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Problem of the Day

Problem of the Day. If f ''(x) = x(x + 1)(x - 2) 2 , then the graph of f has inflection points when x =. a) -1 only b) 2 only c) -1 and 0 only d) -1 and 2 only e) -1, 0, and 2 only. Problem of the Day. If f ''(x) = x(x + 1)(x - 2) 2 ,

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Problem of the Day

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  1. Problem of the Day If f ''(x) = x(x + 1)(x - 2)2, then the graph of f has inflection points when x = a) -1 only b) 2 only c) -1 and 0 only d) -1 and 2 only e) -1, 0, and 2 only

  2. Problem of the Day If f ''(x) = x(x + 1)(x - 2)2, then the graph of f has inflection points when x = a) -1 only b) 2 only c) -1 and 0 only d) -1 and 2 only e) -1, 0, and 2 only

  3. Area A = ½ b .h A = b . h A = 2 triangles A = 6 triangles Archimedes utilized the exhaustion method for finding the area of some regions that are not polygons

  4. Area By squeezing a circle between 2 polygons and allowing the number of sides of the polygon to increase would result in a better approximation

  5. 5 5 0 2 Which gives the better approximation of the area under the curve? 0 2

  6. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 5 f(2/5) f(4/5) The area of each rectangle is? 2/5 10/5 4/5 width x height 8/5 6/5

  7. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(2/5) 5 f(4/5) Using right endpoints (area is greater than area of rectangles) 2/5 10/5 6/5 4/5 8/5 Area = f(2/5)(2/5) + f(4/5)(2/5) + f(6/5)(2/5) + f(8/5)(2/5) + f(10/5)(2/5)

  8. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(2/5i) 5 Using right endpoints (area is greater than area of rectangles) Area = b . h Area = Σ f(2/5i)(2/5) 2/5 10/5 6/5 5 4/5 8/5 i = 1

  9. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 Area = b . h Area = Σ f(2/5i)(2/5) f(2/5i) 5 5 i = 1 = 2/5 - (2/5i)2 + 5 2/5 10/5 6/5 4/5 8/5

  10. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 Area = b . h Area = Σ f(2/5i)(2/5) f(2/5i) 5 5 i = 1 = 2/5 - (2/5i)2 + 5 2/5 10/5 6/5 4/5 8/5 = 2/5 - (4/25)i2 + 5

  11. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 Area = b . h Area = Σ f(2/5i)(2/5) f(2/5i) 5 5 i = 1 = 2/5 - (2/5i)2 + 5 2/5 10/5 6/5 4/5 8/5 = 2/5 - (4/25)i2 + 5 = 2[- ( 4 )(5(5+1)(2(5)+1) + 5(5)] 5 25 6 = 2/5(405/25) = 6.48

  12. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(0) f(2/5) 5 Using left endpoints (area is less than area of rectangles) 0 4/5 8/5 6/5 2/5 Area = f(0)(2/5) + f(2/5)(2/5) + f(4/5)(2/5) + f(6/5)(2/5) + f(8/5)(2/5)

  13. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 Area = b . H Area = Σ f(2/5(i - 1))(2/5) f(2/5(i - 1)) 5 5 i = 1 0 4/5 8/5 6/5 2/5 Using left endpoints (area is less than area of rectangles)

  14. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(2/5(i - 1)) Area = b . h Area = Σ f(2/5(i - 1))(2/5) 5 5 i = 1 = 2/5 - (2/5(i - 1))2 + 5 0 4/5 8/5 6/5 2/5

  15. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(2/5(i - 1)) Area = b . h Area = Σ f(2/5(i - 1))(2/5) 5 5 i = 1 = 2/5 - (2/5(i - 1))2 + 5 0 4/5 8/5 6/5 2/5 = 2/5 - (4/25(i2 - 2i + 1)) + 5

  16. Summation notation can be used to generalize the area calculation process Find the area between f(x) = -x2 + 5 and x = 2 f(2/5(i - 1)) Area = b . h Area = Σ f(2/5(i - 1))(2/5) 5 5 i = 1 = 2/5 - (2/5(i - 1))2 + 5 0 4/5 8/5 6/5 2/5 = 2/5 - (4/25(i2 - 2i + 1)) + 5 = 2 [- ( 4 (5(5+1)(2(5)+1) - 2(5(5+1)) + 1(5)) + 5(5)] 5 25 6 2 = 2/5(101/5) = 8.08

  17. What if we let the number of subintervals approach ∞?

  18. Finding Upper and Lower Sums as number of subintervals approaches infinity 1) partition inteval into n subintervals Δx = upper boundary - lower boundary n 2) determine if using left endpoints or right endpoints i or (i - 1) 3) evaluate summation and find the limit as n approaches infinity

  19. Summary Area of circumscribed rectangles Area of inscribed rectangles Area of region on [a, b] < < Upper Sum Lower Sum n n S(n) = f(Mi)Δx s(n) = f(mi)Δx Mi = a + Δx(i or i-1) mi = a + Δx(i or i-1)

  20. Determining mi and Mi interval [1, 3] thus a = 1 and b = 3 Δx = 3 - 1 = 2 n n Lower sum - right endpoint Upper sum - left endpoint Mi = 1 + 2(i - 1) n Mi = a + Δx(i or i-1) mi = 1 + 2i n mi = a + Δx(i or i-1)

  21. Determining mi and Mi interval [6, 11] Δx = 11 - 6 = 5 n n Lower sum - left endpoint Upper sum - right endpoint mi = 6 + 5(i - 1) n Mi = 6 + 5i n

  22. Using our previous example Using left endpoints Find the area between f(x) = -x2 + 5 and x = 2 f(2/n(i - 1)) Back to original problem - Let's do it in general terms with left endpoints 5 0 4/n 8/n 6/n 2/n Area = b . h Area = Σ f(2/n(i - 1))(2/n) n lim n ∞ i = 1 = 2/n - (2/n(i - 1))2 + 5

  23. Using our previous example Using left endpoints Find the area between f(x) = -x2 + 5 and x = 2 Area = b . h Area = Σ f(2/n(i - 1))(2/n) f(2/n(i - 1)) 5 n lim n ∞ i = 1 2 0 Δx=(2-0) n = 2/n - (2/n(i - 1))2 + 5 = 2/n - (4/n2(i2 - 2i + 1)) + 5

  24. Using our previous example Using left endpoints Find the area between f(x) = -x2 + 5 and x = 2 f(2/n(i - 1)) 5 = 2/n - (4/n2(i2 - 2i + 1)) + 5 2 0 Δx=(2-0) n = 2 [- ( 4 (n(n+1)(2(n)+1) - 2(n(n+1)) + 1(n)) + 5(n)] n n26 2 = 2 [- ( 4 ((2n3 + 3n2 + n) - 2(n2 + n) + 1(n)) + 5(n)] n n26 2

  25. Using our previous example Using left endpoints Find the area between f(x) = -x2 + 5 and x = 2 = 2 [- ( 4 ((2n3 + 3n2 + n) - 2(n2 + n) + 1(n)) + 5(n)] n n26 2 = -16n3 - 24n2 - 8n + 8n2 + 8n - 8n + 10n 6n3 n3 n3 n = -16 + 10 6 = 7 1/3

  26. Using our previous example Using right endpoints Find the area between f(x) = -x2 + 5 and x = 2 5 f(2/ni) Back to original problem - Let's do it in general terms with right endpoints 0 2 Δx = (2-0) n Area = b . h Area = Σ f(2/ni)(2/n) n i = 1 = 2/n - (2/ni)2 + 5

  27. Using our previous example Using right endpoints Find the area between f(x) = -x2 + 5 and x = 2 f(2/ni) 5 Area = b . h Area = Σ f(2/ni)(2/n) n i = 1 0 2 = 2/n - (2/ni)2 + 5 Δx = (2-0) n = 2 - ( 4 )i2 + 5 n n2 = 2[- ( 4 )(n(n+1)(2(n)+1) + 5(n)] n n2 6

  28. Using our previous example Using right endpoints Find the area between f(x) = -x2 + 5 and x = 2 = 2[- ( 4 )(n(n+1)(2(n)+1) + 5(n)] n n2 6 = -16n3 - 24n2 - 8n + 10n 6n3 n = -16 + 10 6 = 7 1/3

  29. So . . . . limit process with upper sums gives 7 1/3 limit process with lower sums gives 7 1/3 We can conclude . . . . ?

  30. Theorem 4.3 If f is continuous and nonnegative on the interval [a, b], the limit as n approaches infinity of the lower and upper sum exists and are the same. * the choice of the number of subintervals does not affect the limit and by the squeeze value theorem we are free to choose an arbitrary x value in the subinterval Area = f(Ci)Δx where Δx= (b - a)/n and Ci = a + Δx(i or i-1)

  31. Example Find the area of the region bounded by f(x) = 4 - x2, the x-axis, x = 1, and x = 2 1) it is continuous and nonnegative 2) subintervals = 2 -1/n = 1/n using right endpoints determine the summation

  32. Example Find the area of the region bounded by f(x) = 4 - x2, the x-axis, x = 1, and x = 2 subintervals = 2 -1/n = 1/n using right endpoints determine the summation Area = f(Ci)Δx

  33. Example Find the area of the region bounded by f(x) = 4 - x2, the x-axis, x = 1, and x = 2 Area = f(Ci)Δx ] ( ] 2 ( ( ( 4 - 1 + i1 n n ] ( ] ( 4 - 1 - 2i - i21 n n2 n ] ] ( 3 - 2i - i21 n n2 n (

  34. Example Find the area of the region bounded by f(x) = 4 - x2, the x-axis, x = 1, and x = 2 ] ] ( ( 3 - 2i - i21 n n2 n 1 n 2 n2 1 n3 3 - i - i2 [3 - (1 + 1) - (1 + 1 + 1 ) n 3 2n 6n2 5/3

  35. If the region is bounded by the y axis Δx becomes Δy

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