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Aim: An introduction to the 2 nd central Idea of Calculus.

Aim: An introduction to the 2 nd central Idea of Calculus. Do Now:. h. A = bh. h. b. b. Area. Rectangle. Triangle. A = 1/2 bh. Area. Life Gets Complex. Lake Wallawalla. Life Gets Complex. Lake Wallawalla. f (1). f (2). Approximating the Area of a Plane Region.

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Aim: An introduction to the 2 nd central Idea of Calculus.

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  1. Aim: An introduction to the 2nd central Idea of Calculus. Do Now:

  2. h A = bh h b b Area Rectangle Triangle A = 1/2 bh

  3. Area

  4. Life Gets Complex Lake Wallawalla

  5. Life Gets Complex Lake Wallawalla

  6. f(1) f(2) Approximating the Area of a Plane Region How would we approximate the area of the shaded region under the curve f(x) = -x2 + 5 in the interval [0, 2]? Approximate area is sum of areas of 2 rectangles having equal widths Arect = w·l height – value of f(x) inscribed rectangles 1 width – value of x Arect = xf(x) 2 Aunder curve A1 + A2 A  (1)f(1) + (1)f(2) = (1)[-(1)2 + 5] + (1)[-(2)2 + 5] A  4 + 1 = 5

  7. 5 intervals of equal width - 2/5 units f(2/5) f(4/5) f(6/5) f(8/5) height of f(x) f(2) 2/5 4/5 6/5 8/5 2 Approximating the Area of a Plane Region How would we approximate the area of the shaded region under the curve f(x) = -x2 + 5 in the interval [0, 2]? right endpoints inscribed rectangles 1 2 3 A = w·l 4 5 A = Δxf(x) right endpoints are length of each rectangle lof rectangle 1 = f(2/5) = -(2/5)2 + 5 = 4.84 w of rectangle 1 = 2/5 Arect 1 = (2/5)4.84 = 1.936

  8. Approximating the Area of a Plane Region right endpoints f(2/5) right endpoints are length of each rectangle f(4/5) f(6/5) inscribed rectangles f(8/5) 1 2 l 1 = f(2/5) = -(2/5)2 + 5 = 4.84 3 f(2) 4 5 Arect 1 = 1.936 2/5 4/5 6/5 8/5 2 l2 = f(4/5) = -(4/5)2 + 5 = 4.36 Arect 2 = 1.744 l3 = f(6/5) = -(6/5)2 + 5 = 3.56 Arect 3 = 1.424 Arect 4 = .976 l4 = f(8/5) = -(8/5)2 + 5 = 2.44 l5 = f(2) = -(2)2 + 5 = 1 Arect 5 = .4 Sum of 5 areas = 6.48

  9. f(0) f(2/5) f(4/5) f(6/5) f(8/5) Approximating the Area of a Plane Region How would we approximate the area of the shaded region under the curve f(x) = -x2 + 5 in the interval [0, 2]? left endpoints 5 intervals of equal width - 2/5 units circum-scribed rectangle 1 height of f(x) 2 3 4 A = w·l 5 A = Δxf(x) 2/5 4/5 6/5 8/5 2 left endpoints are length of each rectangle lof rectangle 1 = f(0) = -(0)2 + 5 = 5 w of rectangle 1 = 2/5 Arect 1 = (2/5)5 = 2

  10. f(0) f(2/5) f(4/5) f(6/5) f(8/5) 2/5 4/5 6/5 8/5 2 Approximating the Area of a Plane Region left endpoints left endpoints are length of each rectangle circum-scribed rectangle l 1 = f(0) = -(0)2 + 5 = 5 Arect 1 = 2 l2 = f(2/5) = -(2/5)2 + 5 = 4.84 Arect 2 = 1.936 l3 = f(4/5) = -(4/5)2 + 5 = 4.36 Arect 3 = 1.744 Arect 4 = 1.424 l4 = f(6/5) = -(6/5)2 + 5 = 3.56 l5 = f(8/5) = -(8/5)2 + 5 = 2.44 Arect 5 = .976 Sum of 5 areas = 8.08

  11. right endpoints left endpoints f(2/5) f(0) f(2/5) f(4/5) f(4/5) f(6/5) f(6/5) f(8/5) f(8/5) f(2) 2/5 4/5 6/5 8/5 2 2/5 4/5 6/5 8/5 2 Approximating the Area of a Plane Region How would we approximate the area of the shaded region under the curve f(x) = -x2 + 5 in the interval [0, 2]? Approximate area is sum of areas of rectangles lower sum upper sum 6.48 < area of region < 8.08 increasing number of rectangles – closer approximations.

  12. width Devising a Formula • Let a be left endpoint of the interval of area to be found • Let b be right endpoint of interval of area partition into n intervals lower sum yn - 1 • height of rectangle 1 isy0 • height of rectangle 2 is y1 • height of rectangle 3 is y2 etc. y1 yn - 1 y0 yn - 2 2 • height of last rectangle is yn - 1 1 a b

  13. Devising a Formula • Using left endpoint to approximate area under the curve is the more rectangles the better the approximation lower sum yn - 1 the exact area? take it to the limit! y1 yn - 1 y0 yn - 2 2 1 a b left endpoint formula

  14. Right Endpoint Formula • Using right endpoint to approximate area under the curve is yn upper sum yn - 1 right endpoint formula y2 y1 midpoint formula y0 a b

  15. y3 y2 y1 y0 Model Problem Approximate the area under the curve y = x3 from x = 2 to x = 3 using four left-endpoint rectangles. width: height: y0 = y1 = y2 = y3 = 2 3

  16. y3 y2 y1 y0 Model Problem Approximate the area under the curve y = x3 from x = 2 to x = 3 using four left-endpoint rectangles. 2 3

  17. Model Problem Approximate the area under the curve y = x3 from x = 2 to x = 3 using four right-endpoint rectangles. y4 y3 y2 y1 2 3

  18. Homework Approximate the area under the curve y = x3 from x = 2 to x = 3 using four midpoint rectangles.

  19. Model Problem Approximate the area under the curve y = 6 + 2x - x3 for [0, 2] using 8 left endpoint rectangles. Sketch the graph and regions.

  20. Model Problem Approximate the area under the curve y = 4 – x2 for [-1, 1] using 4 inscribed rectangles.

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