Successfully reported this slideshow.

5.4 more areas

580 views

Published on

Published in: Technology
  • Be the first to comment

  • Be the first to like this

5.4 more areas

  1. 1. More on Areas
  2. 2. We are to find the area of a given enclosed region R. R More on Areas
  3. 3. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, R More on Areas x
  4. 4. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. R a=x b=x x
  5. 5. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. R a=x b=x x
  6. 6. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. R L(x) a=x b=x x x
  7. 7. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] R a=x b=x x L(x) x
  8. 8. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] x1 x2 xi–1 xi R a=x0 b=xn x
  9. 9. More on Areas We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x, x, x, .. x=b} be a regular partition of 012n[a, b] and select an arbitrary point x* in each sub-interval i [x , x]. i–1ix1 x2 xi–1 xi R a=x0 b=xn x
  10. 10. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x, x, x, .. x=b} be a regular partition of 012n[a, b] and select an arbitrary point x* in each sub-interval i [x , x]. i–1ia=xxxb=x 0 1 2 x1 * xi–1 xi R More on Areas x
  11. 11. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. x1 * x2 * a=xxxb=x 0 1 2 xi–1 xi R More on Areas * x
  12. 12. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. x1 * x2 * x3 * xi * a=xxxb=x 0 1 2 xi–1 xi R More on Areas * x
  13. 13. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. x1 * x2 * x3 * xi * a=xxxb=x 0 1 2 xi–1 xi R More on Areas L(xi) * * x
  14. 14. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. x1 * x2 * x3 * xi * cross–sectional length at xi * a=xxxb=x 0 1 2 xi–1 xi R More on Areas L(xi) * * x
  15. 15. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. Let Δx = the width of each subinterval. x1 * x2 * x3 * xi * cross–sectional length at xi * a=xxxb=x 0 1 2 xi–1 xi R More on Areas * L(xi*) x
  16. 16. We are to find the area of a given enclosed region R. Take a ruler x and measure R from one end to the other end, and assume that R spans from x = a to x = b. Let L(x) = cross–sectional length at a generic x. Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. Let Δx = the width of each subinterval. The rectangle with L(xi) as height and Δx as width approximates the area in R that is spanned from xi–1 to xi. x1 * x2 * x3 * xi * cross–sectional length at xi * a=xxxb=x 0 1 2 xi–1 xi R More on Areas * * Δx L(xi*) x
  17. 17. More on Areas R x1 * x2 * x3 * xi * a=x0 x b=x 1 x2 xi–1 xi The Riemann sum x
  18. 18. More on Areas R Δx L(x1*) x1 * x2 * x3 * xi * a=x0 x b=x 1 x2 xi–1 xi L(x1The Riemann sum *)Δx x
  19. 19. More on Areas R Δx Δx L(x1) L(x2* *) x1 * x2 * x3 * xi * a=x0 x b=x 1 x2 xi–1 xi L(x1*)Δx+ L(x2The Riemann sum *)Δx+ … x
  20. 20. More on Areas Δx * L(xi) R Δx Δx L(x1) L(x2* *) x1 * x2 * x3 * xi * Δx L(xn* ) xn* x a=x0 x b=x 1 x2 xi–1 xi L(x1*)Δx+ L(x2*)Δx+ … L(xnThe Riemann sum *)Δx
  21. 21. More on Areas Δx L(xi) R x1 * x2 * x3 * xi * Δx L(xn* ) a=x0 x b=x 1 x2 xi–1 xi n The Riemann sum L(x*)Δx+ L(x *)Δx+ … L(x*)Δx = Σ L(x*)Δx 12nii=1 of all such rectangles approximates the area of R. L(x1) L(x2* *) * Δx Δx xn* x
  22. 22. More on Areas Δx * L(xi) R Δx Δx L(x1) L(x2* *) x1 * x2 * x3 * xi * Δx L(xn* ) xn* a=x0 x b=x 1 x2 xi–1 xi n The Riemann sum L(x*)Δx+ L(x *)Δx+ … L(x*)Δx = Σ L(x*)Δx 12nii=1 of all such rectangles approximates the area of R. In fact the mathematical definition of the area of R is the limit n Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, x
  23. 23. More on Areas Δx * L(xi) Δx Δx L(x1) L(x2* *) x1 * x2 * x3 * xi * Δx L(xn* ) xn* a=x0 x b=x 1 x2 xi–1 xi n Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx = of all such rectangles approximates the area of R. In fact the mathematical definition of the area of R is the limit the area of R is A = ∫ b L(x) dx. x=a R L(x1*)Δx+ i=1 n Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and by the FTC x
  24. 24. More on Areas Δx * L(xi) Δx Δx L(x1) L(x2* *) x1 * x2 * x3 * xi * Δx L(xn* ) xn* a=x0 x b=x 1 x2 xi–1 xi x n Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx = of all such rectangles approximates the area of R. In fact the mathematical definition of the area of R is the limit the area of R is A = ∫ b L(x) dx. x=a R L(x1*)Δx+ i=1 n Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and by the FTC Theorem. The area of a 2D region is the definite integral of its cross–section (length) function.
  25. 25. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2
  26. 26. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 y = x2 y = –x2 + 2x x
  27. 27. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. y = x2 y = –x2 + 2x x
  28. 28. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x x
  29. 29. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x
  30. 30. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x x
  31. 31. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x x
  32. 32. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. x
  33. 33. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. L(x) =(–x2 + 2x) – x2 x x L(x) =(–x2 + 2x) – x2
  34. 34. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. 1 (–x2 + 2x) – x2 dx = Hence the area is ∫ x=0 L(x) =(–x2 + 2x) – x2 x x L(x) =(–x2 + 2x) – x2
  35. 35. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. 1 (–x2 + 2x) – x2 dx = ∫ Hence the area is ∫ x=0 L(x) =(–x2 + 2x) – x2 x x 1 –2x2 + 2x dx 0 L(x) =(–x2 + 2x) – x2
  36. 36. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. 1 (–x2 + 2x) – x2 dx = ∫ Hence the area is ∫ x=0 1 –2x2 + 2x dx = –2 3 1 x3 + x2 | 0 0 L(x) =(–x2 + 2x) – x2 x x L(x) =(–x2 + 2x) – x2
  37. 37. More on Areas Example A. a. Find the area bounded by y = –x2 + 2x and y = x2 We need to find the span of the area. They are the x–coordinates of the intersections of the curves. y = x2 y = –x2 + 2x Set the equations equal to find the intersection points. x2 = –x2 + 2x 2x2 = 2x 2x(x – 1) = 0 so x = 0, 1. 1 (–x2 + 2x) – x2 dx = ∫ Hence the area is ∫ x=0 1 –2x2 + 2x dx = –2 3 1 = x3 + x2 | 0 0 1 3 L(x) =(–x2 + 2x) – x2 x x L(x) =(–x2 + 2x) – x2
  38. 38. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
  39. 39. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 f(x) = 2x – x3 g(x) = –x2
  40. 40. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. f(x) = 2x – x3 g(x) = –x2
  41. 41. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). f(x) = 2x – x3 g(x) = –x2
  42. 42. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. f(x) = 2x – x3 g(x) = –x2
  43. 43. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2
  44. 44. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3
  45. 45. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3 x3 – x2 – 2x = 0
  46. 46. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3 x3 – x2 – 2x = 0 x(x2 – x – 2) = 0
  47. 47. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3 x3 – x2 – 2x = 0 x(x2 – x – 2) = 0 x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
  48. 48. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3 x3 – x2 – 2x = 0 x(x2 – x – 2) = 0 x(x – 2)(x + 1) = 0 so x = –1, 0, and 2. Therefore the span for the 1st area is from x = –1 to x = 0
  49. 49. More on Areas b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2 The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is f(x) on top. We set the equations equal to solve for the span. f(x) = 2x – x3 g(x) = –x2 –x2 = 2x – x3 x3 – x2 – 2x = 0 x(x2 – x – 2) = 0 x(x – 2)(x + 1) = 0 so x = –1, 0, and 2. Therefore the span for the 1st area is from x = –1 to x = 0 and the span for the 2nd area is from x = 0 to x = 2.
  50. 50. 0 –x2 – (2x – x3) dx ∫ x= –1 More on Areas f(x) = x – x3 2 2x – x3 – (–x2) dx g(x) = –x2 Hence the total bounded area is + ∫ x= 0 –1 0 2
  51. 51. 0 –x2 – (2x – x3) dx ∫ x= –1 More on Areas f(x) = x – x3 2 2x – x3 – (–x2) dx g(x) = –x2 Hence the total bounded area is + ∫ x= 0 = –1 0 2 0 x3 – x2 – 2x dx + ∫0 ∫–1 2 – x3 + x2 + 2x dx
  52. 52. 0 –x2 – (2x – x3) dx ∫ x= –1 More on Areas f(x) = x – x3 2 2x – x3 – (–x2) dx 2 g(x) = –x2 Hence the total bounded area is + ∫ x= 0 = –1 0 2 0 x3 – x2 – 2x dx + ∫0 ∫–1 2 – x3 + x2 + 2x dx x4 4 = – x3 3 – x2 | 0 –1 + –x4 4 + x3 3 + x2 | 0
  53. 53. 0 –x2 – (2x – x3) dx ∫ x= –1 More on Areas f(x) = x – x3 2 2x – x3 – (–x2) dx 2 g(x) = –x2 Hence the total bounded area is + ∫ x= 0 = –1 0 2 0 x3 – x2 – 2x dx + ∫0 ∫–1 2 – x3 + x2 + 2x dx x4 4 = – x3 3 – x2 | 0 –1 + –x4 4 + x3 3 + x2 | 0 = 5 12 + 8 3 = 37 12
  54. 54. 0 –x2 – (2x – x3) dx ∫ x= –1 More on Areas f(x) = x – x3 2 2x – x3 – (–x2) dx 2 g(x) = –x2 Hence the total bounded area is + ∫ x= 0 = –1 0 2 0 x3 – x2 – 2x dx + ∫0 ∫–1 2 – x3 + x2 + 2x dx x4 4 = – x3 3 – x2 | 0 –1 + –x4 4 + x3 3 + x2 | 0 = 5 12 + 8 3 = 37 12 A type I region R is a region that is bounded on top by a curve y = f(x) and bounded below by y = g(x) from x = a to x = b. The above examples are regions of type I.
  55. 55. More on Areas A type II region R is a region that is bounded to the right by a curve x = f(y) and to the left by x = g(y) from y = a to y = b.
  56. 56. More on Areas A type II region R is a region that is bounded to the right by a curve x = f(y) and to the left by x = g(y) from y = a to y = b. y = b y = a x = g(y) x = f(y)
  57. 57. More on Areas A type II region R is a region that is bounded to the right by a curve x = f(y) and to the left by x = g(y) from y = a to y = b. Its cross–sectional length is L(y) = f(y) – g(y). y = b y = a x = g(y) x = f(y)
  58. 58. More on Areas A type II region R is a region that is bounded to the right by a curve x = f(y) and to the left by x = g(y) from y = a to y = b. Its cross–sectional length is L(y) = f(y) – g(y). y = b y = a x = g(y) x = f(y) L(y) = f(y) – g(y)
  59. 59. More on Areas A type II region R is a region that is bounded to the right by a curve x = f(y) and to the left by x = g(y) from y = a to y = b. Its cross–sectional length is L(y) = f(y) – g(y). b Therefore the area of R = ∫ f(y) – g(y) dy y = b y = a y=a x = g(y) x = f(y) L(y) = f(y) – g(y)
  60. 60. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.
  61. 61. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. y = x – 2 y = √x
  62. 62. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. y = x – 2 y = √x
  63. 63. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x
  64. 64. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4
  65. 65. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4 0 = x2 – 5x + 4
  66. 66. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4.
  67. 67. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.
  68. 68. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. 4
  69. 69. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. As a type I region, the span is from x = 0 to x = 4. 4
  70. 70. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x 4 x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function.
  71. 71. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x 4 x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function.
  72. 72. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 y = x – 2 y = √x 4 x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function. Therefore to find the area of the region, we have to split it into two pieces, I and II as shown.
  73. 73. More on Areas Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x. The shaded region is the area in question. For the span of the region, set √x = x – 2 I II y = x – 2 y = √x 4 x = x2 – 4x + 4 0 = x2 – 5x + 4 0 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution. As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function. Therefore to find the area of the region, we have to split it into two pieces, I and II as shown.
  74. 74. More on Areas I II y = x – 2 y = √x 2 4 The total area is (area I) + (area II) i.e.
  75. 75. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4
  76. 76. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 4
  77. 77. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 2 | + 0 4
  78. 78. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 + 2x3/2 3 ( – x2 2 2 | + 2x ) 0 4 | 2 4
  79. 79. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 + 2x3/2 3 ( – x2 2 2 | + 2x ) 0 4 | 2 = 2(2)3/2 3 + 4
  80. 80. More on Areas I II y = x – 2 y = √x 4 The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 + 2x3/2 3 ( – x2 2 2 | + 2x ) 0 4 | 2 = 2(2)3/2 3 + 2(4)3/2 3 [( – 42 2 + 8 ) – 2(2)3/2 – 3 22 2 ( + 4 ) ]
  81. 81. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 + 2x3/2 3 ( – x2 2 2 | + 2x ) 0 4 | 2 = 2(2)3/2 3 + 2(4)3/2 3 [( – 42 2 + 8 ) – 2(2)3/2 – 3 22 2 ( + 4 ) ] = 2(2)3/2 + 3 16 3 – 2(2)3/2 3 – 2 4
  82. 82. More on Areas I II y = x – 2 y = √x The total area is (area I) + (area II) i.e. 2 √x dx + ∫ x= 0 2 4 √x – (x – 2) dx ∫ x= 2 = 2x3/2 3 + 2x3/2 3 ( – x2 2 2 | + 2x ) 0 4 | 2 = 2(2)3/2 3 + 2(4)3/2 3 [( – 42 2 + 8 ) – 2(2)3/2 – 3 22 2 ( + 4 ) ] = 2(2)3/2 + 3 16 3 – 2(2)3/2 3 – 2 = 10 3 4
  83. 83. More on Areas y = x – 2 y = √x However, we may view this as a type II region.
  84. 84. More on Areas y = x – 2 y = √x y = 2 However, we may view this as a type II region. It spans from y = 0 to y = 2.
  85. 85. More on Areas y = x – 2 y = √x y = 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions.
  86. 86. More on Areas y = x – 2 y = √x y = 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2
  87. 87. More on Areas y = x – 2 y = √x y = 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2
  88. 88. More on Areas y = 2 y = √x so x = y2 y = x – 2 so x = y + 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2
  89. 89. More on Areas y = 2 y = √x so x = y2 y = x – 2 so x = y + 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2
  90. 90. More on Areas L(y) = y + 2 – y2 y = x – 2 so x = y + 2 y = √x so x = y2 y = 2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
  91. 91. More on Areas L(y) = y + 2 – y2 y = x – 2 so x = y + 2 y = √x so x = y2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2. Therefore the area is 2 ∫ y + 2 – y2 dy y = 0 y = 2
  92. 92. More on Areas L(y) = y + 2 – y2 y = x – 2 so x = y + 2 y = √x so x = y2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2. Therefore the area is 2 ∫ y + 2 – y2 dy y = 0 y2 2 = + 2y – y3 | 2 3 0 y = 2
  93. 93. More on Areas L(y) = y + 2 – y2 y = x – 2 so x = y + 2 y = √x so x = y2 However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2. Therefore the area is 2 ∫ y + 2 – y2 dy y = 0 y2 2 = + 2y – y3 | 2 3 0 = 6 – 8 3 = 10 3 y = 2

×