Bob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk

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1 Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions of ech other, much in the sme sense tht division nd multipliction re inverse opertions of ech other. PreClculus Clculus Differentition slope of secnt line = slope of tngent line Definite Integrtion re of rectngle = re of region

2 Bo Brown Mth Clculus Chpter, Section CCBC Dundlk Theorem: Formlly, the Fundmentl Theorem of Clculus sys the following: If function f is continuous on the closed intervl nd if F is n ntiderivtive of f on the intervl, then f ( d = Written nother wy: = F() F() Eercise : Compute d using two different methods. (i) geometriclly (ii) using the FUNdmentl Theorem of Clculus f ( F( F( = Then, y the FTC, d = L L d = Are of Trpezoid = W = Question: Why don t we hve to worry out which constnt, c, tht we use when we pply the FTC? Suppose tht F( c. Then F() F() = Eercise : Evlute the following definite integrls. (i) 6 d Let F (. Then F( =. Therefore, 6 d = (ii) Let 6 d F ( 6. Then F( = Therefore, 6 d =.

3 Bo Brown Mth Clculus Chpter, Section CCBC Dundlk (iii) 6 d Let F (. Then F( =. Therefore, 6 d = (iv) 6 7 d If F ( 6, from Eercise (iii) on pge of Hndout., we lredy determined tht n ntiderivtive is F( 6. Thus, 6 7 d = 8 (v) d If F (, from Eercise (ii) on pge of Hndout., we lredy determined tht n ntiderivtive is 8 Therefore, d = F(. (vi) sin( ) d Let 0 F (. Then F( = Therefore, sin( ) d = 0

4 Bo Brown Mth Clculus Chpter, Section CCBC Dundlk Men Vlue Theorem for Integrls There is theorem tht sttes tht the re of the region under curve (relly, the re etween the curve nd the -is) is greter thn or equl to the re of n inscried rectngle nd less thn or equl to the re of circumscried rectngle. The Men Vlue Theorem for Integrls uilds on this theorem nd sttes tht, somewhere etween the inscried rectngle nd the circumscried rectngle, there is rectngle with se whose re is precisely equl to the re of the region under the curve. Theorem: If function f is continuous on the closed intervl numer c stisfying c such tht, then there eists f ( d = Averge Vlue of Function Def.: The vlue, f(c), given in the Men Vlue Theorem for Integrls, is clled the verge vlue of f on the intervl. Dividing oth sides of the eqution in the Theorem ove y, we rrive t the verge vlue of f on. Averge vlue = f(c) =

5 Bo Brown Mth Clculus Chpter, Section CCBC Dundlk Eercise : Determine the verge vlue of f ( over the intervl 8. crude pproimtion, using only smll, finite numer of function (y) vlues ectly, y the definition 8 Eercise : Determine the verge vlue of g( = + 6 over the intervl 7. Eercise c: Determine the verge vlue of h( = sin( over the intervl 0.

6 Bo Brown Mth Clculus Chpter, Section 6 CCBC Dundlk The Second Fundmentl Theorem of Clculus Erlier we sw tht the definite integrl of f on the intervl ws defined using the constnt s the upper limit of integrtion nd s the vrile of integrtion. However, slightly different sitution my rise in which the vrile is used s the upper limit of integrtion. In order to void the confusion of using in two different wys, t is temporrily used s the vrile of integrtion. Rememer tht the definite integrl is not function of its vrile of integrtion. Definite Integrl s Numer Definite Integrl s Function of f ( d F ( f ( t) dt Eercise : Evlute the function F( t dt t = -, 0,, nd. F ( ) t dt = F ( 0) t dt = F ( ) t dt = F ( ) t dt = 0 Eercise : Wht is the formul for F(? Tke guess sed on the ove clcultions, nd then use integrtion to verify your guess. (Note tht this F( is type of function tht is often referred to s n ccumultion function.)

7 Bo Brown Mth Clculus Chpter, Section 7 CCBC Dundlk Eercise c: F ( ) is formul for the etween nd the from to. Eercise d: F ( = = = Theorem: (Second Fundmentl Theorem of Clculus) If f is continuous on n open intervl I contining, then, for every vlue in the intervl I, Eercise : Determine the derivtive. (i) y t dt (ii) y t dt

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