solutionshw6-555-2011

# solutionshw6-555-2011 - b-c . Page 149 Problem 3. Using the...

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Solutions homework 6. Page 141 Problem 2. If f g , then m k ( f ) = inf { f ( x ) : x k - 1 x x k } ≤ m k ( g ) = inf { g ( x ) : x k - 1 x x k } and thus s f ( σ ) s g ( σ ) for any subdivision σ .. By deﬁnition this implies that Z a b f Z a b g. The corresponding inequalities for the upper integral are the same. Page 141 Problem 3. For any subdivision σ we have that m k = 0, so that all lower sums s ( σ ) = 0. Hence the lower integral R a b f = 0. For the upper integral we use a subdivision σ = { a = x 0 < x 1 = c - ±/ 2 < x 3 = c + ±/ 2 < x 4 = b } to get an upper sum S ( σ ) = ± . Hence the upper integral R b a f = 0. For the second part using a subdivision σ containing the interval [ c - ±,c + ± ], we get that s ( σ ) = b - c - ± , so the lower integral is b - c . For the same subdivisions we get that S ( σ ) = b - c + ± . Hence the upper integral is b - c . Combining the two inequalities we see that both are equal to
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Unformatted text preview: b-c . Page 149 Problem 3. Using the above problem (replace b by x for c < x b ) we get that F ( x ) = 0 for a x c and F ( x ) = x-c for c < x b . Page 153 Problem 5. True, as by the rst problem of this assignment, we get 0 R a b f R b a f = 0. Hence the lower and upper Riemann integral of f are equal, so f is Riemann integrable. Page 165 Problem 1. Take g ( x ) = 0 for all x [ a,b ]. Let { x n : n 1 } = Q [ a,b ] and dene f ( x n ) = 1 for all n and f ( x ) = 0 for all other x [ a,b ]. Then f is not Riemann integrable, but f ( x ) = g ( x ) except for x = x n 1...
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## This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.

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