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Properties of Integrals

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A.) Prove: [tex]m_{i}[/tex](f) + [tex]m_{i}[/tex](g) [tex]\leq[/tex] [tex]m_{i}[/tex](f+g) [tex]\leq[/tex] [tex]M_{i}[/tex](f+g) [tex]\leq[/tex] [tex]M_{i}[/tex](f) + [tex]M_{i}[/tex](g)

    B.) Prove: [tex]M_{i}[/tex]([tex]\alpha[/tex]f) = {[tex]\alpha[/tex][tex]M_{i}[/tex](f) [tex]\alpha[/tex] [tex]\succ[/tex] 0 or [tex]\alpha[/tex][tex]m_{i}[/tex](f) [tex]\alpha[/tex] [tex]\prec[/tex] 0
    [tex]m_{i}[/tex]([tex]\alpha[/tex]f) = {[tex]\alpha[/tex][tex]m_{i}[/tex](f) [tex]\alpha[/tex] [tex]\succ[/tex] 0 or [tex]\alpha[/tex][tex]M_{i}[/tex](f) [tex]\alpha[/tex] [tex]\prec[/tex] 0

    C.) Take P = [tex]P_{1}[/tex] = {a,b}, and write [tex]m_{i}[/tex] = m, [tex]M_{i}[/tex] = M (i=1). Give an example for each of the following:
    (a) m(f) + m (g) [tex]\prec[/tex] m(f+g)
    (b) M(f+g) [tex]\prec[/tex] M(f) + M(g)
    (c) both (a) and (b) hold

    2. Relevant equations

    P = {[tex]x_{0}[/tex],...,[tex]x_{n}[/tex]} and is a partition of [a,b]
    [tex]M_{i}[/tex] = {sup {f(x) : x [tex]\in[/tex] [[tex]x_{i-1}[/tex], [tex]x_{i}[/tex]]}}
    [tex]m_{i}[/tex] = {inf {f(x) : x [tex]\in[/tex] [[tex]x_{i-1}[/tex], [tex]x_{i}[/tex]]}}
    [tex]\alpha[/tex] is just any multiplier

    3. The attempt at a solution

    For A.) and B.) i was going to take different sets and show it that way, but it will be a long drawn out process, i can't put my finger on the "shortcut/abbreviated" version. Any suggestions on what method to use??

    And for C.) (a) f = [-2,-1) g = [-1,0]
    (b) f = [0,1) g = [1,2]
    (c) ??
  2. jcsd
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