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

2. Aim: Finding Area Course: Calculus Area A = bh b h Rectangle A = 1/2 bh b h Triangle

3. Aim: Finding Area Course: Calculus Area

4. Aim: Finding Area Course: Calculus Life Gets Complex Lake Wallawalla

5. Aim: Finding Area Course: Calculus Life Gets Complex Lake Wallawalla

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

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

8. Aim: Finding Area Course: Calculus Approximating the Area of a Plane Region 2/5 4/5 6/5 8/5 2 f(2/5) f(4/5) f(6/5) f(8/5) f(2) right endpoints are length of each rectangle right endpoints inscribed rectangles 1 2 3 4 5 l 2 = f(4/5) = -(4/5) 2 + 5 = 4.36 l 3 = f(6/5) = -(6/5) 2 + 5 = 3.56 l 4 = f(8/5) = -(8/5) 2 + 5 = 2.44 l 5 = f(2) = -(2) 2 + 5 = 1 l 1 = f(2/5) = -(2/5) 2 + 5 = 4.84 A rect 1 = 1.936 A rect 2 = 1.744 A rect 3 = 1.424 A rect 4 =.976 A rect 5 =.4 Sum of 5 areas = 6.48

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

10. Aim: Finding Area Course: Calculus Approximating the Area of a Plane Region 2/5 4/5 6/5 8/5 2 f(4/5) f(6/5) f(2/5) f(0) f(8/5) left endpoints circum- scribed rectangle left endpoints are length of each rectangle l 1 = f(0) = -(0) 2 + 5 = 5 A rect 1 = 2 l 2 = f(2/5) = -(2/5) 2 + 5 = 4.84 l 3 = f(4/5) = -(4/5) 2 + 5 = 4.36 l 4 = f(6/5) = -(6/5) 2 + 5 = 3.56 l 5 = f(8/5) = -(8/5) 2 + 5 = 2.44 A rect 2 = 1.936 A rect 3 = 1.744 A rect 4 = 1.424 A rect 5 =.976 Sum of 5 areas = 8.08

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

12. Aim: Finding Area Course: Calculus 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 lower sum partition into n intervals a b height of rectangle 1 is y 0 1 2 y n - 1 y n - 2 height of rectangle 2 is y 1 height of rectangle 3 is y 2 etc. height of last rectangle is y n - 1 y0y0 y1y1 y n - 1 width

13. Aim: Finding Area Course: Calculus Devising a Formula Using left endpoint to approximate area under the curve is lower sum a b 1 2 y n - 1 y n - 2 y0y0 y1y1 y n - 1 the more rectangles the better the approximation the exact area? take it to the limit! left endpoint formula

14. Aim: Finding Area Course: Calculus Using right endpoint to approximate area under the curve is Right Endpoint Formula upper sum a b y0y0 y1y1 y n - 1 ynyn right endpoint formula midpoint formula y2y2

15. Aim: Finding Area Course: Calculus Model Problem Approximate the area under the curve y = x 3 from x = 2 to x = 3 using four left-endpoint rectangles. width: 23 height: y 0 = y0y0 y1y1 y2y2 y3y3 y 1 = y 2 = y 3 =

16. Aim: Finding Area Course: Calculus Model Problem Approximate the area under the curve y = x 3 from x = 2 to x = 3 using four left-endpoint rectangles. 23 y0y0 y1y1 y2y2 y3y3

17. Aim: Finding Area Course: Calculus Model Problem Approximate the area under the curve y = x 3 from x = 2 to x = 3 using four right-endpoint rectangles. 23 y1y1 y2y2 y3y3 y4y4

18. Aim: Finding Area Course: Calculus Homework Approximate the area under the curve y = x 3 from x = 2 to x = 3 using four midpoint rectangles.

19. Aim: Finding Area Course: Calculus Model Problem Approximate the area under the curve y = 6 + 2x - x 3 for [0, 2] using 8 left endpoint rectangles. Sketch the graph and regions.

20. Aim: Finding Area Course: Calculus Model Problem Approximate the area under the curve y = 4 – x 2 for [-1, 1] using 4 inscribed rectangles.