INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).
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1 INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely mny poins, we cn ssume wihou loss of generliy h Q hs jus more poin hn P, sy Q = P {s}. Since he heory of lower sums is prllel o he heory of upper sums (jus reverse he inequliies nd swp sup nd inf), i suffices o show h U(P ) U(Q). Le i be he unique elemen of {, 2,..., n} such h Pu Then We hve i < s < i. M i = sup{f() : i s} M i = sup{f() : s i } i = s i i = i s. M i M i, M i M i, nd i = i + i. U(P ) = M j j = j= M j j + M i i + j<i M j j, j>i nd so U(P ) U(Q). U(Q) = j<i M j j + M i i + M i i + M j j, j>i M i i = M i i + M i i M i i + M i i, De: Ocober 23, 25.
2 2 JOHN QUIGG Eercise 2. Le f : [, b] R be inegrble. Prove h for ll ɛ > here eiss priion P of [, b] such h U(P ) L(P ) < ɛ. Soluion: Le ɛ >. Choose priion P of [, b] such h boh Then U(P ) L(P ) < ɛ. U(P ) < f + ɛ 2 nd L(P ) > f ɛ 2. Eercise 3. Le f : [, b] R be bounded. Prove h if for ll ɛ > here eiss priion P of [, b] such h U(P ) L(P ) < ɛ, hen f is inegrble. Soluion: Pu c = sup{l(p ) : P is priion of [, b]} d = sup{u(p ) : P is priion of [, b]}. For ll priions P nd Q of [, b] we hve L(P ) U(Q), becuse, leing R = P Q, we hve L(P ) L(R) since P R, nd similrly U(R) U(Q) since Q R, nd of course L(R) U(R). Thus c d, so o show h f is inegrble, i suffices o show h for ll ɛ > we hve d c < ɛ. Le ɛ >. Choose priion P of [, b] such h U(P ) L(P ) < ɛ. Since U(P ) d nd L(P ) c, we hve d c < ɛ, s desired. Eercise 4. Le f : [, b] R be bounded, nd le v R. Prove h if for ll ɛ > here eiss priion P of [, b] such h for every Riemnn sum u ssocied o P we hve u v < ɛ, hen f is inegrble nd f = v. Soluion: Assume he condiion in involving Riemnn sums. Le ɛ >. Choose priion P such h for every ssocied Riemnn sum u we hve u v < ɛ/3. Then for ll ssocied Riemnn sums u, w we hve u w u v + v w < ɛ 3 + ɛ 3 = 2ɛ 3. Then U(P ) L(P ) = sup{ u w : u nd w re Riemnn sums ssocied o P }, 2ɛ 3 < ɛ. Thus f is inegrble. For he oher pr, we rgue by conrdicion: suppose f v. By symmery i suffices o ssume h f < v. Pu ɛ = ( v 2 ) f.
3 INTEGRALS 3 Then ɛ >. Choose priion P such h for every ssocied Riemnn sum u we hve u v < ɛ, hence v ɛ < u. Since L(P ) is he inf of he ssocied Riemnn sums, we hve v ɛ L(P ). Bu hen which is conrdicion. L(P ) v ɛ = f + ɛ > Eercise 5. Le f : [, b] R be inegrble. Prove h for ll ɛ > here eiss priion P of [, b] such h for every Riemnn sum u ssocied o P we hve u f < ɛ. Soluion: Le ɛ >. Choose priion P of [, b] such h U(P ) L(P ) < ɛ. Le u be Riemnn sum ssocied o P. Then We lso hve Therefore L(P ) u U(P ). L(P ) f U(P ). u f U(P ) L(P ) < ɛ. f, Eercise 6. (Dirichle funcion) Prove h he funcion f : [, ] R defined by { if Q f() = if / Q is no inegrble. Soluion: Le P = {,..., n } be ny priion of [, ]. Le i {,..., n}. By Densiy of Rionls nd Densiy of Irrionls, here eis Q nd y / Q such h, y [ i, i ], nd hen f() = nd f(y) =. Thus M i = nd m i =. Hence U(P ) = M i i = i = L(P ) = i= m i i = i= i= i =, so U(P ) L(P ) =. Since he priion P ws rbirry, we conclude h f is no inegrble. i=
4 4 JOHN QUIGG Eercise 7. Le p R. Prove h he improper inegrl converges if nd only if p >. p d Soluion: Firs le p. For ech (, ) we hve d = ( p ). p p If p > hen p <, so lim p =. On he oher hnd, if p < hen p >, so lim p =. Thus / p d converges if p > nd diverges if p <. I remins o show h / p d diverges for p =. For ech (, ) we hve d = log, nd lim log =, so he inegrl diverges for =. Eercise 8. Prove h Hin: Drbou s Theorem. lim n ( ) i sin n n = 2. i= Soluion: Define f : [, ] R by f() = sin(). Then f is coninuous, hence inegrble. For ech n N, pu P n = { i n } { : i =,,..., n =, n, 2 n,..., n } n,. Then P n is priion of he inervl [, ]. We hve P n = /n. Also, for ech i =,..., n we hve i/n [(i )/n, i/n]. Thus n i= sin(i/n)(/n) is Riemnn sum of f ssocied o P n. By Drbou s Theorem we hve ( ) i lim sin n n n = f = sin() d = 2. i= Eercise 9. Prove h lim n i= n + i = log 2. Soluion: Define f : [, 2] R by f() = /. Then f is coninuous, hence inegrble.
5 For ech n N, pu { } n + i P n = n : i =,..., n. Then P n is priion of he inervl [, 2]. We hve INTEGRALS 5 P n = /n. Also, for ech i =,..., n we hve (n + i)/n [(n + i )/n, (n + i)/n]. Thus we ge Riemnn sum of f ssocied o P n by ( ) n + i f n n = n n + i n = n + i. i= By Drbou s Theorem we hve lim n i= n + i = 2 i= f = 2 i= d = log 2. Eercise. Find ll vlues of p R mking he improper inegrl converge. p d Soluion: Firs le p. For ech (, ) we hve d = ( ) p. p p If p > hen p <, so lim p =. On he oher hnd, if p < hen p >, so lim p =. Thus / p d converges if p < nd diverges if p >. I remins o show h / p d diverges for p =. For ech (, ) we hve d = log, nd lim log =, so he inegrl diverges for =. Eercise. Show h d =. + 2 Soluion: The inegrnd /( + 2 ) is coninuous on R, bu boh endpoins of he inervl (, ) re infinie, so we pick poin beween nd, sy, nd consider he improper inegrls /( + 2 ) d nd /( + 2 ) d seprely.
6 6 JOHN QUIGG Firs, we hve d = lim d = lim n = 2. On he oher hnd, since /(+( ) 2 ) = /(+ 2 ), he chnge of vribles convers he improper inegrl /( + 2 ) d ino /( + 2 ) d. Thus + 2 d = =. Eercise 2. Prove h he improper inegrl converges. sin Hin: use Inegrion by Prs on sin / d. Soluion: Le >. We pply Inegrion by Prs: sin d cos d = cos cos 2 We hve lim cos / = since cos is bounded nd /. Thus i remins o show h he improper inegrl cos /2 d converges. For ll [, ) we hve cos 2, 2 nd he improper inegrl /2 d converges becuse 2 >. Therefore cos / 2 d converges by he Comprison Theorem. Eercise 3. Le f : [, ) [, ) be uniformly coninuous nd improperly inegrble (i.e., f converges). Prove h lim f() =. Eercise 4. Find n emple of funcion f : [, ) [, ) which is coninuous nd no improperly inegrble (i.e., f diverges). Soluion: Define f : [, ) R by f() =. Then f is coninuous since i is consn. We hve f = lim d = lim =. d. Eercise 5. In ech pr below, find he derivive of he funcion f : R R defined by he given formul:
7 () (b) (c) f() = f() = f() = INTEGRALS 7 e 3 log( 2 +) e 3 log( 2 +) d d d. Soluion: Firs noe h he funcion is coninuous on R since by elemenry clculus hs is minimum =, nd his minimum vlue is 5. () By he Fundmenl Theorem of Clculus nd he Chin Rule we hve f () = ( ) d (e 3 ) 8 + e d e3 = 3e 3 e 24 + e (b) Applying he rule f = f, we hve b f () = d log( 2 +) d d (c) We hve e 3 log( 2 +) = (log( 2 + )) 3 + log( 2 + ) + 5 = 2 (log( 2 + )) 3 + log( 2 + ) d = so by prs () nd (b) we hve log( 2 +) d + ( ) d d log(2 + ) e d, f () = 3e 3 e 24 + e (log( 2 + )) 3 + log( 2 + )
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