 = 7/8 k=1 34th task: 34 k=1 (1/2) k = 1 (1/2) 34 = 17179869183 17179869184 0.9999999999417923391 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 12 / 47

Geometric series Even after completing only" 34 tasks, Homer is pretty darn close to the bus stop! Zeno said Homer had to complete an infinite amount of tasks to get to the bus Zeno maintained this is impossible Mathematically speaking, he was mistaken Resolution to the paradox occurs when we sum up an infinite amount of terms! Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 13 / 47

Geometric series After a finite amount of tasks completed, Homer has traveled Let N tend to infinity: k=1 N (1/2) k = 1 (1/2) N. k=1 (1/2) k = lim N ( 1 (1/2) N ) Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 14 / 47

Geometric series How do we know what k=1 (1/2) k = lim N ( 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. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 15 / 47

Geometric series The key statement is the following: By choosing N big enough", the difference between the finite sum N (1/2) k k=1 and 1 can be made arbitrarily small. Textbook definition: The limit of a quantity S N as N equals L if ɛ > 0, N such that L S N < ɛ. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 16 / 47

Geometric series So, (1/2) k = 1 k=1 and Homer makes it to the bus after all! In general, for r < 1, the sum k=0 is called an infinite geometric series. r k = 1 1 r Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 17 / 47

Basic notions of length To answer our original question: is it possible to divide a strip of length 1 into an infinite amount of pieces such that the sum of the length of the pieces is still equal 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. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 18 / 47

Area under parabolic arch Let s consider a slightly different problem: how to calculate the area of a parabolic segment? More specifically, let s consider the following: let f (x) = 1 x 2. How do we calculate the area under the arch"? That is, the area between the graph and the x-axis? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 19 / 47

Area under parabolic arch Archimedes idea was to use the area of a triangle to approximate the area under the arch. We call this approximation by simple geometric shapes a very powerful idea. Area of triangle = 1 2 bh We break up" the parabola into a bunch of different triangles as follows. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 20 / 47

Area under parabolic arch Algorithm for dividing area under arch into triangles: First: Identify base of parabolic segment Next: Draw line that is parallel to that base, and also tangent to the parabola Lastly: Draw triangle that connects two endpoints of the base with point of tangency We will use this algorithm repeatedly to draw more and more triangles. There is an important relationship between the triangles, expressed in the following theorem. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 21 / 47

Area under parabolic arch Procedure for estimating area under the arch: Step 0: first draw one triangle Step 1: draw two more triangles Step 2: draw four more triangles Step 3: draw eight more triangles... Can you guess the general pattern? Our objective is to now add up the areas of all the triangles at Step N, using the theorem stated. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 22 / 47

Area under parabolic arch At Step 0, our triangle T 0 has base (or width) of length 2, height of length 1. Hence area of T 0 is A 0 = (1/2)(2)(1) = 1. At Step 1, we draw two more triangles called triangle α and triangle β. Our theorem tells us that their areas are equal. It also tells us the width of α is equal to the 1/2 the width of T 0, and that the height of α equals 1/4 the height of T 0. Hence Area of α = 1 ( ) ( ) 1 1 2 2 2 4 1 = 1 8 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 23 / 47

Area under parabolic arch total area of triangles after Step 1 = A 0 + (area of α) + (area of β) = A 1 = 1 + 1/8 + 1/8 = 5/4 What we really want is the total area of all triangles drawn at Step N. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 24 / 47

Area under parabolic arch To get the total area of all triangles drawn at Step N, we should observe: At Step N, we draw 2 N more triangles All of these new triangles have the same area The area of one of these triangles is 1/8 the area of the previous triangle Exercise: Compute the total area A N at Step N. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 25 / 47

Area under parabolic arch To get the total area of all triangles drawn at Step N, we should observe: At Step N, we draw 2 N more triangles All of these new triangles have the same area The area of one of these triangles is 1/8 the area of the previous triangle Exercise: Compute the total area A N at Step N. ( ) ( ( )) ( (1 ) ) 1 1 1 N A N = A 0 + 2 1 8 A 0 + 2 2 8 8 A 0 +... + 2 N A 0 8 A N = N ( 1 k 2 8) k A 0 k=0 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 25 / 47

Area under parabolic arch This sum can be simplified to A N = N k=0 ( ) 1 k A 0. 4 Now, it should be clear that at each step, we don t quite capture of ALL of the area 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 infinite number of steps and draw an infinite number of triangles. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 26 / 47

Area under parabolic arch Let A = k=0 k=0 ( ) 1 k A 0. 4 This is an infinite geometric series! We saw how to compute this earlier. ( ) 1 k 1 A = A 0 = A 0 4 1 1/4 = 4 3 A 0. Since A 0 = 1, we conclude the area under f (x) = 1 x 2 is equal to 4/3. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 27 / 47

Area under parabolic arch The intuition is essentially identical to the one behind the resolution to Zeno s paradox. 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 of triangles. Take the limit, however, and consider an infinite number of triangles. Then we get the exact area under the arch. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 28 / 47

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 a finite number of triangles. Consider, however, a large enough" number of triangles Difference between the exact area under the arch, and sum of the areas of the triangles will be arbitrarily small. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 29 / 47

Area under a general curve What if we want to calculate the area under a more general curve? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 30 / 47

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 crowded and you re not in a rush, and the car is set to run on cruise control let s suppose it s set at 70mph. You re bored in the backseat, but instead of asking mom or dad are we there yet" for 5th time in the last 20 minutes, you decide to calculate yourself how far you have traveled. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 31 / 47

Some motivation For simplicity, let s say the cruise control was turned on at noon, and that the highway is straight, with little to no twists and turns. What would a graph of your car s speed versus time look like? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 32 / 47

Some motivation Of course the graph is a constant function, with constant value 70 miles per hour. Now what if you wanted to compute how far you ve traveled in two and a half hours? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 33 / 47

Some motivation Of course, since we know our speed is a constant 70mph, and we want to know how far we ve traveled in 2.5 hours time, we simply multiply: distance = (70 mph) (2.5 hours) = 175 miles Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 34 / 47

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 changing Your car has to accelerate/decelerate to keep up How would you calculate your distance traveled now? For example... Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 35 / 47

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 on the 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!). Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 36 / 47

Area under a general curve In general your car s velocity as a function of time might not look so nice." Traffic and weather conditions, for instance, might cause nonlinear accelerations. For example, what if your car s velocity as a function of time looked like this: Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 37 / 47

Area under a general curve How would you go about finding the total distance traveled, i.e. area under this curve? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 38 / 47

Riemann sums Recall Archimedes method for finding area under a parabolic arch Used simple geometric shapes, who area was known, to approximate the area Riemann sums use this idea to approximate area under a curve We ll approximate our areas with rectangles Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 39 / 47

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 will approximate the area under a curve. Just like what Archimedes expected as he increased the number of triangles he used to approximate the area under the parabolic arch. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 40 / 47

Riemann sums A general Riemann sum, with N rectangles, gives the following approximation to the area under a curve y = f (x). Area under f (x) N x f (x i ). i=1 It should be stressed that x is simply the base of our rectangles, while f (x i ) is the height of each rectangle. Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 41 / 47

Riemann sums Can anyone guess how we are going to define the definite integral? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 42 / 47

The definite integral Just like we did with the infinite geometric sum and the area under a parabolic arch, to define the definite integral, we take the limit as N tends to infinity of our finite Riemann sums. definite integral of f (x) from a to b = lim N N x f (x i ) where a = x 0 and b = x N. We have a special notation for the definite integral: b a f (x)dx. i=1 Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 43 / 47

Application of the definite integral Now let s investigate how we can apply this thing. Recall the volume of a cone V cone = (1/3)πr 2 h How is this formula obtained? Recall also volume of a cylinder V cyl = πr 2 h Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 44 / 47

Estimate volume of a cone For simplicity suppose that the radius of the cone equals its height (r = h). And consider the function y = x. We will rotate this line about the x-axis This should give us a three-dimensional object a cone To get the volume of this cone, we approximate it by volumes of cylinders Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 45 / 47

Volume of a cone Exercise: calculate the volume of a cone (with radius = height) by revolving y = x around the x-axis. First break the interval 0 to R into N pieces uniform pieces. What is x? What is x i? For N pieces, you should have N cylinders. What is the volume of each cylinder? What is the the height of each cylinder? The radius? 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 the cylinders. 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 :-) Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 46 / 47

Thanks for listening! Questions? Sean Carney (University of Texas at Austin) Integration Made Easy October 25, 2015 47 / 47