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Unformatted text preview: Honors Introduction to Analysis I Homework X Solution April 26, 2009 Problem 1 Prove the integral mean value theorem: if f is continuous on [ a,b ] , then there exists y ∈ ( a,b ) such that R b a f ( x ) dx = ( b a ) f ( y ) . Solution. Let F ( x ) = R x a f ( t ) dt , for a ≤ x ≤ b . By theorem 6.1.2, F is C 1 and F = f . We can apply the mean value theorem for F , and we get that there is a y ∈ ( a,b ) such that F ( y ) = F ( b ) F ( a ) b a . But F ( b ) F ( a ) = R b a f ( x ) dx and F ( y ) = f ( y ). Hence there is a y ∈ ( a,b ) such that Z b a f ( x ) dx = ( b a ) f ( y ) Problem 2 Let g be continuous on [ a,b ] , and let f ( x ) = R x a ( x t ) g ( t ) dt . Prove that f is a solution of the differential equation f 00 = g and the initial conditions f ( a ) = f ( a ) = 0 . Solution. First, just by the definition of f , we have that f ( a ) = R a a ( x t ) g ( t ) dt = 0. Also, since ( x t ) g ( t ) is continuous in x , then f is at least C 1 . By theorem 6.1.7, for a ( x ) = a , b ( x ) = x , and instead of g ( x,t ) consider ( x t ) g ( t ), we have that f ( x ) = b ( x )( x b ( x )) g ( b ( x )) a ( x )( x a ) g ( a )+ Z x a ∂ [( x t ) g ( t )] ∂x dt = ( x x ) g ( x ) · ( x a ) g ( a )  {z } =0 + Z x a g ( t ) dt Hence f ( x ) = R x a g ( t ) dt , from which it follows that f ( a ) = R a a g ( t ) dt = 0. Since g is continuous, we get that f is at least C 1 , so f is at least C 2 , which means we can take the second derivative. All we have to do is to apply theorem 6.1.2 to f , and get that f 00 ( x ) = g ( x ). Problem 3 Suppose you want to compute a table of arctangents for values of x in [0 , 1] using the formula arctan x = R x 1 / (1 + t 2 ) dt . What size must you take for δ if you want the error to be at most 1 / 1000 using the midpoint rule....
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This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell.
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 PROTSAK
 Mean Value Theorem

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