MATH 140 Lecture Notes - Lecture 30: Antiderivative, Mean Value Theorem

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Math140 lecture 30 fundamental theorem of calculus: note 2: if f is greater than or equal to 0, then g(x) is the area below the graph of f and the x-axis. Theorem 5. 12: (partial) proof, let f be continuous on [a, b]. (cid:1856)(cid:1872) for (cid:1853) (cid:1854). = (cid:4666)(cid:4667) (cid:3276) (cid:3276) (cid:3031) (cid:3031) (cid:4666)(cid:4667: note that: (cid:1863)(cid:4666)(cid:4667) (cid:1856)+ (cid:1863)(cid:4666)(cid:4667) (cid:1856)= (cid:1863)(cid:4666)(cid:4667) (cid:3029)(cid:3028) (cid:3030)(cid:3029) (cid:3030)(cid:3028) =(cid:4666)(cid:4667)(cid:4666) (cid:4667) (cid:3031: thus, (cid:4666)(cid:4667, let (cid:1833)(cid:4666)(cid:4667)= (cid:1872)(cid:2871)(cid:1856)(cid:1872) , then (cid:1833) (cid:4666)(cid:4667): if y = h(x) and h"(x) exists, then if (cid:1833)(cid:4666)(cid:4667)= (cid:1858)(cid:4666)(cid:1872)(cid:4667)(cid:1856)(cid:1872) (cid:1857)(cid:3118)(cid:4666)(cid:884)(cid:4667)+(cid:1857)(cid:3119)(cid:4666)(cid:885)(cid:2870)(cid:4667: generally, if h and k are differentiable on [a, b] and (cid:1833)(cid:4666)(cid:4667)= (cid:1858)(cid:4666)(cid:1872)(cid:4667)(cid:1856)(cid:1872) (cid:4666)(cid:4667) ,(cid:1853) (cid:1872), then (cid:1833) (cid:4666)(cid:4667)=(cid:1858)(cid:4666)(cid:4667) for (cid:1853) (cid:1854): using the chain rule: Fundamental theorem of calculus: let f be continuous on [a, b]. Mean value theorem for integrals: proof, proof, for first part: see theorem 5. 12, for second part: =(cid:1832)(cid:4666)(cid:1854)(cid:4667) (cid:1832)(cid:4666)(cid:1853)(cid:4667): if f is any antiderivative of f on [a, b], then (cid:1858)(cid:4666)(cid:1872)(cid:4667)(cid:1856)(cid:1872) (cid:3029)(cid:3028, (cid:1858)(cid:4666)(cid:1872)(cid:4667)(cid:1856)(cid:1872)

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