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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (1)

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Post by Weston Lee on February 16, 2015

When working on midpoint Riemann sum, and you have sections of three, would you use the middle component?

e.g. t    0     2     4
    w(t) 4     6     7.2

Reimann Sums

  • Slightly different from rectangle rule: Reimann sums can have irregular x intervals
  • Apply just like rectangle rule
  • Reimann sums uses a table of values rather than the function form

Reimann Sums

Given the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫010 f(x) dx
f(0) = 2        f(3) = 5        f(4) = 5.5 f(7) = 3        f(10) = 7
  • 010 f(x) dx ≈ 5(3 − 0) + 5.5 (4 − 3) + 3(7 − 4) + 7(10 − 7)
  • 010 f(x) dx ≈ 15 + 5.5 + 9 + 21
010 f(x) dx ≈ 50.5
Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫010 f(x) dx
f(0) = 2        f(3) = 5        f(4) = 5.5 f(7) = 3        f(10) = 7
  • 010 f(x) dx ≈ 2(3 − 0) + 5(4 − 3) + 5.5(7 − 4) + 3(10 − 7)
  • 010 f(x) dx ≈ 6 + 5 + 16.5 + 9
010 f(x) dx ≈ 36.5
Given the values below, use right-hand Riemann Sum with 2 intervals to approximate ∫04 f(x) dx
f(0) = 2        f(3) = 5        f(4) = 5.5 f(7) = 3        f(10) = 7
  • 04 f(x) dx ≈ 5(3 − 0) + 5.5(4 − 3)
  • 04 f(x) dx ≈ 15 + 5.5
04 f(x) dx ≈ 20.5
Given the values below, use right-hand Riemann Sum with 5 intervals to approximate ∫−55 f(x) dx
f(−5) = 25        f(−1) = 1        f(0) = 0 f(2) = 4        f(4) = 16        f(5) = 25
  • −55 f(x) dx ≈ 1(−1 − (−5)) + 0(0 − (−1)) + 4(2 − 0) + 16(4 − 2) + 25(5 − 4)
  • −55 f(x) dx ≈ 4 + 0 + 8 + 32 + 25
−55 f(x) dx ≈ 69
Given the values below, use right-hand Riemann Sum with 3 intervals to approximate ∫05 f(x) dx
f(−5) = 25        f(−1) = 1        f(0) = 0 f(2) = 4        f(4) = 16        f(5) = 25
  • 05 f(x) dx ≈ 4(2 − 0) + 16(4 − 2) + 25(5 − 4)
  • 05 f(x) dx ≈ 8 + 32 + 25
05 f(x) dx ≈ 65
Given the values below, use left-hand Riemann Sum with 2 intervals to approximate ∫−50 f(x) dx
f(−5) = 25        f(−1) = 1        f(0) = 0 f(2) = 4        f(4) = 16        f(5) = 25
  • −50 f(x) dx ≈ 25(−1 − (−5)) + 1(0 − (−1))
  • −50 f(x) dx ≈ 100 + 1
−50 f(x) dx ≈ 101
Given the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫−22 f(x) dx
f(−2) = −8        f(−1) = −1        f(0) = 0 f(1) = 1        f(2) = 8
  • −22 f(x) dx ≈ −1(−1 −(−2)) + 0(0 − (−1)) + 1(1 − 0) 8(2 − 1)
  • −22 f(x) dx ≈ −1 + 0 + 1 + 8
−22 f(x) dx ≈ 8
Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫−22 f(x) dx
f(−2) = −8        f(−1) = −1        f(0) = 0 f(1) = 1        f(2) = 8
  • −22 f(x) dx ≈ −8(−1 − (−2)) + −1(0 − (−1)) + 0(1 − 0) + 1(2 − 1)
  • −22 f(x) dx ≈ −8 − 1 + 0 + 1
−22 f(x) dx ≈ −8
Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫012 f(x) dx
f(−2) = −8        f(−1) = −1        f(0) = 0 f(1) = 1        f(2) = 8        f(4) = 64 f(8) = 512        f(10) = 1000        f(12) = 1728
  • 012 f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4) + 1000(10 − 8) + 1728(12 − 10)
  • 012 f(x) dx ≈ 1 + 8 + 128 + 2048 + 2000 + 3456
012 f(x) dx ≈ 7641
Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫08 f(x) dx
f(−2) = −8        f(−1) = −1        f(0) = 0 f(1) = 1        f(2) = 8        f(4) = 64 f(8) = 512        f(10) = 1000        f(12) = 1728
  • 08 f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4)
  • 08 f(x) dx ≈ 1 + 8 + 128 + 2048
08 f(x) dx ≈ 2185

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Reimann Sums

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Reimann Sums 0:08
    • Definition
  • Example 1 2:48
  • Example 2 5:38
  • Example 3 7:21
  • Example 4 9:14