Integration and Sums

Return to List of Lectures

Previous Lecture

• Misc
  • Make appointments for this week soon
  • Hour exam 2
    • Next Tuesday (Nov. 18)
    • Covers material from 1st exam through optimization (e.g., chain rule, implicit differentiation, related rates, Mean Value Theorem, curve sketching, extrema and optimization, etc.)
    • 6 - 9 short answer questions, including word problems
    • Rules and format otherwise similar to 1st exam, especially open references rule
    • Review of your questions/issues in 2nd half of Thursday’s class
• Questions?
• Integration and Sums
  • Section 5.1
    • Area under curve
      • Upper sum (over-estimate), lower sum, midpoint rule
    • Distance traveled as area under curve
      • Antiderivative (s(t)) of velocity (v(t)) is distance traveled
      • Displacement includes changes in direction, could be negative, unlike total distance traveled
    • Average value of function calculated as upper sums, etc.
  • Estimating antiderivatives
    • Much like distance-from-velocity example in book, the value of any antiderivative of f′(t) can be calculated by approximating f′ as constant over small t intervals and adding up products of approximate values times widths of intervals
      
    • For example, find F(2) where F is antiderivative of f′(t) = t²
      • f′(t) = t² from t = 0 to t = 2
      • 4 intervals, use midpoints

    Endpoints	f′(t)	f′(midpoint)
    0.0	0.0
    0.5	0.25	0.0625
    1.0	1.0	0.5625
    1.5	2.25	1.5625
    2.0	4.0	3.0625

    • • Sum of f′(midpoint) times width of interval
        • = 0.0625 × 0.5 + 0.5625 × 0.5 + 1.5625 × 0.5 + 3.0625 × 0.5
        • = 2.625
      • Actual should be 8/3 = 2.666...
      • Use R to look at this for a serious number of intervals (but first for the 4 intervals we just did)
        

        > x = seq( 0, 2, length = 5 )
        > x
        [1] 0.0 0.5 1.0 1.5 2.0
        > x[2]
        [1] 0.5
        > x[2:5]
        [1] 0.5 1.0 1.5 2.0
        > x[1:4]
        [1] 0.0 0.5 1.0 1.5
        > m = ( x[1:4] + x[2:5])/2
        > m
        [1] 0.25 0.75 1.25 1.75
        > y = m^2
        > y
        [1] 0.0625 0.5625 1.5625 3.0625
        > y * 0.5
        [1] 0.03125 0.28125 0.78125 1.53125
        > sum( y * 0.5 )
        [1] 2.625
        > x = seq( 0, 2, length = 5001 )
        > m = ( x[1:5000] + x[2:5001])/2
        > y = m^2
        > dx = 2/5000
        > sum( y * dx )
        [1] 2.666667

  • Sums
    • Notation
      
    • We’ll want to consider what happens to sums of rectangle areas as the number of rectangles increases towards infinity
    • Sums with variable upper bounds are good for this; such sums often have rules for their numeric values
      
    • Describe the process of summing areas of rectangles with heights taken in any of various ways from function values with each rectangle using summation notation
      
  • Problem set
• Next
  • Summation notation, limits, and Riemann sums
  • Read section 5.2 from “limits of finite sums” through end of section

Next Lecture