Lesson 4.2 day 2

4.2 Area . . . Day 2Origins of Calculus ‐ two classicproblems:the tangent line problemPQthe area problemabThe Area of a Plane RegionDef.Arearectangle= b . h1

Example:Use 5 rectangles (equal width) to "squeeze" the exact area between twoapproximations of the area of the region lying between the graph off (x) = x 2 and the x­axis between x = 0 and x = 5.width = total length = 5 ­ 0# rectangles 5 = 1intervals: [0, 1] , [1, 2] , [2, 3] , [3, 4] , [4, 5]0 52

left endpoints:| | | |0 1 2 3 4 5| | | |0 1 2 3 4 53

ight endpoints:| | | |0 1 2 3 4 5| | | |0 1 2 3 4 54

Theory and Terminology:y = f (x)(a, f (a))(b, f (b))ab5

subdivide [a, b] into n subintervalsy = f (x)(a, f (a))(b, f (b))a(x 0(x 1 x 2 x 3. . . x i. . . x n­1b( x n(width = x= total length =# rectanglesb ­ an6

y = f (x)(a, f (a))(b, f (b))a x 1 x 2 x 3. . . x i . . . x n­1b(x 0x x x( x n(}}}(Notice:x 0 = ax 1 = a + 1 . xx 2 = a + 2 . xx 3 = a + 3 . x. . . . . .x i = a + i . xx n = a + n . x = b7

ithsubintervalf (M i ){3.1}f (m i )ExtremeValue Thm.8

Defs.Area of the i th inscribed rectangle = f (m i ) .xArea of the i th circumscribed rectangle =f (M i ).xLower sum = s(n) =ni = 1.f (m i ) xtotal areaof inscribedrectanglesUpper sum = S(n) =nf (M i ). xi = 1total areaof circumscribedrectangless(n)Area ofRegionS(n)9

Def. of the Area of a Region in the PlaneLet f be continuous and nonnegative on [a, b].The area of the region bounded by the graph of f ,the x­axis, and the vertical lines x = a and x = b isArea =limnnf (c i ) .xi = 1wherexb ­ a= n12

Assignmentp262-263 #23-29 odd,#47, 49, 51, 53,#57, 59, 72limit processhorizontal rectangles(see Ex. 7 p261)14

p263 #60Use the limit process to find the area of the region betweenthe graph of the function and the y­axis over the indicatedy­interval. Sketch the region.f ( y) = 4y ­ y 2 , 1 y 2y2 _1x15