The Basic Properties of the Integral

Transcription

1 The Bsic Properties of the Integrl When we compute the derivtive of complicted function, like + sin, we usull use differentition rules, like d [f()+g()] d f()+ d g(), to reduce the computtion d d d to tht of finding the derivtives of the simple prts, like nd sin, of the complicted function. Ectl the sme technique is used in computing integrls. Here re unch of integrtion rules. Theorem 1 (Arithmetic of Integrtion). Let, nd A,B,C e rel numers. Let the functions f() nd g() e integrle on n intervl tht contins nd. Then () () (c) (d) [f()+g()] d [f() g()] d [Cf()] d C [Af()+Bg()] d A + +B g() d g() d g() d Tht is, integrls depend linerl on the integrnd. (e) d Proof. The first three formule re ll specil cses of formul (d). For emple, formul () is just formul (d) with A B 1. So to get the first four formule it s good enough to just prove formul (d), which we ll do just compring the definitions of the left nd right hnd sides. Let s introduce the nottion h() Af()+Bg() for the integrnd on the left hnd side. Then, Definition 4 in the notes Definition of the Integrl, the left hnd side is the limit s n of h( i,n ) n Af( i,n )+Bg( i,n ) n Af( i,n) n +Bg( i,n) n Af( i,n ) n + Bg( i,n ) n ( prt () of Theorem in the notes Definition of the Integrl ) c Joel Feldmn. 15. All rights reserved. 1 Jnur 9, 15

2 A f( i,n) n +B g( i,n) n ( prt () of Theorem in the notes Definition of the Integrl ) Sustituting the definitions of nd g() d into A + B g() d gives tht the right hnd side is ectl the limit s n of the lst line of (1). Geometricll, formul (e) just ss tht the re of the rectngle with running from to nd running from to 1 is, which is ovious. It is lso es to see formul (e) lgericll since, if we use f() 1 to denote the integrnd of the left hnd side, then, Defin ition 4 in the notes Definition of the Integrl, d lim n f( i,n ) n lim n 1 n lim ( ) n (1) Emple In Emple 1 of the notes Definition of the Integrl, we sw tht e d e. So [ e +7 ] d e d+7 d ( Theorem 1.d with A 1, f e, B 7, g 1) (e)+7 (1 ) e+6 ( Emple 1 in the notes Definition of the Integrl nd Theorem 1.e) Emple Theorem 3 (Arithmetic for the Domin of Integrtion). Let,,c e rel numers. Let the function f() e integrle on n intervl tht contins, nd c. Then () () (c) c + c c Joel Feldmn. 15. All rights reserved. Jnur 9, 15

3 Proof. For nottionl simplicit, let s ssume tht c nd f() for ll. The identities, tht is Are (,), f() nd c +, tht is, c Are (,), f() Are (,) c, f() +Are (,) c, f() re intuitivel ovious. See the figures elow. We won t give forml proof. f() f() c So we concentrte on the formul. The midpoint Riemnn sum pproimtion to with 4 suintervls is ( f + 1 +f + 3 +f + 5 +f + 7 ) ( 7 f f f f ) 8 () 4 We re now going to write out the midpoint Riemnn sum pproimtion to with 4 suintervls. Note tht is now the lower limit on the integrl nd is now the upper limit on the integrl. This is likel to cuse confusion when we write out the Riemnn sum, so we ll temporril renme to A nd to B. The midpoint Riemnn sum pproimtion to B with 4 suintervls is A ( f A+ 1 B A +f A+ 3 B A +f A+ 5 B A +f A+ 7 B A ) B A ( 7 f 8 A B +f 8 A B +f 8 A B +f 8 A+ 7 ) B A 8 B 4 Nowreclling tht A nd B, we hve tht themidpoint Riemnn sum pproimtion to with 4 suintervls is ( 7 f f f f ) 8 (3) 4 The curl rckets in () nd (3) re equl to ech other the terms re just in the reverse order. The fctors multipling the curl rckets in () nd (3), nmel nd, re 4 4 negtives of ech other, so (3) (). The sme computtion with n suintervls shows tht the midpoint Riemnn sum pproimtions to nd with n suintervls re negtives of ech other. Tking the limit n gives. c Joel Feldmn. 15. All rights reserved. 3 Jnur 9, 15

4 Emple 4 We sw, in Emple 8 of the notes Definition of the Integrl, tht, when B >, B d B nd d. B Theorem 3, B d nd B B d d B B We m comine the three sttements B d B when B > into the single sttement d B d B when B > d for ll rel numers (When >, set B nd when <, set B.) Appling Theorem 3 et gin, we hve d d+ d d d Emple 4 Emple 5 Recll tht So if if f ( ) d f ( ) d+ f( ) d+ f ( ) d Emple 5 c Joel Feldmn. 15. All rights reserved. 4 Jnur 9, 15

5 Emple 6 Here is concrete emple of how the method of Emple 5 is used. We regoing to compute [ ] 1 d gin. But this time we re going to use onl the properties of Theorems 1 nd 3 nd the fcts tht d d Tht d is prt (e) of Theorem 1. We sw tht d in Emple 4. The purpose of this emple is to show how the properties of Theorems 1 nd 3 cn e used to rewrite [ 1 ] d in terms of d nd d. First we re going to get rid of the solute vlue signs. Reclling tht whenever nd whenever, we hve, Theorem 3.c, [ ] [ ] 1 [ ] 1 d 1 d+ 1 d [ ] 1 [ ] 1 ( ) d+ 1 d [ ] 1 [ ] 1+ d+ 1 d Now we ppl prts () nd () of Theorem 1, nd then (4). [ ] 1 d d+ d+ d d (4) [ ()]+ () 1 +[1 ] 1 Emple 6 More Properties of Integrtion Even nd Odd Functions (Optionl) Recll tht A function f() is sid to e even if f( ) f() for ll. Two emples of even functions re f() cos nd f(). In fct, if f() is n even power of, then f() is n even function. The prt of the grph f() with, m e constructed drwing the prt of the grph with (s in the figure on the left elow) nd then reflecting it in the is (s in the figure on the right elow). elow. c Joel Feldmn. 15. All rights reserved. 5 Jnur 9, 15

6 1 1 In prticulr, if f() is n even function nd >, then the two sets (,) nd is etween nd f() (,) nd is etween nd f() re reflections of ech other in the is nd so hve the sme signed re. Tht is A function f() is sid to e odd if f( ) f() for ll. Two emples of odd functions re f() sin nd f() 3. In fct, if f() is n odd power of, then f() is n odd function. The prt of the grph f() with, m e constructed drwing the prt of the grph with (like the solid line in the figure on the left elow) nd then reflecting it in the is (like the dshed line in the figure on the left elow) nd then relecting the result in the is (i.e. flipping it upside down, like in the figure on the right, elow). 1 1 In prticulr, if f() is n odd function nd >, then the signed res of two sets (,) nd is etween nd f() (,) nd is etween nd f() c Joel Feldmn. 15. All rights reserved. 6 Jnur 9, 15

7 re negtives of ech other to get from the first set to the second set, ou flip it upside down, in ddition to reflecting it in the is. Tht is Theorem 7 (Even nd Odd). Let >. () If f() is n even function, then () If f() is n odd function, then Proof. For n function + When f is even, the two terms on the right hnd side re equl. When f is odd, the two terms on the right hnd side re negtives of ech other. c Joel Feldmn. 15. All rights reserved. 7 Jnur 9, 15

8 More Properties of Integrtion Inequlities for Integrls (Optionl) Theorem 8 (Inequlities for Integrls). Let e rel numers nd let the functions f() nd g() e integrle on the intervl. () If f() for ll, then () If there re constnts m nd M such tht m f() M for ll, then m( ) (c) If f() g() for ll, then M( ) g() d (d) We hve f() d Proof. () Just ss tht if the curve f() lies ove the is nd, then the signed re of (,), f() is t lest zero, which is ovious. Alterntivel, we could just oserve tht, in the definition of (Definition 4 in the notes Definition of the Integrl ), ever term f( i,n ). n () Since f() M, we hve M f() so tht [ ] M M d M( ) which implies tht M( ). The rgument showing m( ) is similr. (c) Since f() g(), we hve g() f() so tht [ ] g() g() d g() d (d) For n, f() is either f() or f(depending on whether f() is positive or negtive), so we certinl hve f() f() f() f() c Joel Feldmn. 15. All rights reserved. 8 Jnur 9, 15

9 Appling prt (c) to ech of those inequlities gives f() d f() d Since f() d is equivlent to f() d (move the to the right hnd side of the inequlit nd the f() d to the left hnd side of the inequlit), we hve Tht s just nother w to s f() d f() f() d f() d Emple 9 ( /3 cos d) For etween nd, the function cos tkes vlues etween 1 nd 1 3 cos tkes vlues etween 1 nd 1. Tht is nd so the function 3 1 cos 1 Consequentl, Theorem 8. with, 3, m 1 nd M 1, /3 3 cos d 3 Emple 9 c Joel Feldmn. 15. All rights reserved. 9 Jnur 9, 15