The Fundamental Theorem of Calculus May 2 2010 The fundamental theorem of calculus has two parts Theorem Part I Let f be a continuous function on a b and define a function g a b R by Z x g x f a Then g is differentiable on a b and for every x a b g 0 x f x At the end points g has a one sided derivative and the same formula holds That is the right handed derivative of g at a is f a and the left handed derivative of f at b is f b Proof This proof is surprisingly easy It just uses the definition of derivatives and the following properties of the integral Rb 1 If f is continuous on a b then a f exists 2 If f is continous on a b and c a b then Z c f Z b a c f Z b f a 3 If m f M on a b then b a m Z b a 1 f b a M Let x be a point in a b We just treat the case of x a b since the endpoints can be treated similarly If x a b we shall show that g 0 x g 0 x f x Knowing that the two one sided derivatives exist and are equal we can conclude that the derivative exists and has this value By definition g x h g x g 0 x lim h 0 h Property 1 assures us that g is well defined provided that h b x Property 2 allows us to simplify the numerator since it implies that g x h g x Z x h a f Z x f a Z x h f 1 x This is already great since we only need to worry about f over the small interval x x h A picture is helpful here but I don t have time to include one in thise notes Draw one yourself Now recall the definition of a limit We have to show that given any 0 there is a 0 such that g x h g x f x h 2 whenever 0 h This is where we use the continuity of f at x We know from this that there is a such that f x0 f x whenever x0 x This means that f x f x0 f x for all such x0 We use this same our criterion for the limit in equation 2 Let us verify that this works Suppose that 0 h Then on the interval x x h we know that f is between f x and f x By property 3 of integrals it follows that f x h Z x h f f x h x Since h 0 we can divide both sides by h to conclude that f x 1 h Z x h f f x i e x f x g x h g x f x h 2 This is exactly what we needed The left handed derivatives are done in essentially the same way Theorem Part II Let f be a continuous function on a b Suppose that F is continuous on a b and that F 0 f on a b Then Z b f F b F a a Proof Consider the function g in the previous theorem Since g is differentiable on a b it is continuous there including at the end points where the one sided deritaives exist We also know that g and F are differentiable on a b and that there derivatives are equal Recall that we had as a consequence of the mean value theorem for derivatives that F and g differ by a constant That is there is a number C such that g x F x for all x a b Then F b F a g b C g a C g b g a Z b a 3 f Z a a f Z b a f
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