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Lower and Upper Riemann sums of sin(x)

  1. Oct 27, 2014 #1

    Task in real analysis:
    P is a uniform partition on [0, π] and is divided into 6 equal subintervals. Show that the lower and upper riemann sums of sin (x) over P is lesser than 1.5 and larger than 2.4 respectively.

    My attempt at the solution:
    The greates value and the least value of sin x over an subinterval (xi- xi-1) is 1 and -1. The upper and lower riemann sums is then:
    upper sum: S(P) = ∑i=16 (xi- xi-1) = x6 - x0 = π
    Lower sum: s(P) = - i=16 (xi- xi-1) = -π


    From this one could say that S(P) > 2.4 and s(P) < 1.5, but i dont feel like this is a full answer to the problem and i dont see another approach to solving the problem, so if anyone could give me some clue or tips it would be much appreciated.
     
  2. jcsd
  3. Oct 27, 2014 #2

    pasmith

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    Homework Helper

    Those are not the correct values for the upper and lower sums.

    Let [itex]M_i = \sup\{ f(x) : x_{i-1} \leq x \leq x_i\}[/itex] and [itex]m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}[/itex]. Then [tex]
    S(P) = \sum_{i=1}^6 M_i(x_i - x_{i-1}), \\
    s(P) = \sum_{i=1}^6 m_i(x_i - x_{i-1}).
    [/tex] What are the maximum and minimum values of [itex]\sin x[/itex] when [itex]\frac{(i-1)\pi}6 = x_{i-1} \leq x \leq \frac{i\pi}6 = x_i[/itex]? They aren't all 1 and -1 respectively (aside from anything else, [itex]\sin(x) \geq 0[/itex] for [itex]x \in [0, \pi][/itex]).
     
  4. Oct 27, 2014 #3
    Thank you for replying. I saw that myself a while ago and found what the maximum and minimum value of sin x over each of the 6 subintervals such that the upper sum sin x over partition P should be:
    S(P) = ∑i=16 Δx ⋅max(sinx)i = (π/6) ⋅(1/2 + √3/2 + 1 + 1 + √3/2 + 1/2) = (π/6) ⋅ (3 + √3) ≈ 2,47 > 2,4
    and similar for the lower sum:
    s(P) = (π/6) ⋅ (1 + √3) ≈ 1,43 < 1,5
     
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