1. Tutorial 6 The Definite Integral
MT129 – Calculus and Probability

2. Outline Anti-differentiation Areas
Definite Integrals and the Fundamental Theorem
Areas in the xy-Plane
Applications of the Definite Integral
MT129 – Calculus and Probability

3. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5
Anti-differentiation
Definition
Example
Anti-differentiation: The process of determining f (x) given f ΄(x)
If f ΄(x) = 2x, then f (x) = x2
EXAMPLE
Find all antiderivatives of the function f (x) = 9x8
SOLUTION
The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x9  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.
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5

4. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7
Anti-differentiation and The Indefinite Integrals
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7

5. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9
Rules of Integration
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9

6. Finding Anti-derivatives
EXAMPLE
Determine the following
SOLUTION
Using the rules of indefinite integrals, we have
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10

7. Finding Anti-derivatives
EXAMPLE
Find the function f (x) for which and f (1) = 3.
SOLUTION
The unknown function f (x) is an anti-derivative of One
Anti-derivative is Therefore, by Theorem I,
Now, we want the function f (x) for which f (1) = 3.
So, 3 = 1 + C and therefore, C = 2. Therefore, our function is
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11

8. Anti-derivatives 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 anti-derivative of v(t) for which s(0) = The anti-derivative of v(t) is
Therefore, C = So, our antiderivative is s(t) = − 16t
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13

9. Anti-derivatives 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.
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14

10. Anti-derivatives 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).
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15

11. Areas 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
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #18

12. Calculating Definite Integrals
EXAMPLE
Calculate the following integral.
SOLUTION
The figure shows the graph of the function f (x) = x Since f (x) is nonnegative for 0 ≤ x ≤ 1, the definite integral of f (x) equals the area of the shaded region in the figure below.
The region consists of a rectangle and a triangle.
By geometry, 1
Thus the area under the graph is = 1, and hence 1
0.5
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #32
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13. The Definite Integral MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #34

14. Calculating Definite Integrals
EXAMPLE
Calculate the following integral
SOLUTION
The figure shows the graph of the function f (x) = x on the interval −1 ≤ x ≤ 1. The area of the triangle above the x-axis is 0.5 and the area of the triangle below the x-axis is 0.5. Therefore, from geometry we find that
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #35

15. The Fundamental Theorem of Calculus
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #37

16. 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,
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #38

17. The Fundamental Theorem of Calculus
EXAMPLE
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.
(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.
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #39

18. Area Under a Curve as an Anti-derivative
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #41

19. Properties of Definite Integrals
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #44

20. Area Between Two Curves
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Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #45

21. Finding the Area Between Two Curves
EXAMPLE
Let R be the region enclosed by the graphs of f (x) = x2 + 5x  7 and g (x) = x2 + 5x + 1. Find the area of the region R.
SOLUTION
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46

22. Finding the Area Between Two Curves
EXAMPLE
Find the area of the region between y = x2 – 3x and the x-axis (y = 0) from x = 0 to x = 4.
SOLUTION
Upon sketching the graphs we can see that the two graphs cross; and by setting x2 – 3x = 0, we find that they cross when x = 0 and when x = 3. Thus one graph does not always lie above the other from x = 0 to x = 4, so that we cannot directly apply our rule for finding the area between two curves. However, the difficulty is easily surmounted if we break the region into two parts, namely the area from x = 0 to x = 3 and the area from x = 3 to x = 4. For from x = 0 to x = 3, y = 0 is on top; and from x = 3 to x = 4, y = x2 – 3x is on top. Consequently,
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46

23. Finding the Area Between Two Curves
CONTINUED
Thus the total area is =
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #47

24. Finding the Area Between Two Curves
CONTINUED
y = x2 – 3x
y = 0
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #48

25. 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
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #49

26. Finding the Area Between Two Curves
CONTINUED
Therefore, to represent all the shaded regions, we have
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #50

27. 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
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #51

28. 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).
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #52

29. Average Value of a Function Over an Interval
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #55

30. 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.
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #56

31. Average Value of a Function Over an Interval
EXAMPLE
(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
MT129 – Calculus and Probability
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #57