The Fundamental Theorem of Calculu1

The Fundamental Theorem of Calculu1 - f mh ) ( + h x x h v...

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The Fundamental Theorem of Calculus (Part 1) If f is continuous on [ a, b ], then the function g defined by = x a dt t f x g ) ( ) ( a x b Is continuous on [ a, b ] and differentiable on ( a, b ), and g’ ( x ) = f ( x ). Proof: + - = - + h x a x a dt t f dt t f x g h x g ) ( ) ( ) ( ) ( = + - + x a h x x x a dt t f dt t f dt t f ) ( ) ) ( ) ( ( = + h x x dt t f ) ( And so, for h ≠ 0, Eq 2: h x g h x g ) ( ) ( - + = h 1 + h x x dt t f ) ( For now let us assume that h > 0. Since f is continuous on [ x, x+h ], the Extreme Value Theorem says that there are numbers u and v in [ x, x+h ] such that f ( u ) = m and f ( v ) = M , where m and M are the absolute minimum and maximum values of f on [ x, x+h ]. + h x x Mh dt t
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Unformatted text preview: f mh ) ( + h x x h v f dt t f h u f ) ( ) ( ) ( Since h > 0, we can divide this inequality by h : + h x x v f dt t f h u f ) ( ) ( 1 ) ( Use equation 2 (Eq 2) to replay the middle part of this inequality: ) ( ) ( ) ( ) ( v f h x g h x g u f -+ Now we let h 0. Then u x, and v x , since u and v lie between x and x + h . ) ( ) ( lim ) ( lim x f u f u f x u h = = ) ( ) ( lim ) ( lim x f v f v f x v h = =...
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