1. Unit: Integration Scott Pauls Department of Mathematics
Version 1, August 2016
Scott Pauls
Department of Mathematics
Dartmouth College

2. Instructor’s overview - 1
These slides are not meant to be prescriptive, but serve as a skeletal outline of the course syllabus using active learning methods throughout the class.
In these slides we assume that most classes have the same general format:
(5-10 minutes) A brief introductory mini-lecture, often recapitulating and extending the end of the previous class, leading into the first group activity.
(10-15 minutes) Group activity meant to explore a new topic or extend and old one.
(5-10 minutes) A short follow up on the results and lead-in to a second piece which is sometimes lecture/discussion and sometimes another period for group work.
(10-15 minutes) Whatever the second piece is.
(5-10 minutes) Follow up on second piece
(5-10 minutes) A short introduction to the homework, preparation, and next piece of material for the next class.
This format is the result of tinkering with a mixed lecture format for 3 terms. As such, it is still a work in progress and we will discuss adaptations both to the general format as well as for individual topics throughout the term.

3. Instructor’s overview – 2
The problems used in the slides below are indicative what level we expect throughout the course. Instructors will want to supplement these with both easier and harder examples as class progress dictates. One goal in providing these templates is to help ensure a uniform level of instruction to which we can test. For each set of group work, there are a sequence of questions, ordered by difficulty. Each one is aimed to build up to the next with the goal of completing the last one, which is the most abstract. Those that are circled are candidates for write-up and grading by group.
As we continue to refine MATH 3, we want to standardize the course a bit more than in the past. These templates are meant to help keep all instructors roughly on the same page so that when we get to exams, there are few surprises for both instructors and students. As you’ll see, this is a skeleton, there is still lots of room for your own take on the material.

4. Average final mastery by topic (F15)
We’ve highlighted the topics that students had the most trouble with. Consequently, we now spend more time on these topics.
For integration topics, most students master these just fine, the lower averages simply reflect the shorter time available to practice.
Midterm 1
Midterm 2

5. Anti-differentiation and indefinite integrals
𝑑 𝑑𝑥 𝐹 𝑥 = 𝐹 ′ 𝑥 =𝑓 𝑥 ∫𝑓 𝑥 𝑑𝑥=𝐹 𝑥 +𝐶 Examples:
To introduce class 19, we discuss indefinite integrals in terms of the process of anti-differentiation. A key point to stress is that for every differentiation identity they know, there is a corresponding anti-differentiation identity.
This helps, for example, to demonstrate where the constant of integration comes from as the derivative of a constant is zero.
This also foreshadows u-substitution.
Remind students of our earlier example which hinted at anti-differentiation – our falling boulder problem for related rates.
Function
Antiderivative 𝑥
𝑥 𝐶
sin⁡(𝑥)
− cos 𝑥 +𝐶
cos⁡(𝑥)
sin 𝑥 +𝐶
𝑒 𝑥
𝑒 𝑥 +𝐶

6. Rectilinear motion
If 𝑝(𝑡) gives the position of an object in motion at time 𝑡, then we know that 𝑣 𝑡 =𝑝′(𝑡) is its velocity and 𝑎 𝑡 = 𝑣 ′ 𝑡 =𝑝′′(𝑡) is its acceleration. Using anti-differentiation, we can start with the acceleration due to gravity and work our way backwards. If 𝑎 𝑡 =−9.8 𝑚/ 𝑠 2 then 𝑣 𝑡 =−9.8𝑡+𝐶. What is the constant? If we know 𝑣 0 = 𝑣 0 then 𝐶= 𝑣 0 . Continuing, we anti-differentiate again yielding 𝑝 𝑡 =−− 𝑡 2 + 𝑣 0 𝑡+𝐷. Again, if we know 𝑝 0 = 𝑝 0 then 𝐷= 𝑝 0 .
Contextualize this computation with the idea of differential equations and their solutions. This is a simple application but is at the heart of kinematics.
𝑣 0
𝑝 0

7. Group Work: The Siege of Syracuse
If the trajectory of a boulder is given by (𝑥 𝑡 ,𝑦 𝑡 ) with 𝑥 0 =𝑦 0 =0 and we set 𝑦 ′′ 0 =−9.8 𝑚/ 𝑠 2 and 𝑥 0 =0, what are the functions 𝑥(𝑡) and 𝑦 𝑡 ?
If, as in the diagram, the catapult launches with a fixed speed but variable angle 𝜃, what are 𝑥′(0) and 𝑦 ′ 0 ?
Given all of this information, where does the boulder hit the x-axis?
Suppose a ship sits at (100 𝑚,0) and you have a catapult that launches with an initial speed of 𝑠=25. How do we set up the catapult to hit the ship? 𝑠
𝑦 ′ 𝑡 𝜃
This is a long problem, yet again with Archimedes – this time defending Syracuse from Roman attack in 214 BCE. Wikipedia summary: The Siege of Syracuse by the Roman Republic took place in 214–212 BC, at the end of which the Magna Graecia Hellenistic city of Syracuse, located on the east coast of Sicily, fell. The Romans stormed the city after a protracted siege giving them control of the entire island of Sicily. During the siege, the city was protected by weapons developed by Archimedes. Archimedes, the great inventor and polymath, was slain at the conclusion of the siege by a Roman soldier, in contravention of the Roman general Marcellus' instructions to spare his life.
One of the (many) techniques Archimedes devised to protect Syracuse was the computation of the trajectory of boulders launched catapults, which allowed the soldiers of Syracuse to accurately aim at the ships against them. In this group work, we’ll use the notion of a parameterized curve in the plane to work out what the trajectory of a boulder would look like given initial conditions on velocity.
Setup:
We think of the curve traced out by (𝑥 𝑡 ,𝑦 𝑡 ) with 𝑥 0 =𝑦 0 =0 (for convenience). In the usual coordinate system, gravity acts in the y direction: 𝑦 ′′ 𝑡 =−9.8 𝑚 𝑠 Moreover, we assume that there are no forces impacting travel in the x direction: 𝑥 ′′ 𝑡 =0.
The group work leads student through the following computation. The computation for 𝑦 is the same as the last slide, yielding 𝑦 𝑡 =− 𝑡 2 + 𝑦 ′ 0 𝑡. The computation for 𝑥 yields 𝑥 𝑡 = 𝑥 ′ 0 𝑡. From this we can compute the place where the curve hits the x-axis again. First, find 𝑦 𝑡 =− 𝑡 2 + 𝑦 ′ 0 𝑡=0 yielding 𝑡=0, 2𝑦 ′ Plugging the latter into the formula for x gives 𝑥 2 𝑦 ′ =2 𝑥 ′ 0 𝑦 ′
From the diagram, we have 𝑥 ′ 0 =𝑠 𝑐𝑜𝑠 𝜃 and 𝑦 ′ 0 =𝑠 sin 𝜃 making the x coordinate at landing 2 𝑠 2 sin(𝜃)cos(𝜃)/9.8. So, if the ship is at 𝑥= 𝑥 0 then we have sin 2𝜃 = 9.8 𝑥 0 2 𝑠 2 or 𝜃= sin − 𝑥 0 2 𝑠
𝑥 ′ 𝑡
( 𝑥 ′ 𝑡 , 𝑦 ′ 𝑡 )

8. Challenge problems
In the catapult problem, we fixed the trajectory on a plane. Can you generalize to all of three-dimensional space?
What is the maximum range of a catapult with speed 𝑠? How is that range achieved?
What are the benefits and drawbacks of placing catapult higher or lower than the sea level?
As with the other challenge problems, these are good to throw out there but probably shouldn’t take up much class time.

9. Finding areas
What is the area of an object in the plane? Simplification: What is the area between a curve 𝑦=𝑓(𝑥) and the 𝑥-axis? A question we can answer: What is the area of a rectangle?
At the end of class 19, introduce the application of finding areas under curves as a step towards finding areas of objects generally. We focus on the basic idea of breaking up areas into approximating rectangles. Here’s a good place to remind them of Archimedean exhaustion in the calculation of 𝜋 as an example of breaking things up into simpler approximating objects.
While we give the formal definition at the beginning of next class, this is a good time to show how to define the rectangles and motivate the sum that will arise in the general formula.

10. Riemann sums 𝐴=ℎ𝑤= 𝑒 −2 ⋅1 ℎ=𝑓 2 = 𝑒 −2 w=Δ𝑥=1
𝑦= 𝑒 −𝑥
𝐴=ℎ𝑤= 𝑒 −2 ⋅1
ℎ=𝑓 2 = 𝑒 −2
After the recap at the beginning of class 20, we give the formal definition of Riemann sums. In this slide, we give the left endpoint formulation. It is worth covering this in detail as students tend to have difficulty in several areas: the amount of abstraction, finding the left endpoints from a list of abstractly defined points, and converting to summation notation.
With four rectangles (as in the figure) the left endpoint Riemann sum is
𝑒 −0 ⋅1+ 𝑒 −1 ⋅1+ 𝑒 −2 ⋅2+ 𝑒 −3 ⋅1 = 𝑖=1 4 𝑒 − 𝑖−1 Δ𝑥
w=Δ𝑥=1

11. Riemann sums
For 𝑛 rectangles, Δ𝑥= 𝑏−𝑎 𝑛 . Sample points are given by: {𝑎,𝑎+Δ𝑥, 𝑎+2Δ𝑥, …,𝑎+𝑛Δ𝑥=𝑏} Left endpoints: 𝑎,𝑎+Δ𝑥, 𝑎+2Δ𝑥, …,𝑎+ 𝑛−1 Δ𝑥 Right endpoints: {𝑎+Δ𝑥, 𝑎+2Δ𝑥, …,𝑎+𝑛Δ𝑥=𝑏} Area of 𝒌 𝒕𝒉 box: Left endpoints: 𝑓 𝑎+ 𝑘−1 Δx Δ𝑥= 𝑓 𝑎+ 𝑘−1 𝑏−𝑎 𝑛 𝑏−𝑎 𝑛 Right endpoints: 𝑓 𝑎+kΔx Δ𝑥= 𝑓 𝑎+ 𝑘 𝑏−𝑎 𝑛 𝑏−𝑎 𝑛 Riemann sums: Left endpoints: 𝑘=1 𝑛 𝑓 𝑎+ 𝑘−1 𝑏−𝑎 𝑛 𝑏−𝑎 𝑛 Right endpoints: 𝑘=1 𝑛 𝑓 𝑎+ 𝑘 𝑏−𝑎 𝑛 𝑏−𝑎 𝑛
This slide gives all the pieces for putting together the Riemann sums. We should accompany this with several concrete examples to help students get a feel for the notation. For most of them, summation notation is new.

12. Riemann sums Group work Let 𝑓 𝑥 = 𝑥 2 −𝑥+1.
Estimate 1 3 𝑓 𝑥 𝑑𝑥 using Riemann sums with right endpoints with 3 rectangles.
Write down the Riemann sum with left endpoints for 1 3 𝑓 𝑥 𝑑𝑥 with 𝑛 rectangles.
Write down the Riemann sum with right endpoints for 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 with 6 rectangles.
Write down the Riemann sum with left endpoints for 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 with 𝑛 rectangles.
This set of group exercises is meant to help students familiarize themselves with the notation and manipulations necessary to write down Riemann sums. This is generally difficult for our students – it is rarely covered in high school calculus classes and is a fairly large conceptual jump.
It is good to emphasize that this is the beginning of numerical integration and that the techniques we learn to do exact integration are very limited – not many integrals can be directly evaluated. We’ll return to this next class with the integral of the Gaussian distribution.

13. Evaluating definite integrals
Find I= 1 4 𝑥 3 −4𝑥 𝑑𝑥. Left hand endpoints: 𝐼= lim 𝑛→∞ 𝑖=1 𝑛 1+(𝑖−1) 3 𝑛 3 −4 1+(𝑖−1) 3 𝑛 3 𝑛 Right hand endpoints: 𝐼= lim 𝑛→∞ 𝑖=1 𝑛 1+𝑖 3 𝑛 3 −4 1+𝑖 3 𝑛 3 𝑛
In the second portion of lecture in class 20, we should walk through a problem of moderate difficulty where we can use the summation formulae to the right to explicitly evaluate a definite integral. This is one such example – choose a different one if you like. This is a good point to again emphasize the connection to limits.

14. Definite integrals Group work
We’ll assign each group one of three simple
functions: 𝑓 𝑥 =𝑥, 𝑥 2 , or 𝑥 3 . For your function, complete the following problems:
Write down the definition of 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 using Riemann sums.
Find Δ𝑥 for the sum using 𝑛 rectangles and compute the heights of the rectangles using right or left-hand endpoints (your choice).
Simplify the summands using algebra and the formulae above.
Compute the resulting limits.
Compute 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 using Riemann sums.
In class 21, we’ll continue our discussion of Riemann sums. This is the first group work – there’s a lot of time between the last slide and this one – we’ll want to fill it with examples and more informal group work.
This set of problems will help solidify the skills needed to use Riemann sums effectively and to set the stage for the link between antiderivatives and definite integrals given by the FTC. This is a good set of problems to have different groups present different pieces, continuing to emphasize clear writing and presentation as well as demonstrating the various calculations. Be sure to point out the similarities between the answers and the anti-derivatives of these functions.
This problem, and its support, can easily take up most of this class.

15. The Fundamental Theorem of Calculus
Unless we need more time (class 22 is built as a catch-up day of sorts) at the end of class 21 we’ll introduce the Fundamental Theorem of Calculus to bring together our a priori different investigations of definite and indefinite integrals.
There’s a lot of ways to approach this but we should all emphasize some common themes:
The connection between anti-derivatives and areas at first seems somewhat miraculous. There are some nice proofs of these two in the text and, if your class seems up for it, exploring the proofs isn’t a bad idea as they again link together limits, the definition of the derivative, and the definition of the integral.
The FTC places a lot of our attention on finding anti-derivatives – we’ll spend the rest of the term thinking about this. But, many functions don’t have elementary anti-derivatives. This tension points to the need for robust approximation methods, of which Riemann sums is the beginning.

16. The Fundamental Theorem of Calculus
Group Work
Pick on of the graphs to the right to be 𝑓(𝑥) and let 𝑔 𝑥 = 0 𝑥 𝑓 𝑡 𝑑𝑡 .
At what values of 𝑥 do local maxima and minima of 𝑔(𝑥) occur?
Where does 𝑔(𝑥) obtain its global maximum on this interval?
On which intervals is 𝑔 𝑥 concave up and concave down?
Sketch a graph of 𝑔 𝑥 .
By class 23, we should move entirely to the FTC. This group work is meant to help develop the basic skills in using the FTC effectively.

17. Calculating areas with Riemann sums
Consider the Gaussian function 𝑓 𝑥 = 1 2 𝜎 2 𝜋 𝑒 𝑥−𝜇 2 2𝜎 2 1.
This is a nice problem to show the necessity of Riemann sums in the context of the (failure of) the FTC. This example also provides a link to probability and statistics.
To motivate this slide, we will introduce the Gaussian 𝑓 𝑥 = 𝜎 2 𝜋 𝑒 𝑥−𝜇 𝜎 2 as the function that helps describe the normal distribution. While we don’t have time to go into a huge amount of detail, the main points to cover are:
The idea of a probability distribution and, in particular, the normal distribution. Give (general) examples where the normal distribution comes up (either explicitly or as a model): height distributions, SAT scores, least squares parameter fitting, etc.
How we use the pdf to compute the cdf and the relation to computing probabilities: 𝑃 𝑥≤𝜏 = −∞ 𝜏 𝜎 2 𝜋 𝑒 𝑥−𝜇 𝜎 2 𝑑𝑥
There is no simple anti-derivative for 𝑓(𝑥) so we can’t easily compute an indefinite integral. For the definite integral, we need to use approximation.

18. Integration techniques: 𝑢-substitution
FTC applied to the chain rule: 𝑓 𝑔 𝑥 ′ = 𝑓 ′ 𝑔 𝑥 𝑔 ′ 𝑥 ⟹ 𝑎 𝑏 𝑓 ′ 𝑔 𝑥 𝑔 ′ 𝑥 𝑑𝑥=𝑓 𝑔 𝑏 −𝑓(𝑔 𝑎 )
To set up class 24, we discuss using the FTC on differentiation rules to produce integration rules. Our first application is transforming the chain rule into u-substitution.

19. Substitution
Group work

20. Integration techniques: integration by parts
FTC applied to the product rule: 𝑓 𝑥 𝑔 𝑥 ′ = 𝑓 ′ 𝑥 𝑔 𝑥 +𝑓 𝑥 𝑔 ′ 𝑥 ⟹ 𝑎 𝑏 𝑓 ′ 𝑥 𝑔 𝑥 +𝑓 𝑥 𝑔 ′ 𝑥 𝑑𝑥=𝑓 𝑏 𝑔 𝑏 −𝑓 𝑎 𝑔(𝑎) Rewritten: 𝑎 𝑏 𝑓 𝑥 𝑔 ′ 𝑥 𝑑𝑥=𝑓 𝑥 𝑔 𝑥 | 𝑎 𝑏 − 𝑎 𝑏 𝑓 ′ 𝑥 𝑔 𝑥 𝑑𝑥
We end class 24 by introducing integration by parts.

21. Integration by parts
Group work