Math 220 Calculus I

3. Math 220Calculus I Section 6.4 Areas in the x-y Plane Example F

4. 6.4 Example A-1 Find the area of the geometric figure pictured below.

5. 6.4 Example A-1 Find the area of the geometric figure pictured below. The figure consists of a rectangle and a semi-circle.

6. 6.4 Example A-1 Find the area of the geometric figure pictured below. The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.)

7. 6.4 Example A-1 Find the area of the geometric figure pictured below. The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.) The total area is the sum of the two areas.

8. 6.4 Example A-1 Find the area of the geometric figure pictured below. The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.) The total area is the sum of the two areas.

9. 6.4 Example A-1 Find the area of the geometric figure pictured below. The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.) The total area is the sum of the two areas.

10. 6.4 Theory In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b: Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

11. 6.4 Theory In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b: Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

12. 6.4 Theory In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b: Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

13. 6.4 Theory In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b: Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

14. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

15. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

16. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas. “above” “below”

17. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

18. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

19. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

20. 6.4 Example B Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3. As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

21. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

22. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3. Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same.

23. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3. Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same. However, in this case some of the area lies below the x-axis, and would thus be a negative area (as noted in Lecture 6.3).

24. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3. Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same. However, in this case some of the area lies below the x-axis, and would thus be a negative area (as noted in Lecture 6.3). Does this affect the calculation of area between the two curves?

25. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3. “above” “below”

26. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

27. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

28. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

29. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

30. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

31. 6.4 Example C Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3. So the integral calculation gives us the same result as in Example B: Area = 6 – ln3. It doesn’t matter whether the two curves lie above or below the x-axis, only that we subtract the “higher/above” curve minus the “lower/below”.

32. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 0 0 5

33. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 Looking at the graph, y = x2 – 4x + 12 lies above y = x2 for a portion of the interval, but is below for the rest of the interval. 0 0 5

34. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 Looking at the graph, y = x2 – 4x + 12 lies above y = x2 for a portion of the interval, but is below for the rest of the interval. We need to determine where the two intersect, then set up two integrals: one for each portion. 0 0 5

35. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 We need to determine where the two intersect, then set up two integrals: one for each portion. 0 0 5

36. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 We need to determine where the two intersect, then set up two integrals: one for each portion. 0 0 5

37. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 We need to determine where the two intersect, then set up two integrals: one for each portion. x = 3 0 0 5

38. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. “below” “above” 15 x = 3 “above” 0 “below” 0 5

39. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 x = 3 0 0 5

40. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 x = 3 0 0 5 Combining like terms is not an option—the two polynomials have different boundaries to evaluate.

41. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 x = 3 0 0 5

42. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 x = 3 0 0 5 = –18 + 36 + 0 – 0 + 32 – 48 – 18 + 36

43. 6.4 Example D Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4. 15 x = 3 0 0 5 = –18 + 36 + 0 – 0 + 32 – 48 – 18 + 36 = 20

44. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 0 0 5

45. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection. 0 0 5

46. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection. Between these two points the area is bounded, and lies between the two curves. 0 0 5

47. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection. Between these two points the area is bounded, and lies between the two curves. We need to solve for the intersections to find the boundaries (or limits) of integration. 0 0 5

48. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 We need to solve for the intersections to find the boundaries (or limits) of integration. 0 0 5

49. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 We need to solve for the intersections to find the boundaries (or limits) of integration. 0 0 5

50. 6.4 Example E Find the area between the curves y = 2x and y = x2 – 4x + 8. 10 We need to solve for the intersections to find the boundaries (or limits) of integration. 0 0 5