MATH 249 Lecture Notes - Lecture 1: Function Composition, Product Rule, Antiderivative
15 Oct 2016
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Suppose that f is continuous on [a; b] : F (x) =z x a f (t) dt; a (cid:20) x (cid:20) b: Then f is continuous on [a; b] and di erentiable on (a; b) : In fact, f is an antiderivative of f on (a; b) : F 0 (x) = f (x) for all x in (a; b) : Fix a number x in the interval [0; 5] such that x 6= 0: then. = the area under the graph of f on the interval [0; x] = the area of a rectangle with base x and height 3. By de(cid:133)nition of the de(cid:133)nite integral, f (0) =r 0. F (x) = 3x for all x in [0; 5] : The fundamental theorem of calculus tells us that f is an antiderivative of f on the interval (0; 5) : this can be veri(cid:133)ed easily: F 0 (x) = d dx (3x) = 3 = f (x) :
