Fundamental Theorem of Calculus

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1 Funmentl Teorem of Clculus Liming Png 1 Sttement of te Teorem Te funmentl Teorem of Clculus is one of te most importnt teorems in te istory of mtemtics, wic ws first iscovere by Newton n Leibniz inepenently. Tis teorem revels te unerlying reltion between ifferentition n integrtion, wic glues te two subjects into uniform one, clle clculus. Teorem 1. (Te Funmentl Teorem of Clculus) Suppose f() is continuous function on [, b]. Ten: i). f(t) t = f() ii). b f() = F (b) F (), were F is n ntierivtive of f Remrk 2. We sometimes lso write F (b) F () in te wy F () b, so Prt (ii) of te teorem cn lso be written s b b f() = F () Now we re going to look into ec of tese two prts of te teorem. 2 Prt I Let g() = f(t) t, note g() is function of. Prt (i) tells us tt g() = f(), i.e. g() is n ntierivtive of f(). First we see geometric eplntion: f(t) t is te signe re boune by f(t) n te intervl [, ]. If we increse to +, te increse in te 1

2 re is + f(t) t f() wen is smll number. Tis inictes f() + wen is smll, so it implies f(t) t g( + ) g() g () = lim g( + ) g() = f() Now let s formulte n lgebric proof. Proof. Define g() in te sme wy s bove, we ve g( + ) g() = + f(t) t f(t) t = + f(t) t We my ssume >, te oter cse cn be iscusse in similr fsion. Consier te continuous function f on te close intervl [, + ]. We know it s bsolute mimum M n bsolute minimum m on [, + ], so m + f(t) t M. Assume f(u) = M n f(v) = m, were u, v re in [, + ]. We get: f(v) f(v) f(v) + + f(t) t f(u) f(t) t f(u) g( + ) g() f(u) 2

3 If we tke, since u, v re in [, + ], it follows u n v, so f(u) f() n f(v) f(), te lst line of te inequlities implies g( + ) g() lim f(v) lim i.e. f() g () f(), we see g () = f(). Emple 3. Fin 2 e 3t t 1 lim f(u) Let g() = 1 e3t t, we see 2 e 3t t = g( 2 ). Te Funmentl Teorem 1 of Clculus sys g () = e 3. By te cin rule we get g(2 ) = g ( 2 )( 2 ) = e 3(2) (2) = 2e 32 Emple 4. Compute t2 t Observe tt t2 t = 3 5 t 2 t 2+1 t 2 t, we see t 2 t = 3 5 t 2 t 2+1 t 2 t = (3 5) 2 (3 5) (2 + 1) 2 (2 + 1) = (3 5) 2 3 (2 + 1) 2 2 = 3(3 5) 2 2(2 + 1) 2 Definition 5. f is integrble on [, b], efine te verge of f on [, b] to be te number f [,b] = 1 b f() b Emple 6. If v(t) is velocity function, ten te verge pf v(t) on [, b], v [,b] = 1 v(t) t, is efine to be te verge velocity on [, b]. b b 3 Prt II We first give proof. 3

4 Proof. By Prt (i), f(t) t = f(), so g() = f(t) t is n ntierivtive of f(). If F () is noter ntierivtive of f(), we know tt F () = g() + C for some constnt C. Tis implies F (b) F () = g(b) g() = b f(t) t f(t) t = b f(t) t Remrk 7. Tere is n intuitive eplntion of Prt (ii) s well. We know tt b f() = lim n m i f( i ) i, so wen we tke m very smll, we ve b f() f( i ) i Now on ec [ i 1, i ], since tis is smll intervl, we ve so b f( ) = F ( ) F ( i) F ( i 1 ) i i 1 = F ( i) F ( i 1 ) i f() F ( i ) F ( i 1 ) f( ) i f( i ) i F ( i ) F ( i 1 ) = F (b) F () finlly, wen we tke te limit m i, we get te equlity. Emple 8. Evlute 3 1 e An ntierivtive of f() = e Teorem of Clculus, is F () = e, so by te Funmentl 3 1 e = F (3) F (1) = e 3 e Definition 9. Te inefinite integrl of function f is efine to be f() = F () + C, were F () = f(), i.e. F () is n ntierivtive of f(), n C enotes constnt. 4

5 Emple 1. Evlute 3 (3 4) Since ( 3 4) = C, we ve 3 ( 3 4) = ( ) ( ) = Applictions Emple 11. We re going to prove te re formul for circles. First, recll te efinition of te number π: π is te rtio of te circumference n imeter of circle. By tis efinition, we know tt te circumference of circle of rius R is 2πR, since 2R is te imeter. Now given circle of rius R, we re going to fin its re. We ivie [, R] into n subintervls of equl lengt r = R n, wit enpoints r =, r 1,..., r n 1, r n = R, n by te following picture, we see tt wen n is getting big, te re of te circle cn be pproimte by te following: R n = (2πr i ) r Tking te limit, we get te re of te circle is lim R n = n R R 2πr r = πr 2 = πr 2 5

6 Emple 12. We cn lso use te ie bove to obtin te volume formul for bll of rius R. A bll of rius R cn be escribe to be te region 2 + y 2 + z 2 R 2 on Crtesin coorinte system. We cn subivie te bll into n orizontl pieces of equl eigt z = 2R, n let r n =, r 1 = + 2R,..., r n n = R. Wen n is big, te volume of te i-t piece is close to cyliner wit rius R2 ri 2 n eigt z, wic is R n = π( R 2 ri 2)2 z = We get te volume for te bll to be lim R n = lim n n π(r 2 ri 2 ) z π(r 2 ri 2 ) z = An ntierivtive of 1 r 2 is R 2 r r3 3, so R π(r 2 r 2 ) = π R R π(r 2 r 2 ) (R 2 r 2 ) = π(r 2 r r3 3 ) = 4 3 πr3 Emple 13. In tis emple we will get te surfce re of spere of rius R. Given bll of rius R, tere is noter wy to compute its volume: ivie te intervl [, R] into n subintervls of equl lengt r to ecompose te bll into sells. If we enote te surfce re of spere of rius r to be S(), ten we see te volume of te bll cn be pproimte by R R n = f(r i ) r 6

7 Tking te limit s n, we get te volume: R f(r) r By Prt (i) of te teorem, we know R f(r) r = f(r), n on te oter R n, we ve lrey compute in te previous emple tt R f(r) r = 4 3 πr3, so f(r) = R (4 3 πr3 ) = 4πR 2 7

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