1. Definite Integrals
Finney Chapter 6.2

2. Riemann Sums
The sums you studied in the last section are called Riemann Sums
When studying area under a curve, we consider only intervals over which the function has positive values because area must be positive
Riemann Sums are more general and we may use any interval over which a function is continuous
Intervals over which a function has negative values give the negative value of the area under the curve for that interval
What follows is a description of how a Riemann sum is created

3. Riemann Sums
Creating a Riemann Sum is similar to what you practiced in the previous section
For a function 𝑓 over a continuous interval [𝑎,𝑏], we divide the interval into 𝑛 subintervals by choosing 𝑛−1 points between 𝑎 and 𝑏, 𝑥 1 , 𝑥 2 , …, 𝑥 𝑛−1 such that 𝑎= 𝑥 0 < 𝑥 1 < 𝑥 2 <…< 𝑥 𝑛−1 < 𝑥 𝑛 =𝑏
We can think of this as a set, 𝑃={ 𝑥 0 , 𝑥 1 , 𝑥 2 ,…, 𝑥 𝑛 } that is called a partition of [𝒂,𝒃]

4. Riemann Sums
The 𝑛 subintervals are indexed by the right endpoint; that is, the first subinterval is [ 𝑥 0 , 𝑥 1 ], the second subinterval is [ 𝑥 1 , 𝑥 2 ], and the 𝑘th subinterval is [ 𝑥 𝑘−1 , 𝑥 𝑘 ]
In a Riemann Sum, each subinterval can have any width
Hence, the first subinterval has width Δ 𝑥 1 , the second subinterval has width Δ 𝑥 2 , the 𝑘th subinterval has width Δ 𝑥 𝑘 (again, these values need not be equal)
A value is selected from anywhere in each subinterval; the value from the 𝑘th subinterval is denoted 𝑐 𝑘 ; that is, 𝑐 𝑘 is contained in [ 𝑥 𝑘−1 , 𝑥 𝑘 ]

5. Riemann Sums
At each subinterval we construct a rectangle such that the height of the rectangle is 𝑓( 𝑐 𝑘 ) and the width is Δ 𝑥 𝑘
The area of each such rectangle is 𝑓 𝑐 𝑘 𝛥 𝑥 𝑘 ; this value can be positive, negative, or zero
Finally we take the sum of these products 𝑆 𝑛 = 𝑘=1 𝑛 𝑓( 𝑐 𝑘 ) Δ 𝑥 𝑘
The value of the sum depends on the partition 𝑃 and the choice of numbers 𝑐 𝑘 and is called the Riemann Sum for 𝒇 on [𝒂,𝒃]

6. Riemann Sums
Since the subintervals may have different widths, rather than allowing the number of rectangles to approach infinity, we take the longest subinterval and allow it to approach zero
This longest subinterval is called the norm of the partition 𝑃 and denoted by | 𝑃 |
Regardless of the way we choose 𝑃 and each value of 𝑐 𝑘 , all Riemann sums converge to a common value as all subintervals tend to zero

7. The Definite Integral as a Limit of Riemann Sums
DEFINITION
Let 𝑓 be a function defined on a closed interval [𝑎,𝑏]. For any partition 𝑃 of [𝑎,𝑏], let the numbers 𝑐 𝑘 be chosen arbitrarily in the subintervals [ 𝑥 𝑘−1 , 𝑥 𝑘 ].
If there exists a number 𝐼 such that
lim | 𝑃 |→0 𝑘=1 𝑛 𝑓 𝑐 𝑘 Δ 𝑥 𝑘 =𝐼
no matter how 𝑃 and the 𝑐 𝑘 ’s are chosen, then 𝑓 is integrable on [𝑎,𝑏] and 𝐼 is the definite integral of 𝑓 over [𝑎,𝑏].

8. Theorem 1: Existence of Definite Integrals
THEORM
All continuous functions are integrable. That is, if a function 𝑓 is continuous on an interval [𝑎,𝑏], then its definite integral over [𝑎,𝑏] exists.
(Note that, since the definition allows us to choose any partition, this theorem allows us to go back to using regular partitions for which all subintervals have the same length.)

9. Theorem 1: Existence of Definite Integrals
The Definite Integral of a Continuous Function on [𝑎,𝑏]
Let 𝑓 be continuous on [𝑎,𝑏], and let [𝑎,𝑏] be partitioned into 𝑛 subintervals of equal length, Δ𝑥= 𝑏−𝑎 𝑛 . Then the definite integral of 𝑓 over [𝑎,𝑏] is given by
lim 𝑛→∞ 𝑘=1 𝑛 𝑓 𝑐 𝑘 Δ 𝑥 𝑘
where each 𝑐 𝑘 is chosen arbitrarily in the 𝑘th subinterval. Note that, since the norm of a partition is |𝑃| = 𝑏−𝑎 𝑛 , then
lim 𝑃 →0 𝑏−𝑎 𝑛 =0= lim 𝑛→∞ 𝑏−𝑎 𝑛

10. Terminology and Notation
Recall that we defined the derivative as lim Δ𝑥→0 Δ𝑦 Δ𝑥 = 𝑑𝑦 𝑑𝑥
This notation is due to Leibniz, who switched from Greek letters to Roman letters to denote the derivative
He also created notation for integrals
The Greek capital sigma was chosen to represent sums, and an elongated Roman s represents the integral

11. Terminology and Notation
lim 𝑛→∞ 𝑘=1 𝑛 𝑓( 𝑐 𝑘 ) Δ 𝑥 𝑘 = 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥
We read this as “the integral from a to b of f of x, dx” or as “the integral from a to b of f of x with respect to x”.
Note that 𝑎 and 𝑏 are the endpoints of the interval [𝑎,𝑏] and are written with the greater value on top and the lesser value on the bottom (we will see later than we can reverse this with an appropriate correction)

12. Terminology and Notation
lim 𝑛→∞ 𝑘=1 𝑛 𝑓( 𝑐 𝑘 ) Δ 𝑥 𝑘 = 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥
The greater value is called the upper limit of integration; the lesser value is the lower limit of integration
The symbol ∫ is called the integral sign
The function f is called the integrand
The notation dx denotes that x is the variable of integration

13. Terminology and Notation
lim 𝑛→∞ 𝑘=1 𝑛 𝑓( 𝑐 𝑘 ) Δ 𝑥 𝑘 = 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥
Because the value of a definite integral depends only on the function, the letter we use as the variable of integration does not matter and is called the dummy variable
Hence, for a given function f over an interval 𝑎,𝑏 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥= 𝑎 𝑏 𝑓(𝑡) 𝑑𝑡= 𝑎 𝑏 𝑓(𝑢) 𝑑𝑢

14. Terminology and Notation
lim 𝑛→∞ 𝑘=1 𝑛 𝑓( 𝑐 𝑘 ) Δ 𝑥 𝑘 = 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥
Finally, a note about the difference between an indefinite integral (or antiderivative) and a definite integral:
Taking an indefinite integral results in a family of functions that differ by a constant
Taking a definite integral results in a real number
You will see how these are related to each other and to derivatives in the next section

15. Example 1
The interval [−1,3] is partitioned into 𝑛 subintervals of equal length, Δ𝑥= 3− −1 𝑛 = 4 𝑛 . Let 𝑚 𝑘 denote the midpoint of the 𝑘th subinterval. Express the limit lim 𝑛→0 𝑘=1 𝑛 (3 𝑚 𝑘 2 −2 𝑚 𝑘 +5) Δ𝑥 as an integral.

16. Example 1
Using lim 𝑛→∞ 𝑘=1 𝑛 𝑓( 𝑚 𝑘 ) Δ𝑥 = 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥 over [−1,3] we get lim 𝑛→∞ 𝑘=1 𝑛 3 𝑚 𝑘 2 −2 𝑚 𝑘 +5 Δ𝑥 = −1 3 (3 𝑥 2 −2𝑥+5) 𝑑𝑥

17. Area Under a Curve DEFINITION
If 𝑦=𝑓(𝑥) is nonnegative and integrable over a closed interval [𝑎,𝑏], then the area under the curve 𝒚=𝒇(𝒙) from 𝒂 to 𝒃 is the integral of 𝑓 from 𝑎 to 𝑏,
𝐴= 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥
(Note that we can use integrals to calculate area, and we can use area to calculate integrals.)

18. Example 2
Evaluate the integral −2 2 4− 𝑥 2 𝑑𝑥.

19. Example 2
The graph is the upper half of a circle (i.e., a semi-circle) with radius 2. Since we know that the area of semi-circle is 𝐴= 𝜋 𝑟 2 2 and that the area under the curve is 𝐴= −2 2 4− 𝑥 2 𝑑𝑥 then −2 2 √(4− 𝑥 2 ) 𝑑𝑥= 𝜋 =2𝜋

20. Definite Integral and Area
The value of an integral over which a continuous function 𝑓 has negative values is a negative number
Of course, this interval also covers an area
But since we want area to be a positive value, if 𝑓 𝑥 <0
Area=− 𝑎 𝑏 𝑓 𝑥 𝑑𝑥

21. Definite Integral and Area
The value of integrable functions that have both positive and negative values on [𝑎,𝑏] is
𝑎 𝑏 𝑓(𝑥) 𝑑𝑥= area above 𝑥-axis − area below 𝑥-axis

22. Theorem 2: Constant Functions
If 𝑓 𝑥 =𝑐, where 𝑐 is a constant, over [𝑎,𝑏], then
𝑎 𝑏 𝑓(𝑥) 𝑑𝑥= 𝑎 𝑏 𝑐 𝑑𝑥=𝑐(𝑏−𝑎)
PROOF
Using a Riemann Sum with regular partitions, we have Δ𝑥= 𝑏−𝑎 𝑛 . The definite integral is then
𝑎 𝑏 𝑐 𝑑𝑥= lim 𝑛→∞ 𝑘=1 𝑛 𝑐 ⋅ 𝑏−𝑎 𝑛

23. Theorem 2: Constant Functions
PROOF
Note that 1 𝑛 is a constant with respect to summation, so
lim 𝑛→∞ 𝑘=1 𝑛 𝑐 ⋅ 𝑏−𝑎 𝑛 =𝑐(𝑏−𝑎) ⋅ 1 𝑛 ⋅ lim 𝑛→∞ 𝑘=1 𝑛 1 =𝑐 𝑏−𝑎 ⋅ 1 𝑛 ⋅𝑛=𝑐(𝑏−𝑎)
We conclude that
𝑎 𝑏 𝑐 𝑑𝑥=𝑐(𝑏−𝑎)

24. Theorem 2 Viewed Geometrically
It’s easy to show that theorem 2 holds by using geometry

25. Example 3: A Train Problem
A train moves along a track at a constant velocity of 75 mph from 7:00 AM to 9:00 AM. Express its total distance traveled as an integral. Evaluate the integral using Theorem 2.

26. Example 3: A Train Problem
By Theorem 2, with 𝑐=75, 𝑎=7, and 𝑏= 𝑑𝑥=75 9−7 =75⋅2=150 More importantly, note the units: 75 miles 1 hour ⋅2 hours=150 miles

27. Example 3: A Train Problem

28. Example 3: A Train Problem
Let’s try to make some connections among the things you have learned
The integral is defined as the limit of a Riemann Sum when the number of rectangles approaches infinity
A Riemann Sum gives the area under a curve (or the negative value of that area if the curve lies under the x-axis)
In the train problem, the integral is a distance, specifically, the distance that the train traveled from 7:00 AM to 9:00 AM
But distance traveled over a given period of time is given by the position function, not the velocity function
On the other hand, we know that the velocity function is obtained as derivative of the position function
What seems to be the relation between the definite integral ( and hence, the area under the curve) of a velocity function and the position of the train?

29. Integrals on a Calculator
As we proceed through this chapter, and later on the AP exam, there will be instances where it is either very difficult or impossible to calculate a definite integral
You must learn how to use your calculator to determine integrals
On a TI84, finding a definite integral is intuitive; just remember that this function is accessed through the MATH key

30. Integrals on a Calculator
Directions for using the TI89 are below (from the TI89 manual, which can be downloaded from the internet)

31. Example 4: Evaluating Definite Integrals on a Calculator
Use a calculator to evaluate the following integrals numerically. (a) −1 2 𝑥 sin 𝑥 𝑑𝑥 (b) 𝑥 2 𝑑𝑥 (c) 0 5 𝑒 − 𝑥 2 𝑑𝑥

32. Example 4: Evaluating Definite Integrals on a Calculator
Evaluate the following integrals numerically. (a) −1 2 𝑥 sin 𝑥 𝑑𝑥≈2.043 (b) 𝑥 2 𝑑𝑥≈ (c) 0 5 𝑒 − 𝑥 2 𝑑𝑥≈0.886

33. Example 4: Evaluating Definite Integrals on a Calculator
Integrals can also be evaluated from the Graph screen. This will not only give the value of the integral between the limits of integration, but will also shade the area. Note that 𝑒 − 𝑥 2 is the basic form of the Gaussian curve (or the normal curve). It turns out that it is impossible to find an explicit value for this integral. If we want to know the area under the curve, we must approximate it using a Riemann Sum (and the calculator does this much more efficiently that any human ever could). You will encounter many more functions like this, so knowing how to use the calculator is essential.

34. Discontinuous Integrable Functions
By Theorem 1, a function that is continuous over an interval is guaranteed to be integrable
That is, continuity implies integrability (compare this with what we learned about derivatives: differentiability implies continuity)
But it is sometimes the case that a function that has a discontinuity over some interval can still be integrated
In general, some functions with jump discontinuities or point discontinuities (but not unbounded discontinuities) are integrable

35. Example 5: Integrating a Discontinuous Function
Find −1 2 𝑥 𝑥 𝑑𝑥

36. Example 5: Integrating a Discontinuous Function
We use the fact that the area (actually, the net area) under a curve over a given interval is equal to the definite integral evaluated over the interval. The given function is equal to 1 for 𝑥>0 and to −1 for 𝑥<0; it is not defined at 𝑥=0 and this is where the discontinuity occurs.

37. Example 5: Integrating a Discontinuous Function

38. Example 5: Integrating a Discontinuous Function
Since the net area (as determined by 𝐴=𝑏ℎ and knowing that the area between −1 and 0 is negative) is 2+ −1 =1, then −1 2 𝑥 𝑥 𝑑𝑥=1

39. Definite Integral as an Accumulator Function
Example 1 from section 6.1 asked you to find the amount of snow that had fallen from 3 A.M. to 10:30 A.M.
This was solved by finding the area under the graph showing the rate of snowfall over this interval of time
If 𝑆(𝑇) represents the amount of snow that has fall by time 𝑇, then we can express this as an integral 𝑆 𝑇 = 3 𝑇 𝑟(𝑡) 𝑑𝑡 where 𝑟(𝑡) is the rate of snowfall at time 𝑡

40. Definite Integral as an Accumulator Function
The definite integral makes it possible to build new functions called accumulator functions
Note that this is a function of the value of 𝑇 𝑆 𝑇 = 3 𝑇 𝑟(𝑡) 𝑑𝑡
That is, 𝑆 is a function of 𝑇, which is the upper limit of integration in the integral
In this case, we use a different variable of integration, 𝑡, which is a dummy variable

41. Definition: Accumulator Function
If the function 𝑓 is integrable over the closed interval [𝑎,𝑏], then the definite integral of 𝑓 from 𝑎 to 𝑥 defines a new function for 𝑎≤𝑥≤𝑏, the accumulator function
𝐴 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡

42. Example 4: An Accumulator Function
Let 𝑓 𝑡 =1+2𝑡 for 1≤𝑡≤6. Write an accumulator function 𝐴(𝑥) as a polynomial in 𝑥 where 𝐴 𝑥 = 1 𝑥 (1+2𝑡) 𝑑𝑡, 1≤𝑥≤6

43. Example 4: An Accumulator Function

44. Example 4: An Accumulator Function
The area over the interval is a trapezoid with height 𝑥−1 and bases =3 and 1+2𝑥. The area of the trapezoid is the accumulation from 1 to 𝑥 over the interval 1,𝑥 . This area is 𝐴 𝑥 = 𝑥− 𝑥 2 = 𝑥−1 𝑥+2 = 𝑥 2 +𝑥−2

45. Exercise 6.2
Online