solutionshw6-555-2011

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online