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5.6 The Definite Integral as the Limit of a SumGoal:  To find the definite integral when you cannot find the anti-derivative. Recall the Reimann Sum for integration:
Use four rectangles to approximate the area of the region bounded by the graph of f(x) = x2 + 1, the x-axis, x = 0 and x = 2
After dividing the interval [0,2] into 4 equal rectangles, find the midpoint of the new rectangle to find a better approximation of the area.  Evaluate the function at that midpoint for the length of the triangle and use the width of the interval as the width of the rectangle. = 37/ 8 ≈ 4.625 We can use this to approximate any definite integrate just not those representing area.  This is called the Midpoint Rule.
Use the Midpoint Rule with n = 4  to approximate: First break [0,1] into 4 intervals and find the midpoint of each. 0 - ¼       ¼ - ½         ½ - ¾         ¾ - 1         Δx = ¼    1/8          3/8          5/8            7/8         are the midpoints for each interval ≈ .4358
Use the Midpoint Rule with n = 4  to approximate: First break [2,4] into 4 intervals and find the midpoint of each. 2 – 2.5       2.5-3        3-3.5        3.5-4         Δx = ½     9/4          11/4          13/4         15/4        are the midpoints for each interval ≈ 5.6419
Use your calculator to find ≈ 1.4627 See Page 466 for program.  We will write this program for our calculators. Assignment:  368-369/2-32 even due Monday

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Calculus 5 6 Notes

  • 1. 5.6 The Definite Integral as the Limit of a SumGoal: To find the definite integral when you cannot find the anti-derivative. Recall the Reimann Sum for integration:
  • 2. Use four rectangles to approximate the area of the region bounded by the graph of f(x) = x2 + 1, the x-axis, x = 0 and x = 2
  • 3. After dividing the interval [0,2] into 4 equal rectangles, find the midpoint of the new rectangle to find a better approximation of the area. Evaluate the function at that midpoint for the length of the triangle and use the width of the interval as the width of the rectangle. = 37/ 8 ≈ 4.625 We can use this to approximate any definite integrate just not those representing area. This is called the Midpoint Rule.
  • 4.
  • 5. Use the Midpoint Rule with n = 4 to approximate: First break [0,1] into 4 intervals and find the midpoint of each. 0 - ¼ ¼ - ½ ½ - ¾ ¾ - 1 Δx = ¼ 1/8 3/8 5/8 7/8 are the midpoints for each interval ≈ .4358
  • 6. Use the Midpoint Rule with n = 4 to approximate: First break [2,4] into 4 intervals and find the midpoint of each. 2 – 2.5 2.5-3 3-3.5 3.5-4 Δx = ½ 9/4 11/4 13/4 15/4 are the midpoints for each interval ≈ 5.6419
  • 7. Use your calculator to find ≈ 1.4627 See Page 466 for program. We will write this program for our calculators. Assignment: 368-369/2-32 even due Monday