18A.1: #2,4,5 (Area under curve) 18A.2: #1 3 (Definite integral)

18A.1: #2,4,5 (Area under curve) 18A.2: #1 3 (Definite integral) 1. Understand integration as the area under a curve. QB #36,44d 2. Understand and calculate left and right Riemann sums 3. Determine upper and lower estimates of area using Riemann sums A common problem in mathematics is finding the area beneath a curve. This is easy if the curve is a line. But how can we do this for a more complex function? The sum of finite rectangles is called a Riemann Sums which is an approximation to the area under the curve. We limit our discussion here to f(x) > 0! Riemann Sums If the area under the curve f(x) over the interval a x b is split into n rectangles of equal width an approximation to the are beneath the curve is given by The left hand Riemann sum or by The right hand Riemann sum Book uses "Upper & Lower" rectangles which is unfortunate. Left and right are more precise. Will the estimate be an underestimate or an overestimate of the true area? Sketch a diagram and explain your answer to your neighbor. A left hand Riemann sum will: Underestimate the area under an increasing curve Overestimate the area of a decreasing curve. A right hand Riemann sum will: Overestimate the area under an increasing curve Underestimate the area of a decreasing curve. Alternatively: The left hand Riemann sum is: the lower sum for an increasing curve the upper sum for a decreasing curve. The right hand Riemann sum will: the upper sum for an increasing curve the lower sum for a decreasing curve. The more rectangles we use, the smaller their width and thus the more accurate the estimate will be. We can define the exact area by taking a limit. The exact area under the curve f(x) over the interval a x b is given by: Note that as n, x 0 The Definite Integral of f(x) from a to b: Is the area under the curve f(x) between x= a and x = b Notice this is a number. Working with sums is tedious without the power of your calculator (or Excel ish software) [ALPHA][WINDOW][SUMMATION] (Option 2) or [MATH][SUMMATION] (Option 0) will give you a summation expression: Example: Find the upper and lower sums for n = 10, 30, and 50 for the area under x 2 for 1 < x < 2. 1. Define and store n: > 10 [STO] N 2. Define and store Δx: > (2 1)/N [STO] D 3. Create the summation expression: > [ALPHA][WINDOW][2] 4. Using I as the index variable, the Left hand sum starts at I = 0 and goes to N 1 5. The summation expression is f(2 + I*D)*D. Since our function is x 2, enter (2 + I*D) 2 *D. Press enter for the result. 6. For other values of n, revise step 1, then repeat the other steps by pasting the commands to the execution line and pressing enter. It's very efficient to do it this way. 7. For Right hand sums, repeat the process but let the index, I, go from 1 to n. You will need to use your calculator or Excel to do your homework. Do not shortchange this work! 18A.1: #2,4,5 (Area under curve) 18A.2: #1 3 (Definite integral) QB #36,44d 1/12 18A Area under a curve 1

18A.1: #2,4,5 (Area under curve) Present #4,5 18A.2: #1 3 (Definite integral) Present #3 QB #36,44d Present both 18B: #1 3 all (Antiderivatives) QB #39b,42 1. Understand the concept of an antiderivative 2. Find antiderivatives for power functions and exponentials 3. Find antiderivatives given hints about what function to differentiate Elliot is cruising along a straight road at 60 mph (90 ft/sec) when his radar detector beeps. He slams on his brakes, decelerating at a constant rate of 20 ft/sec 2. a) How long does it take him to get to the 20 mph speed limit (30 ft/sec)? b) How far does he travel in that time? We need an equation that describes how the velocity is changing with time. a) v(t) = 90 20t. So we need to solve 30 = 20t + 90 t = 3 sec b) The velocity is changing! So we can't use our usual d = vt. But suppose we had a function that describes position as a function of time. We could find the position at time 0 and the position at time 3. The difference is the distance travelled. Recall that velocity is the derivative of position. So first we need to find a function whose derivative is v(t) = 20t + 90. Can you get it by "reverse differentiating"? Notice the +c. Since the derivative of a constant is zero, there are an infinite number of functions that meet our condition! In other words, Elliot could hit the brakes with any starting position. The antiderivative of a function f(x) is that function whose derivative is f(x). The antiderivative of f is written as F with the property that F'(x) = f(x) We find antiderivatives by a) thinking in reverse from differentiation or b) integration (we will look more at this later) Some examples: In the book HW you are looking for patterns. Therefore, it's important to do all the parts. You should find a pretty simple algorithm at some point. 18B: #1 3 all (Antiderivatives) QB #39b,42 1/17 18B: Antiderivatives 2

18B: #1 3 all (Antiderivatives) Present general rule results of 1 & 2 QB #39b, 42 Present both 18C: #1,2,3def,4,5 (Fundamental theorem) 1. Understand the relationship between antiderivatives, the area under a curve, and the definite integral. 2. Recognize and use the Fundamental Theorem of Calculus So far, our understanding of integrals is as an area Note that is referred to as a definite integral because it has a single value Back to our question. We found that the position of the object is given by: Since Elliot has not backtracked at all, we found the distance he travelled by looking at his position at t = 0 and his position at t = 3: If we take the difference, we have the distance travelled: s(3) s(0) = 180 + c c = 180 feet Let's look at this visually Notice that the distance traveled is the area under the curve since we are summing up bits of rate times time. Areas under curves can represent lots of things! One can calculate these areas by using infinite summations of rectangles or trapezoids but it is rather complex. We will state (without proving: see book) a very important result: The Fundamental Theorem of Calculus (Part I) In other words instead of evaluating an infinite sum using limits, we can simply evaluate the difference of the antiderivative of the function evaluated at the two endpoints of the interval. Proof This is a huge simplification! Let's try a couple: Understanding that definite integrals are just (signed) areas under a curve and using the fundamental theorem, you can develop some important properties: Let's prove You will do the rest of these in your HW. Show all the steps to actually prove them. But don't just prove them, think about what they can do for you. You will need to use these properties. Calculators will integrate functions you can't do analytically! 18C: #1,2,3def,4,5 (Fundamental theorem) 1/19 18C: Fundamental Theorem 3

18C: #1,2,3def,4,5 (Fundamental theorem) Discuss results of #2,3def, #5 fully Mini Quiz you may use a calculator (...this time...) 1. A1 for shape, A1 for labeled intercepts, A1 for labeled extremes, A1 for domain a) A1 for shading in proper domain, A1 for above and below x axis b) M1 for any attempt to find area as opposed to the integral (doubling, absolute value, etc.) A1 for antiderivative of cosx seen anywhere A1 for correct form for area; ie. A = 2( cos(0) ( cos(π/2))) or absolute values A1 for answer (2) c) R1 Integrals are signed areas. Need to take the absolute value or symmetry and double the integral from 0 to π/2 to find actual area. Or equivalent 18D: #2 12 even (Integration) QB #30,32,48b 1. Understand indefinite integrals and the proof of the Fundamental Theorem 2. Find integrals for various familiar functions We have looked at definite integrals as areas under curves over specific intervals. But the Fundamental Theorem of Calculus is more powerful than that. We can use it to relate a function and its antiderivative, even if there is no specific interval under consideration. Let's look at this more deeply: Once way to see this is that the value of a function at x = a represents the instantaneous rate of change of the area underneath it at the place where x = a. We see that there is a fundamental connection between antiderivatives and integrals (area) that can be written as: In words, the indefinite integral of a function is given by the antiderivative of the function plus some constant of integration. The function f(x) is called the integrand. Example: Consider the function The antiderivative of this function is since it's derivative gives f(x) We write this as the integral of f or Notice that the same rules apply as for definite integrals, except that there is always a constant of integration. Another example. Find Check your work! If you differentiate your result, you should get the integrand! Try one: Find In the HW you will practice seeing the relationship between simple integrals and their associated derivatives with some hints given...can you do them without the hints (hint!)? 18D: #2 12 even (Integration) QB #30,32,48b 1/22 18D: Integration 4

18D: #2 12 even (Integration) Present selection QB #30,32,48b Mini Quiz 18E.1: #1,2,5,7 (Indefinite integration) 18E.2: #1bd,2,3 (Finding "c") QB #21,23,28,31,33b,38,47 Coming next Friday 2/2: Integration Quiz 1. Find indefinite integrals for familiar functions 2. Use boundary conditions (a point on the curve) to find the constant of integration. Since we know a lot about differentiation, we can work in reverse to figure out a lot of integrals! Recall the following properties of differentiation: These lead to the following integrals: But beware of x < 0: Notice that because of the chain rule, these rules do not work directly when the argument is a function of x. Don't forget about the constant of integration. Using this in conjunction with what you know about differentiating, you can find many integrals. Let's try some: Always differentiate your result to see that you get the original integrand. Expand parentheses and simplify to get a form you can integrate (when possible). When you know any specific point on the curve that results from the integration, you can find the constant of integration. We may be given a second derivative, f''. To find the original function, integrate twice. You will need a point on f' to be specific on the second integration. Remember, when we see, g is the function whose derivative is f. There is quite a lot of practice here. This is where you 18E.1: #1,2,5,7 (Indefinite integration) will become fluent at integrating basic functions. It's 18E.2: #1bd,2,3 (Finding "c") also where you will really begin to master algebra. Go QB #21,23,28,31,33b,38,47 for speed! Do them all to create fluency! Coming next Friday (2/2): Integration Quiz 1/24 18E: Rules of Integration 5

18E.1: #1,2,5,7 (Indefinite integration) 18E.2: #1bd,2,3 (Finding "c") #2? 3? QB #21,23,28,31,33b,38,47 Any QBs? 18F: #1gh,2cfi,3,4def,5de,6efg,8 13 (Integrating ax + b) QB #24b,45b,46b Quiz Next Friday: 2/2 Integration mechanics 1. Find integrals involving composites of linear functions Do you remember these? Since, it will always be the case that This is a basic extension of integrating f(x). Notice that the derivative of ax + b is just a. So when we integrate a function of ax + b, we will need to divide by a. This is a consequence of the chain rule. Apart from noticing this situation, there is nothing new here. It applies to any function. Notice that integrals involving squares of trig functions cannot be done with the power rule unless there are other factors to offset the effects of the chain rule. However, we can often move forward by rewriting using the double angle formula. Recall that: (These are shown in a different form on your formula sheet) cos 2x = cos 2 x sin 2 x cos 2x = 1 2sin 2 x cos 2x = 2cos 2 x 1 These allow us to tackle integrals like: 18F: #1gh,2cfi,3,4def,5de,6efg,8 13 (Integrating ax + b) QB #24b,45b,46b Quiz Next Friday: 2/2 Integration mechanics 1/26 18F Integrating f(ax + b) 6

18F: #1gh,2cfi,3,4def,5de,6efg,8 13 (Integrating ax + b) Questions as needed QB #24b,45b,46b 1. Find integrals using substitution. We have seen that integrating more complex functions is not straightforward. In a pure calculus class you will learn some techniques for integrating functions that have certain forms. Here we will consider only one of those techniques, using substitution. Consider the integral 18G: #1,2 5 cd each (Substitution) Quiz Fri Integration mechanics Do you notice anything special about it? The second factor is the derivative of the first. This allows us to do the following: This is called, u substitution or just substitution. We use it when there are two factors in the integrand, with one being the derivative of the other (or some constant times the derivative of the other). The general form is: Getting the hang of how this works requires: a) Flexible mastery of the derivatives of common functions b) Careful algebra c) Practice! Try some: Choosing the right substitution is key! You can deal with extra constants floating around. Notice the detail here: Arguments of log functions must be positive, so we use the absolute value. The approach will work with any function. Think ahead to choose the right substitution. 18G: #1,2 5 cd each (Substitution) Quiz Fri Integration mechanics 1/29 18G Substitution 7

Mini Quiz: You may use a calculator give exact answers where possible. 1. 2. 2.31 1.02 2.59 9.96 2. 18G: #2 5 cd each (Substitution) Discuss as needed, after quiz 18H: #1 4,5cfil,6 16even (Definite integrals) Quiz Fri Integration mechanics 1. Practice using definite integrals to reinforce properties and meaning. In general, an integral involves finding the antiderivative. That is, But we can use the fundamental theorem of calculus to evaluate definite integrals. It says that (Note that in order to evaluate this, the function f must be continuous for x [a, b]) a and b are, appropriately, called the limits of integration. The constant of integration cancels out when we do the subtraction. Either way, F(x) is the antiderivative of f(x).* IB uses the notation Some integrals cannot be evaluated using analytical techniques. Consider, for example, Finding a function that differentiates to give xe x (it's antiderivative) is no trivial matter. But if we have access to a good calculator and if the integral is definite, we can approximate the integral with the calculator: MATH/fnInt (option 9) The parameters are: fnint(<function>, <variable>, <lower limit>, <upper limit>) Recall that definite integrals have various properties that we have already proven. Use these liberally (in both directions) to simplify your life. This assignment is another "practice for mastery" assignment. The antiderivatives are not difficult. Rather, the point is to solidify your deeper understanding of properties of integration. You will need to come back to these properties fluidly in the future. Use your calculator as instructed. 18H: #1 4,5cfil,6 16even (Definite integrals) Treat #1 4 as "show that" problems to strengthen your thinking Quiz Fri Integration mechanics 1/31 18H Definite Integrals 8

Quiz today, then... 19A: #1cd,2ghij,3,4 (Area under curve) Do QB #22,25,26,37,40 See QB #2d,9d,27ef,29,34,35de,40d,43,49,50 1. Use integration to find areas under curves. 2. Gain mastery over using the calculator to find difficult definite integrals. Nothing especially new here using what you know to find areas under curves. Identify the function in question Sketch a diagram! Find the limits of integration (might have to solve an equation, maybe twice!) Identify how to do the integration: > Is this a case of f(ax + b)? > Is substitution needed? > Do I need (can I use?) a calculator? Find the antiderivative Use the Fundamental Theorem to evaluate at the limits > Careful with negative signs! Answer the question. See if it makes sense. Include units if appropriate. In some cases, you will need to find the limits of integration. And, as always, be careful if the function goes negative. If it does, the integral will be negative and the area is the opposite of that. Recall how to find a definite integral (approximately) with a calculator: MATH/fnInt (option 9) The parameters are: fnint(<function>, <variable>, <lower limit>, <upper limit>) How can you find area between a function and a curve with your calculator, even if the function changes sign over the interval of integration? Answer: Integrate (or graph) f(x) instead of f(x). Use the MATH/NUM/ABS. 19A: #1cd,2ghij,3,4 (Area under curve) Do QB #22,25,26,37,40 See QB #2d,9d,27ef,29,34,35de,40d,43,49,50 2/5 19A Area under a curve 9

19A: #1cd,2ghij,3,4 (Area under curve) Present 3,4 Do QB #22,25,26,37,40,50 Present all See QB #2d,9d,27ef,29,34,35de,43,49 1. Use integration to find areas between two curves. 19B: #2 16 even (Area between curves) QB #16,17ab The ideas behind finding area under a curve can be extended to find the area between two curves. Consider: The "height" of the area at any value x is the difference between the two functions: f(x) g(x). We need to be careful that this difference has the same sign over the interval being integrated. But in this case the sign of the individual functions is not what's important just that the difference has the same sign. Area Between Two Curves The area between two curves f and g in the interval [a, b] when f(x) g(x) in the interval is given by: f(x) g(x) is sometimes called the difference function. You need to do the subtraction in the correct direction and ensure that f is above or on g over the entire interval. You often need to find intersections to identify limits of integration. Sketch a graph! What would you do if f is not above g over the entire interval? You guessed it break into multiple intervals. Try y = 2x + 6 instead of the x axis To find the area between two curves on a calculator, enter or graph the absolute value of the difference function. 19B: #2 16 even (Area between curves) QB #16,17ab 2/5 19B Area between curves 10

19B: #2 16 even (Area between curves) Present #10,16 QB #16,17ab Mini Quiz 1. Answer the following concisely and precisely. (a) What is the graphical significance of the second derivative of a function? (b) What is the graphical significance of a point of inflection? (c) What are the two mathematical requirements for a point to be a point of inflection? (d) What is true about the second derivative at a maximum? At a minimum? 2. 1. (a) It shows the curvature of a function. f''(a) > 0 means the curve is concave up at x = a f''(a) < 0 means the curve is concave down at x = a (b) It is a point on the curve where the curvature changes from concave up to concave down or vice versa. (c) If x = a is a POI, f''(a) = 0 and; f'' changes sign at x = a (sign diagram) or f'''(a) 0 (d) f''(a) < 0 if x = a is a maximum. f''(a) > 0 if x = a is a minimum 2. 1. Use integration to find distance travelled and displacement from velocity and accelleration. 19C.1: #1 3 (Distance from velocity) QB #5 or 7 or 20 and 11,15 19C.2: #2,4,6,7 (More motion problems) Calculus Unit Exam: Next Fri 2/16 Meanwhile, back to Elliot in his speeding sports car. We know that velocity is the rate of change of position that is: These ideas, in reverse, help us to find distance travelled, change in position, and change in velocity for some function that describes motion. A big idea in physics. Let's look. Be careful with the difference between distance travelled and total displacement (change in position) When the integrand changes sign and you want distance travelled, you need to: a) Use the absolute value on your calculator or b) break the integral into parts The book describes a technique that involves finding different constants of integration for different intervals. I prefer to just break up the integral into different sections. In any case, you need to analyze the sign diagram of the integrand. Let's look first at the idea geometrically (without integration...) You could be clever and integrate the function from 0 to 4, then add twice the integral from 1 to 2. Do you see why? Imagine variants on this idea. Note calculator approach if it's available. In that case, you can use bars! 19C.1: #1 3 (Distance from velocity) QB #5 or 7 or 20 and 11,15 19C.2: #2,4,6,7 (More motion problems) Calculus Unit Exam: Fri 2/16 2/7 19C Kinematics 11

19C.1: #1 3 (Distance from velocity) QB #5 or 7 or 20 and 11,15 Present all 19C.2: #2,4,6,7 (More motion problems) Present #7 Short class: 40 minutes! Calculus Unit Exam: Fri 2/16.1 19D.1: #1bdfh,2,4 8 all (Volume of revolution) 19D.2: #2,3,4,6 (Volume of revolution 2 functions) QB 17c 1. Use integration to find volumes of revolution around the x axis. 2. Use integration to find the volume of revolution between two curves. Before we begin, a little reminder about how to find the volume of a prism. What's a prism? A major use of integration is to calculate and ultimately optimize use of materials in manufacturing. Consider a Hershey's Kiss. The volume of chocolate can be calculated if we know the function, f(x), that describes a cross section that is "spun" an axis. We call these objects radially symmetric. Imagine rotating the curve around the x axis to create a solid of revolution. We can use an integral to sum up successive infinitesimally thin discs between a and b to get the volume created At any point x along the curve, the radius of the kiss is defined by the value of the function at that value of x. In other words, r = f(x). The volume of a single disc is approximately the area of its base times its height. A base π r 2 = π f(x) 2 h = δx So the volume of one disc can be approximated by: V = π r 2 δx and the total volume as the sum. If we take the limit as the thickness (δx) approaches zero, we get: Volume of a solid defined by f(x) rotated around the x axis from a to b Use this idea to find a formula for the volume of a cone of height h and base radius r ⅓π r 2 h The key, of course, is the complexity of the resulting integral which will depend on the defining function. With a little imagination, one can also calculate the volume of a solid created by rotating a function around the y axis. If the original function is given as y = f(x), you will need to find the inverse function, x = f(y) as the discs will be oriented horizontally: Volume of a solid defined by f(x) rotated around the y axis from a to b In some texts this method is called the disk method of finding volumes. Understand where these ideas come from and the formulas will be unnecessary. 2/9 19D.1 Volumes of Revolution 12

.2 1. Use integration to find volumes of revolution between two functions. The plot thickens as the volumes we want to calculate become more complex. To make something hollow, we can use two defining functions. Similarly to what we saw calculating areas between two curves, rotating the upper function will generate one volume, while rotating the lower function will generate another volume to subtract from the first. This can be done in two steps or, sometimes more simply, by subtracting the functions before integrating. As was the case with areas, we need to pay close attention to which curve is the upper one. If that changes over the interval of integration, you must break the integral into different parts and change the direction of the subtraction. In some texts, this method is called the washer method of finding volumes. The primary issue here, like with problem solving, is getting the proper integral set up. Often you will have technology to help evaluate it. So focus on setting up the integrals properly. 19D.1: #1bdfh,2,4 8 all (Volume of revolution) 19D.2: #2,3,4,6 (Volume of revolution 2 functions) QB 17c Mon 2/12: Academic reflections (no classes) Wed 2/14: Review Fri 2/16: Calculus Unit Exam 2/9 19D.2 Volumes w/2 functions 13