1. Chapter 6 The Definite Integral

2. Chapter Outline Antidifferentiation
The Definite Integral and Net Change of a Function
The Definite Integral and Area under a Graph
Areas in the xy-Plane
Applications of the Definite Integral

3. § 6.1
Antidifferentiation

4. Section Outline Antidifferentiation Finding Antiderivatives
Theorems of Antidifferentiation
The Indefinite Integral
Rules of Integration
Antiderivatives in Application

5. Antidifferentiation - Antiderivative
Definition
Example
Antidifferentiation: The process of determining f (x) given f ΄(x)
If , then

6. Finding Antiderivatives
EXAMPLE
Find all antiderivatives of the given function.
SOLUTION
The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant*) as we will see next.
*The derivative of a constant function is zero.

7. Theorems of Antidifferentiation

8. The Indefinite Integral

9. Rules of Integration
Power Rule:
Exponential Rule:
Log Rule:

10. Rules of Integration

11. Finding Antiderivatives
EXAMPLE
Determine the following.
SOLUTION
Using the rules of indefinite integrals, we have

12. Finding Antiderivatives
EXAMPLE
Find the function f (x) for which and f (1) = 3.
SOLUTION
The unknown function f (x) is an antiderivative of One
antiderivative is Therefore, by Theorem I,
Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.

13. Finding Antiderivatives
CONTINUED
So, 3 = 1 + C and therefore, C = 2. Therefore, our function is

14. Antiderivatives in Application
EXAMPLE
A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second.
(a) Find s(t), the height of the rock above the ground at time t.
(b) How long will the rock take to reach the ground?
(c) What will be its velocity when it hits the ground?
SOLUTION
(a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = We can now use this information to find an antiderivative of v(t) for which s(0) = 400.
The antiderivative of v(t) is To determine C,
Therefore, C = So, our antiderivative is s(t) = -16t

15. Antiderivatives in Application
CONTINUED
(b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0.
s(t) = -16t
This is the function s(t).
0 = -16t
Replace s(t) with 0.
-400 = -16t2
Subtract.
25 = t2
Divide.
5 = t
Take the positive square root since t ≥ 0.
So, it will take 5 seconds for the rock to reach the ground.

16. Antiderivatives in Application
CONTINUED
(c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5).
v(t) = -32t
This is the function v(t).
v(5) = -32(5) = -160
Replace t with 5 and solve.
So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign).

17. § 6.2
Definite Integral and Net Change of a Function

18. Section Outline The Definite Integral
Properties of the Definite Integral
The Definite Integral as a Net Change of a Function

19. The Definite Integral

20. The Definite Integral Computing a Definite Integral Evaluate
EXAMPLE
Computing a Definite Integral
Evaluate
SOLUTION
An antiderivative of f(x) = x is

21. Properties of the Definite Integral

22. Using Properties of the Integral
EXAMPLE
Given that
SOLUTION
By (7), we have
Using the given values of the integrals, we find

23. The Definite Integral as a Net Change of a Function
In the definition of the definite integral (1), since f = F, we can rewrite (1) as
(8)
Recall that the derivative is the rate of change of F. Thus, (8) states that the integral of the rate of change of F is its net change as x varies from a to b.

24. § 6.3
Definite Integral and Area under a Graph

25. Section Outline Area Under a Graph Calculating Definite Integrals
The Fundamental Theorem of Calculus
Area Under a Curve as an Antiderivative

26. Area Under a Graph Definition Example
Area Under the Graph of f (x) from a to b: An example of this is shown to the right

27. Theorem I: Area under a Graph

28. Area Under the Graph of a Function
EXAMPLE
Use Theorem 1 to compute the area of the shaded region under the graph of
f(x) = 3x2 + ex, from x = −1 to x = 1.
SOLUTION

29. Riemann Sums
In this section we will learn to estimate the area under the graph of f (x) from x = a to x = b by dividing up the interval into partitions (or subintervals),
each one having width where n = the number of partitions that
will be constructed. In the example below, n = 4.
A Riemann Sum is the sum of the areas of the rectangles generated above.

30. Riemann Sums to Approximate Areas
EXAMPLE
Use a Riemann sum to approximate the area under the graph f (x) on the given interval using midpoints of the subintervals
SOLUTION
The partition of -2 ≤ x ≤ 2 with n = 4 is shown below. The length of each subinterval is x1 x2 x3 x4 -2 2

31. Riemann Sums to Approximate Areas
CONTINUED
Observe the first midpoint is units from the left endpoint, and the midpoints themselves are units apart. The first midpoint is x1 = = = Subsequent midpoints are found by successively adding
midpoints: -1.5, -0.5, 0.5, 1.5
The corresponding estimate for the area under the graph of f (x) is
So, we estimate the area to be 5 (square units).

32. Approximating Area With Midpoints of Intervals
CONTINUED

33. Riemann Sums to Approximate Areas
EXAMPLE
Use a Riemann sum to approximate the area under the graph f (x) on the given interval using left endpoints of the subintervals
SOLUTION
The partition of 1 ≤ x ≤ 3 with n = 5 is shown below. The length of each subinterval is 1
1.4
1.8
2.2
2.6 3 x1 x2 x3 x4 x5

34. Riemann Sums to Approximate Areas
CONTINUED
The corresponding Riemann sum is
So, we estimate the area to be (square units).

35. Approximating Area Using Left Endpoints
CONTINUED

36. The Fundamental Theorem of Calculus

37. The Fundamental Theorem of Calculus
EXAMPLE
Use the Fundamental Theorem of Calculus to calculate the following integral.
SOLUTION
An antiderivative of 3x1/3 – 1 – e0.5x is Therefore, by the fundamental theorem,

38. The Fundamental Theorem of Calculus
EXAMPLE
(Heat Diffusion) Some food is placed in a freezer. After t hours the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where
(a) Compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2.
(b) What does the area in part (a) represent?
SOLUTION
(a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following.

39. The Fundamental Theorem of Calculus
CONTINUED
(b) Since the area under a graph can represent the amount of change in a quantity, the area in part (a) represents the amount of change in the temperature between hour t = 0 and hour t = 2. That change is degrees Fahrenheit.

40. § 6.4
Areas in the xy-Plane

41. Section Outline Properties of Definite Integrals
Area Between Two Curves
Finding the Area Between Two Curves

42. The Definite Integral
Geometric interpretation of the definite integral:

43. Area Between Two Curves

44. Finding the Area Between Two Curves
EXAMPLE
Find the area of the region between y = 2x2 – 4x+6 and y = – x2 + 2x + 1 from x = 1 to x = 2.
SOLUTION
Upon sketching the two graphs (Fig. 7), we see that f(x) = 2x2 − 4x + 6 lies above g(x) = −x2 +2x+1 for
1 ≤ x ≤ 2. Therefore, our formula gives the area of the shaded region as

45. Finding the Area Between Two Curves
EXAMPLE
Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure.
SOLUTION
Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore

46. Finding the Area Between Two Curves
CONTINUED
Therefore, to represent all the shaded regions, we have

47. Finding the Area Between Two Curves
EXAMPLE
Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v1(t) and v2(t), respectively, and v1(t) ≥ v2(t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v1(t) and y = v2(t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A?
SOLUTION
Since v1(t) ≥ v2(t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have

48. Finding the Area Between Two Curves
CONTINUED
But again, since v1(t) ≥ v2(t) for t ≥ 0, we know that So,
this implies that This means that the position of the first rocket is always at least as high (up in the air) as that of the second rocket. That is, the first rocket is always higher up than the second rocket (or at the same height).

49. § 6.5
Applications of the Definite Integral

50. Section Outline Average Value of a Function Over an Interval
Consumers’ Surplus
Future Value of an Income Stream
Volume of a Solid of Revolution

51. Average Value of a Function Over an Interval

52. Average Value of a Function Over an Interval
EXAMPLE
Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1.
SOLUTION
Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to
An antiderivative of 1 – x is Therefore,
So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.

53. Average Value of a Function Over an Interval
EXAMPLE
(Average Temperature) During a certain 12-hour period the temperature at time
t (measured in hours from the start of the period) was degrees. What was the average temperature during that period?
SOLUTION
The average temperature during the 12-hour period from t = 0 to t = 12 is

54. Definition: Consumers’ Surplus

55. Consumers’ Surplus (1 of 2)
EXAMPLE
Find the consumers’ surplus for the following demand curve at the given sales level x.
SOLUTION
Since 20 units are sold, the price must be
Therefore, the consumers’ surplus is

56. Consumers’ Surplus (2 of 2)
CONTINUED
That is, the consumers’ surplus is $20.

57. Future Value of an Income Stream

58. Future Value of an Income Stream
EXAMPLE
(Future Value) Suppose that money is deposited daily into a savings account at an annual rate of $ If the account pays 6% interest compounded continuously, approximately how much will be in the account at the end of 2 years?
SOLUTION
Divide the time interval from 0 to 2 years into daily subintervals. Each
subinterval is then of duration years. Let t1, t2, ..., tn be points chosen
from these subintervals. Since we deposit money at an annual rate of $2000, the amount deposited during one of the subintervals is dollars. If this amount is deposited at time ti, the dollars will earn interest for the remaining 2 – ti years. The total amount resulting from this one deposit at time ti is then

59. Future Value of an Income Stream
CONTINUED
Add the effects of the deposits at times t1, t2, ..., tn to arrive at the total balance in the account:
This is a Riemann sum for the function on the interval ≤ t ≤ 2. Since is very small when compared with the interval, the total amount in the account, A, is approximately
That is, the approximate balance in the account at the end of 2 years is $4250.

60. Volume of a Solid of Revolution

61. Volume of a Solid of Revolution
EXAMPLE
Find the volume of a solid of revolution generated by revolving about the x-axis the region under the following curve.
SOLUTION
Here g(x) = x2, and