1. Chapter 6 The Definite Integral

2.  Antidifferentiation  Areas  Definite Integrals and the Fundamental Theorem  Areas in the xy-Plane  Applications of the Definite Integral Chapter Outline

3. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5 Antidifferentiation DefinitionExample Antidifferentiation: The process of determining f (x) given f ΄(x) If, then

4. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6 Finding AntiderivativesEXAMPLE SOLUTION Find all antiderivatives of the given function. The derivative of x 9 is exactly 9x 8. Therefore, x 9 is an antiderivative of 9x 8. So is x 9 + 5 and x 9 -17.2. It turns out that all antiderivatives of f (x) are of the form x 9 + C (where C is any constant) as we will see next.

5. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7 Theorems of Antidifferentiation

6. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #8 The Indefinite Integral

7. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9 Rules of Integration

8. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10 Finding AntiderivativesEXAMPLE SOLUTION Determine the following. Using the rules of indefinite integrals, we have

9. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11 Finding AntiderivativesEXAMPLE SOLUTION Find the function f (x) for which and f (1) = 3. 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.

10. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #12 Finding Antiderivatives So, 3 = 1 + C and therefore, C = 2. Therefore, our function is CONTINUED

11. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13 Antiderivatives in ApplicationEXAMPLE SOLUTION 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? (a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = 400. 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 = 400. So, our antiderivative is s(t) = -16t 2 + 400.

12. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14 Antiderivatives in Application s(t) = -16t 2 + 400 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. This is the function s(t). 0 = -16t 2 + 400 Replace s(t) with 0. -400 = -16t 2 Subtract. 25 = t 2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground.

13. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15 Antiderivatives in Application v(t) = -32t CONTINUED This is the function v(t). 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). (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(5) = -32(5) = -160

14. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #18 Area Under a Graph DefinitionExample Area Under the Graph of f (x) from a to b: An example of this is shown to the right

15. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #32 Calculating Definite IntegralsEXAMPLE SOLUTION Calculate the following integral. The figure shows the graph of the function f (x) = x + 0.5. 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. 1 0.5 1

16. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #33 Calculating Definite Integrals The region consists of a rectangle and a triangle. By geometry, CONTINUED Thus the area under the graph is 0.5 + 0.5 = 1, and hence

17. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #34 The Definite Integral

18. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #35 Calculating Definite IntegralsEXAMPLE SOLUTION Calculate the following integral. 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

19. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #36 Calculating Definite IntegralsCONTINUED

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

21. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #38 The Fundamental Theorem of CalculusEXAMPLE SOLUTION Use the Fundamental Theorem of Calculus to calculate the following integral. An antiderivative of 3x 1/3 – 1 – e 0.5x is. Therefore, by the fundamental theorem,

22. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #39 The Fundamental Theorem of CalculusEXAMPLE SOLUTION (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? (a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following.

23. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #40 The Fundamental Theorem of Calculus (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 24.533 degrees Fahrenheit. CONTINUED

24. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #41 Area Under a Curve as an Antiderivative

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

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

27. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46 Finding the Area Between Two CurvesEXAMPLE SOLUTION Find the area of the region between y = x 2 – 3x and the x-axis (y = 0) from x = 0 to x = 4. Upon sketching the graphs we can see that the two graphs cross; and by setting x 2 – 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 = x 2 – 3x is on top. Consequently,

28. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #47 Finding the Area Between Two CurvesCONTINUED Thus the total area is 4.5 + 1.833 = 6.333.

29. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #48 Finding the Area Between Two CurvesCONTINUED y = x 2 – 3x y = 0

30. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #49 Finding the Area Between Two CurvesEXAMPLE SOLUTION Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure. Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore

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

32. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #51 Finding the Area Between Two CurvesEXAMPLE SOLUTION Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v 1 (t) and v 2 (t), respectively, and v 1 (t) ≥ v 2 (t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v 1 (t) and y = v 2 (t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A? Since v 1 (t) ≥ v 2 (t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have

33. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #52 Finding the Area Between Two CurvesCONTINUED But again, since v 1 (t) ≥ v 2 (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).

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

35. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #56 Average Value of a Function Over an IntervalEXAMPLE SOLUTION Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1. 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.

36. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #57 Average Value of a Function Over an IntervalEXAMPLE SOLUTION (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? The average temperature during the 12-hour period from t = 0 to t = 12 is

37. Integration by Substitution If u = g(x), then

38. Using Integration by SubstitutionEXAMPLE SOLUTION Determine the integral by making an appropriate substitution. Let u = x 2 + 2x + 3, so that. That is, Therefore, And so we have Rewrite in terms of u. Bring the factor 1/2 outside. Integrate.

39. Using Integration by SubstitutionCONTINUED Replace u with x 2 + 2x + 3.

40. Using Integration by SubstitutionEXAMPLE SOLUTION Determine the integral by making an appropriate substitution. Let u = 1/x + 2, so that. Therefore, And so we have Rearrange factors. Rewrite in terms of u. Integrate. Rewrite u as 1/x + 2.