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Unformatted text preview: 6.2 Properties of the Riemann Integral 87 Since ε was arbitrary, this gives R f dx ≤ R f 1 dx + R f 2 dx . Similarly for the other inequality. For cf , use ∑ cM i ( x i x i 1 ) = c ∑ M i ( x i x i 1 ), etc. (ii) max { f,g } , min { f,g } ∈ R [ a,b ] . Proof. Let h ( x ) := max { f ( x ) ,g ( x ) } . Since f ( x ) ,g ( x ) ≤ h ( x ), Osc ( h,P ) ≤ Osc ( f,P ) + Osc ( g,P ) < ε. (iii) fg ∈ R [ a,b ] . Proof. Use ϕ ( t ) = t 2 to get f 2 ∈ R , then observe 4 fg = ( f + g ) 2 ( f g ) 2 . (iv) f ≤ g on [ a,b ] = ⇒ R b a f ( x ) dx ≤ R b a g ( x ) dx . Proof. HW. Use g f ≥ 0 and part (i). (v)  f  ∈ R [ a,b ] and fl fl fl R b a f ( x ) dx fl fl fl ≤ R b a  f ( x )  dx . Proof. Use ϕ ( t ) =  t  to get  f  ∈ R , then choose c = ± 1 to make c R f ≥ 0 and observe cf ≤  f  = ⇒ fl fl fl fl Z f fl fl fl fl = c Z f = Z cf ≤ Z  f  . (vi) For a < c < b , R c a f ( x ) dx + R b c f ( x ) dx = R b a f ( x ) dx ....
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 Fall '11
 Wong
 Trigraph, Continuous function, dx

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