The Fundamental Theorem of Calculus

Transcription

1 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk Te Funmentl Teorem of Clculus EXAMPLE: If f is function wose grp is sown below n g() = f(t)t, fin te vlues of g(), g(), g(), g(3), g(4), n g(5). Ten sketc roug grp of g. Solution: First we notice tt g() = f(t)t =. From te figure bove we see tt g() is te re of tringle: g() = f(t)t = ( ) = To fin g() we to g() te re of rectngle: g() = f(t)t = f(t)t+ f(t)t = +( ) = 3 We estimte tt te re uner f from to 3 in bout.3, so 3 g(3) = g()+ f(t)t 3+.3 = 4.3 For t > 3, f(t) is negtive n so we strt subtrcting res: 4 g(4) = g(3)+ f(t)t 4.3+(.3) = g(5) = g(4)+ f(t)t 3+(.3) =.7 4

2 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk We use tese vlues to sketc te grp of g: g() = g() = g() = 3 g(3) 4.3 g(4) 3 g(5).7 EXAMPLE: If g() = clculte g (). f(t)t, were = n f(t) = sint, fin formul for g() n Solution: By te Evlution Teorem we ve: g() = Ten g () = ( +) =. sintt = cost] = ( cos) = + REMARK: To see wy tis migt be generlly true we consier continuous function f wit f(). Ten g() = f(t)t Tocomputeg ()fromteefinitionoferivtivewefirstobservett, for >, g(+) g() is obtine by subtrcting res, so it is te re uner te grp of f from to + (te gol re). For smll you cn see tt tis re is pproimtely equl to te re of te rectngle wit eigt f() n wit : Intuitively, we terefore epect tt g(+) g() f() = g(+) g() g () = lim g(+) g() = f() f() Te fct tt tis is true, even wen f is not necessrily positive, is te first prt of te Funmentl Teorem of Clculus.

3 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk THEOREM (Te Funmentl Teorem Of Clculus, Prt I): If f is continuous on [,b], ten te function g efine by g() = f(t)t b is n ntierivtive of f, tt is g () = f() for < < b. Proof: If n + re in te open intervl (,b), ten + g(+) g() = f(t)t f(t)t = f(t)t+ + f(t)t f(t)t = + f(t)t n so, for, g(+) g() = + f(t)t () For now let s ssume tt >. Since f is continuous on [,+], te Etreme Vlue Teorem sys tt tere re numbers u n v in [, + ] suc tt f(u) = m n f(v) = M, were m n M re te bsolute minimum n mimum vlues of f on [,+]. By Property 8 of integrls m(b ) b f() M(b ) we ve + m f(t)t M = f(u) + f(t)t f(v) Since >, we cn ivie tis inequlity by : f(u) + f(t)t f(v) Now we use () to replce te mile prt of tis inequlity: f(u) g(+) g() f(v) () Inequlity () cn be prove in similr mnner for te cse <. Now let. Ten u n v, since u n v lie between n +. Tus limf(u) = limf(u) = f() n limf(v) = limf(v) = f() u v becuse f is continuous t. We conclue, from () n te Squeeze Teorem, tt g () = lim g(+) g() = f() (3) If = or b, ten(3) cn beinterpretes one-sielimit. Weknowtt if f is ifferentible t, ten f is continuous t. If we opt tis teorem for one-sie limits, we obtin tt g is continuous on [,b]. 3

4 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk EXAMPLE: Fin te erivtive of te function g() = t 4 t. Solution: Since f(t) = t 4 is continuous, Prt of te Funmentl Teorem of Clculus gives g () = 4 EXAMPLE: Fin te erivtive of g() = e t t. Solution: Since f(t) = e t is continuous, Prt of te Funmentl Teorem of Clculus gives 3 EXAMPLE: Fin te erivtive of g() = g () = e e t t. Solution: Since f(t) = e t is continuous, Prt of te Funmentl Teorem of Clculus gives In sort, e t t = u e t t = u EXAMPLE: Fin te erivtive of g() = u e t t u = euu e t t = e ( ) ( ) = e 4 3 tt. = e( ) = e 4 EXAMPLE: Fin te erivtive of g() = 4 sectt. 4

5 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk EXAMPLE: Fin te erivtive of g() = 3 tt. Solution: Prt of te Funmentl Teorem of Clculus gives 3 u 3 u tt = tt = 3 tt u u = 3 u u = 3 In sort, 3 tt = 3 () = 3 EXAMPLE: Fin te erivtive of g() = sectt. 4 Solution: By Property of te Definite Integrl n Prt of te Funmentl Teorem of Clculus we ve 4 sectt = sectt = u sectt = u sec tt u u = secuu 4 = sec( 4 ) 4 3 In sort, sectt = 4 4 EXAMPLE: Fin te erivtive of g() = sectt = sec( 4 )( 4 ) = 4 3 sec( 4 ) t t. = 4 3 sec( 4 ) 5

6 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk EXAMPLE: Fin te erivtive of g() = t t. Solution: We ve t t = t t+ t t = t t+ t t terefore t t = t t+ t t = sin () +cos () = sin +cos ( ) = sin cos THEOREM (Te Funmentl Teorem Of Clculus, Prt II): If f is continuous on [,b], ten b f() = F(b) F() were F is ny ntierivtive of f, tt is F = f. Proof: Put g() = f(t)t By te Funmentl Teorem Of Clculus, Prt I, g() is n ntierivtive of f(). Terefore ny oter ntierivtive F() of f() cn be written s It follows tt F() = F() = g()+c = f(t)t+c = +C = C = F(b) = f(t)t+c b f(t)t+c = b f(t)t+f() tus b b F(b) = f(t)t+f() = F(b) F() = f(t)t 6

7 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk THEOREM (Te Men Vlue Teorem for Integrls): If f is continuous on [, b], ten tere eists number c in [,b] suc tt f(c) = f ve = b f() or b b f() = f(c)(b ) EXAMPLE: Fin te verge vlue of te function f() = on te intervl [,4]. Solution: We ve f ve = 4 4 = / = We now fin c: ] /+ 4 = /+ 3 ] 4 3 3/ = ( / ) 3 3/ = 4 9 f(c) = f ve = 4 9 = c = 4 9 = c = ( ) 4 = = 96 8 EXAMPLE: Fin te verge vlue of te function f() = + on te intervl [,]. EXAMPLE: Fin te verge vlue of te function f() = 4 on te intervl [,]. 7

8 Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk EXAMPLE: Fin te verge vlue of te function f() = + on te intervl [,]. Solution: We ve f ve = ( ) (+ ) = 3 ] [+ 3 = 3 We now fin c: f(c) = f ve = = +c = = c = = c = ± EXAMPLE: Fin te verge vlue of te function f() = 4 on te intervl [,]. Solution: We ve We now fin c: f(c) = f ve = π f ve = ( ) = 4 c = π 4 = 4 π = 4 c = π 4 = π = c = ± 4 π 4 ±