b n 2 b n b n 3 b n b n nb n b n b 2 n 2 1 2 3 n b 2 n 2 n n 1 2 1 2 as n

B n 2 b n b n 3 b n b n nb n b n b 2 n 2 1 2 3 n b 2

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· b n + 2 b n · b n + 3 b n · b n + · · · + nb n · b n = b 2 n 2 (1 + 2 + 3 + · · · + n ) = b 2 n 2 · n ( n + 1) 2 1 2 as n → ∞ Therefore, f ( x ) = x is integrable and integraldisplay b 0 x dx = 1 2 . 2
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8. Set f ( x ) = 1 , x rational - 1 , x irrational x [ - 1 , 1] Then, f 2 ( x ) 1 is integrable, integraldisplay 1 - 1 f 2 ( x ) dx = integraldisplay 1 - 1 1 dx = 2. f is not integrable since for any partition P , L ( f, P ) = - 2 and U ( f, P ) = 2. 12. Suppose that f is bounded and f ( x ) 0 for all x [ a, b ]. Then, for any partition P = { a = x 0 , x 1 , x 2 , · · · , x n = b } , m i = inf f ( x ) , x [ x i - 1 ,x i ] 0. Therrefore, L ( f, P ) 0. Since L ( f ) ≥ L ( f, P ), it follows that L ( f ) 0. 13. f is continuous on [ a, b ], f ( x ) 0 for all x , and L ( f ) = integraltext b a f = 0. Prove f ( x ) = 0 for all x . Suppose there is a point c [ a, b ] such that f ( c ) > 0, and suppose, first, that c ( a, b ). By Exercise 21.13 there is a neighborhood N ( c ) (i.e., an interval ( c - δ, c + δ ) ) and a number α > 0 such that f ( x ) α on N ( c ). Let P be the partition { a, c - δ, c + δ, b } . Then L ( { , P ) = inf ( floorright- δ -turnstileright )+inf ( floorright + δ - ( floorright- δ ))+inf ( floorleft-floorright + δ ) + α ·∈ δ + prime· ( floorleft-floorright + δ ) ≥ ∈ δα > prime This is a contradiction since L ( { , P ) ≤ L ( { ) = prime . Additional Problems 1. Suppose that f : ( - 1 , 1) R is infinitely differentiable, and that vextendsingle vextendsingle f ( k ) ( x ) vextendsingle vextendsingle 3 on ( - 1 , 1) for all k . Let p n be the Taylor polynomial of degree n for f in powers of x . (a) Prove that f is uniformly continuous on ( - 1 , 1). Proof: Let epsilon1 > 0. Set δ = epsilon1/ 3. Choose any points x 1 , x 2 I (Assume x 1 < x 2 ). By the Mean Value Theorem, there is a number c ( x 1 , x 2 ) such that f ( x 2 ) - f ( x 1 ) = f prime ( c )( x 2 - x 1 ) .
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