L-idea bażika tal-kunċett tal-integral kienet digà dehret fix-xogħol ta'Arkimedi ta'Siracusa , li għex bejn il-287 u il-212 Q.K, l-ewwel parzjalment, fil-metodu li uża biex jikkalkula l-arja ta'ċirku jew ta' segment ta'parabola magħruf bħala l-metodu ta' l-eżawriment u wara iżjed preċiżament fil-kalkulazzjoni tal-arja tas-superfiċi magħluqa mill-ewwel dawra tal-ispiral (liArkimedi stima minn fuq sa isfel bl-użu ta' każ partikulari ta' dawk li sirna nsejħulhom "sommom ta' Riemann").
Fis-seklu XVII, bosta matematiċi sabu metodi oħra inġenjużi biex jikkalkulaw l-arja taħt il-grafiku ta' funzjonijiet sempliċi, pereżempju:
x α ( α > − 1 ) {\displaystyle x^{\alpha }(\alpha >-1)\;} Fermat 1636 ),1 / x {\displaystyle \displaystyle {1/x}} Nikolaus Merkator ,1668 ).Imma dan kien qabel liNewton uLeibniz skoprew indipendentement it-teorema fundamentali tal-kalkulu integrali li tefgħet id-direzzjoni tal-problema fuq it-tfittix ta' primittiva jew antiderivata tal-funzjoni.
Il-kalkulu kiseb sisien iżjed sodi b'iżvilupp tal-limiti u x-xogħol ta' Cauchy fl-ewwel nofs tas-seklu 19. L-integral kien formalizzat rigorużament għall-ewwel darba, bl-użu tal-limiti minnRiemann f'dak li ngħidulu l-integral ta' Cauchy-Riemann. Għalkemm il-funzjonijiet kollha li huma kontinwi f'biċċiet u limitati fuq intervall limitat huma integrabbli fis-sens ta' Riemann, wara bdew jiġu ikkonsidrati funzjonijiet iżjed ġenerali li għalihom id-definizzjoni ta' Riemann ma tapplikax, u Lebesgue ifformala definizzjoni differenti tal-integral ibbażata fuq it-teorija tal-miżura . Oħrajn ipproponew definizzjonijiet oħra li jestendu l-approċċ ta' Riemann u Lebesgue.
Il-problema oriġinali tal-kalkulu integrali hu dik tad-definizzjoni u l-kalkulazzjoni tal-arja (bis-sinjal) tal-figura li għandha bħala truf, intervall I {\displaystyle \ I} f {\displaystyle \ f} I {\displaystyle \ I} I {\displaystyle \ I} f {\displaystyle \ f} f {\displaystyle \ f} I {\displaystyle \ I}
Jekk il-grafiku tal-funzjoni f {\displaystyle \ f}
Fil-każ ġenerali, l-idea bażika tikkonsisti f'li naqsmu l-figura fi strixxi vertikali dojoq, li jistgħu jitqiesu bħala rettangli, nikkalkulaw l-arja ta' kull rettanglu ċkejken u ngħoddu flimkien ir-riżultati miksuba, u hekk ikollna approssimazzjoni għan-numru li qegħdin infittxu. Billi nibqgħu nissuddividu fi strixxi dejjem idjaq u idjaq, nistgħu niksbu approssimazioni dejjem aħjar għall-integral imfittex: jekk jiġri hekk, ngħidu li l-funzjoni f {\displaystyle \ f} I {\displaystyle \ I} f {\displaystyle \ f} I {\displaystyle \ I}
F'termini iżjed formali, naqsmu l-intervall [ a , b ] {\displaystyle \ [a,b]} n {\displaystyle \ n} [ x s − 1 , x s ] {\displaystyle \ [x_{s-1},x_{s}]} s = 1 , 2 , . . . , n {\displaystyle \ s=1,2,...,n} x 0 = a ; x n = b {\displaystyle \ x_{0}=a;x_{n}=b} t s {\displaystyle \ t_{s}} f ( t s ) {\displaystyle \ f(t_{s})} [ x s − 1 , x s ] {\displaystyle \ [x_{s-1},x_{s}]} f ( t s ) {\displaystyle \ f(t_{s})}
∑ s = 1 n f ( t s ) δ x s := ∑ s = 1 n f ( t s ) ( x s − x s − 1 ) {\displaystyle \sum _{s=1}^{n}f(t_{s})\delta x_{s}:=\sum _{s=1}^{n}f(t_{s})(x_{s}-x_{s-1})} Jekk meta nċekknu l-wisa' tal-intervalli δ x s = x s − x s − 1 {\displaystyle \ \delta x_{s}=x_{s}-x_{s-1}} nħawija dejjem iċken ta' numru i {\displaystyle \ i} f {\displaystyle \ f} [ a , b ] {\displaystyle \ [a,b]} i {\displaystyle \ i}
L-analisi sħieħa tiddependi mill-fatt li kemm il-mod ta' taqsim f'intervalli, u kemm l-għażla tal-punti ġewwa dawk l-intervalli iridujispiċċaw irrilevanti , inkella jiġri li l-arja taħt il-kurva fuq l-intervall ikollha valur li jvarja mal-għażla ta' taqsim f'intervalli u kif il-punti jintgħażlu fl-intervalli.
Il-quddiem nagħtu kundizzjonijiet suffiċjenti biex jiġri dan.
Rappresentazzjoni grafika tal-integral ta' Riemann Ejjew naqsmu l-intervall kompatt [ a , b ] {\displaystyle \ [a,b]} partizzjoni P {\displaystyle \displaystyle {P}} n {\displaystyle \displaystyle {n}} :
P = { a = x 0 < x 1 < x 2 < … < x n − 1 < x n = b } {\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}} Ħalli jkunu
Niddefinixxu s-somma integrali inferjuri (relattiva għall-partizzjoniP {\displaystyle \displaystyle {P}}
s ( f , P ) = ∑ k = 1 n m k ( x k − x k − 1 ) . {\displaystyle s(f,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1}).} Jekk nammettu lif {\displaystyle \displaystyle {f}} s ( f , P ) {\displaystyle \displaystyle {s(f,P)}} R {\displaystyle \mathbb {R} } f {\displaystyle \displaystyle {f}}
Niddefinixxu s-somma integrali superjuri (relattiva għall-partizzjoniP {\displaystyle \displaystyle {P}}
S ( f , P ) = ∑ k = 1 n M k ( x k − x k − 1 ) {\displaystyle S(f,P)=\sum _{k=1}^{n}M_{k}(x_{k}-x_{k-1})} Analogament,S ( f , P ) {\displaystyle \displaystyle {S(f,P)}} R {\displaystyle \mathbb {R} }
Jidher ċar li jekkm ≤ f ( x ) ≤ M , ∀ x ∈ [ a , b ] {\displaystyle m\leq f(x)\leq M,\ \forall x\in [a,b]} P {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}\ }
m ( b − a ) ≤ s ( f , P ) ≤ S ( f , P ) ≤ M ( b − a ) {\displaystyle m(b-a)\leq s(f,P)\leq S(f,P)\leq M(b-a)} Dawn iż-żewġ lemmata wieħed jista' jipprovahom faċilment:
Lemma 1. :
JekkP {\displaystyle \displaystyle {P}} Q {\displaystyle \displaystyle {Q}} [ a , b ] {\displaystyle \displaystyle {[a,b]}} Q {\displaystyle \displaystyle {Q}} P {\displaystyle \displaystyle {P}\ }
s ( f , P ) ≤ s ( f , Q ) ≤ S ( f , Q ) ≤ S ( f , P ) {\displaystyle s(f,P)\leq s(f,Q)\leq S(f,Q)\leq S(f,P)}
Ħalli jkunu
s ( f ) = sup { s ( f , P ) : P {\displaystyle s(f)\displaystyle {=}\sup\{s(f,P):P} [ a , b ] } {\displaystyle \displaystyle {[a,b]}\}} S ( f ) = inf { S ( f , P ) : P {\displaystyle S(f)\displaystyle {=}\inf\{S(f,P):P} [ a , b ] } {\displaystyle \displaystyle {[a,b]}\}} s ( f ) {\displaystyle \displaystyle {s(f)}} integral inferjuri uS ( f ) {\displaystyle \displaystyle {S(f)}} integral superjuri . Mill-lemma preċidenti nistgħu niddeduċu li dawn jissodisfaw
s ( f ) ≤ S ( f ) . {\displaystyle s(f)\leq S(f).} Definizzjoni :Integral skont Riemann
In-numria {\displaystyle \displaystyle {a}} b {\displaystyle \displaystyle {b}} f {\displaystyle \displaystyle {f}\ } a {\displaystyle \displaystyle {a}} b {\displaystyle \displaystyle {b}} varjabbli muta jiġifieri∫ f ( x ) d x {\displaystyle \int \!\!\!f(x){\rm {d}}x} ∫ f ( t ) d t {\displaystyle \int \!\!\!f(t){\rm {d}}t} d x {\displaystyle \displaystyle {{\rm {d}}x}}
Jekk il-funzjoni integrabblif {\displaystyle \displaystyle {f}}
{ ( x , y ) | 0 ≤ y ≤ f ( x ) , x ∈ [ a , b ] } . {\displaystyle \{(x,y)\ |\ 0\leq y\leq f(x),\ x\in [a,b]\}.} Jekk il-funzjonif {\displaystyle \displaystyle {f}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
[ a , b ] {\displaystyle \displaystyle {\ [a,b]}} ( b − a ) / n {\displaystyle \displaystyle {(b-a)/n}} s ( f , P n ) {\displaystyle \displaystyle {s(f,P_{n})}} S ( f , P n ) {\displaystyle \displaystyle {S(f,P_{n})}} n {\displaystyle \displaystyle {n}}
s ( f ) ≥ lim n → ∞ s ( f , P n ) = lim n → ∞ S ( f , P n ) ≥ S ( f ) {\displaystyle s(f)\geq \lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})\geq S(f)} u allura, las ( f ) ≤ S ( f ) {\displaystyle \displaystyle {s(f)\leq S(f)}}
s ( f ) = S ( f ) . {\displaystyle \displaystyle {s(f)=S(f).}} Eżempju 1.
Ħallif ( x ) = x 2 {\displaystyle \displaystyle {f(x)=x^{2}}} [ 0 , 1 ] {\displaystyle \displaystyle {\ [0,1]}}
M k = sup { x 2 | x ∈ [ ( k − 1 ) / n , k / n ] } = ( k / n ) 2 {\displaystyle \displaystyle {M_{k}=\sup\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=(k/n)^{2}}} u
m k = inf { x 2 | x ∈ [ ( k − 1 ) / n , k / n ] } = ( ( k − 1 ) / n ) 2 . {\displaystyle \displaystyle {m_{k}=\inf\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=((k-1)/n)^{2}}.} Mela
S ( f , P n ) = 1 n ∑ k = 1 n ( k n ) 2 = 1 n 3 ∑ k = 1 n k 2 = 1 3 ( n + 1 n ) ( 2 n + 1 2 n ) , {\displaystyle \displaystyle {S(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n}k^{2}={\frac {1}{3}}\left({\frac {n+1}{n}}\right)\left({\frac {2n+1}{2n}}\right),}} fejn użajna l-formula1 2 + 2 2 + … + n 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 {\displaystyle \displaystyle {1^{2}+2^{2}+\ldots +n^{2}=n(n+1)(2n+1)/6}}
Bl-istess mod
s ( f , P n ) = 1 n ∑ k = 1 n ( k − 1 n ) 2 = 1 n 3 ∑ k = 1 n − 1 k 2 = 1 3 ( n − 1 n ) ( 2 n − 1 2 n ) . {\displaystyle \displaystyle {s(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k-1}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n-1}k^{2}={\frac {1}{3}}\left({\frac {n-1}{n}}\right)\left({\frac {2n-1}{2n}}\right).}} Allura
∫ 0 1 x 2 d x = lim n → ∞ s ( f , P n ) = lim n → ∞ S ( f , P n ) = 1 3 . {\displaystyle \int _{0}^{1}x^{2}{\rm {d}}x=\lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})={\frac {1}{3}}.} Eżempju 2.
B'kuntrast mal-eżempju ta' qabel, ejjew nikkunsidraw il-funzjonig : [ 0 , 1 ] ↦ R {\displaystyle \displaystyle {g:[0,1]\mapsto \mathbb {R} }}
g ( x ) = { 1 , j e k k x h i j a r a z z j o n a l i 0 , j e k k x h i j a r r a z z j o n a l i . {\displaystyle {g(x)={\begin{cases}1,\ \mathrm {jekk} \ x\ \mathrm {hija\ razzjonali} \\0,\ \mathrm {jekk} \ x\ \mathrm {hija\ rrazzjonali} .\end{cases}}}} Għal kull partizzjoni tal-intervall[ 0 , 1 ] {\displaystyle \displaystyle {[0,1]}} [ x k − 1 , x k ] {\displaystyle \displaystyle {[x_{k-1},x_{k}]}} M k = 1 {\displaystyle \displaystyle {M_{k}=1}} m k = 0 {\displaystyle \displaystyle {m_{k}=0}}
s ( g , P ) = ∑ k = 1 n m k ( x k − x k − 1 ) = 0 {\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=0} s ( g , P ) = ∑ k = 1 n m k ( x k − x k − 1 ) = ∑ k = 1 n ( x k − x k − 1 ) = 1. {\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=\sum _{k=1}^{n}(x_{k}-x_{k-1})=1.} Melas ( g ) = 0 {\displaystyle \displaystyle {s(g)=0}} S ( g ) = 1 {\displaystyle \displaystyle {S(g)=1}} g {\displaystyle \displaystyle {g}}
La mhux kull funzjoni hi integrabbli, hemm bżonn li niddeċiedu meta l-integral ta' funzjonijeżisti jew le. Il-quddiem nagħtu żewġ klassijiet wiesa' ta' funzjonijiet li huma integrabbli. It-teorema li ġejja hi utli ħafna għal din id-deċizzjoni.
Teorema 1 :Kundizzjoni meħtieġa u biżżejjed għall-integrabbiltà
Hallif {\displaystyle \displaystyle {f}} [ a , b ] {\displaystyle \displaystyle {[a,b]}} f {\displaystyle \displaystyle {f}} ϵ > 0 {\displaystyle \displaystyle {\epsilon \,>\,0}} P {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
S ( f , P ) − s ( f , p ) < ϵ . {\displaystyle \displaystyle {S(f,P)-s(f,p)<\epsilon .}}
Prova : Nissoponu lif {\displaystyle \displaystyle {f}} S ( f ) = s ( f ) {\displaystyle \displaystyle {S(f)=s(f)}} ϵ > 0 {\displaystyle \displaystyle {\epsilon >0}} P 1 {\displaystyle \displaystyle {P_{1}}} [ a . b ] {\displaystyle \displaystyle {[a.b]}}
s ( f , P 1 ) > s ( f ) − ϵ 2 . {\displaystyle \displaystyle {s(f,P_{1})>s(f)-{\frac {\epsilon }{2}}.}} (Din issegwi mid-definizzjoni tas-supremum). Bl-istess mod teżisti partizzjoniP 2 {\displaystyle \displaystyle {P_{2}}} [ a . b ] {\displaystyle \displaystyle {[a.b]}}
S ( f , P 2 ) < S ( f ) + ϵ 2 . {\displaystyle \displaystyle {S(f,P_{2})<S(f)+{\frac {\epsilon }{2}}.}} ĦalliP = P 1 ∪ P 2 {\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}
S ( f , P ) − s ( f , P ) ≤ S ( f , P 2 ) − s ( f , P 1 ) < ( S ( f ) + ϵ 2 ) − ( s ( f ) − ϵ 2 ) = S ( f ) − s ( f ) + ϵ = ϵ . {\displaystyle \displaystyle {S(f,P)-s(f,P)\leq S(f,P_{2})-s(f,P_{1})<\left(S(f)+{\frac {\epsilon }{2}}\right)-\left(s(f)-{\frac {\epsilon }{2}}\right)=S(f)-s(f)+\epsilon =\epsilon .}} ϵ > 0 {\displaystyle \displaystyle {\epsilon >0}} P {\displaystyle \displaystyle {P}} [ a . b ] {\displaystyle \displaystyle {[a.b]}} S ( f , P ) < s ( f , P ) + ϵ {\displaystyle \displaystyle {S(f,P)<s(f,P)+\epsilon }}
S ( f ) ≤ S ( f , P ) < s ( f , P ) + ϵ ≤ s ( f ) + ϵ . {\displaystyle \displaystyle {S(f)\leq S(f,P)<s(f,P)+\epsilon \leq s(f)+\epsilon .}} Laϵ {\displaystyle \displaystyle {\epsilon }} S ( f ) ≤ s ( f ) {\displaystyle \displaystyle {S(f)\leq s(f)}} S ( f ) = s ( f ) {\displaystyle \displaystyle {S(f)=s(f)}} f {\displaystyle \displaystyle {f}}
Proprijetà 1 :Il-monotonija hi biżżejjed għall-integrabbiltà
Prova: Nissoponu li l-funzjoni f {\displaystyle \ f} [ a , b ] {\displaystyle \ [a,b]} ϵ > 0 {\displaystyle \epsilon >0} δ > 0 {\displaystyle \delta >0}
δ < ϵ f ( b ) − f ( a ) . {\displaystyle \delta <{\frac {\epsilon }{f(b)-f(a)}}.} ĦalliP {\displaystyle P} [ a , b ] {\displaystyle \ [a,b]} [ x k − 1 , x k ] {\displaystyle \ [x_{k-1},x_{k}]} δ {\displaystyle \delta } M k = f ( x k ) {\displaystyle \ M_{k}=f(x_{k})} m k = f ( x k − 1 ) {\displaystyle \ m_{k}=f(x_{k-1})}
0 < S ( f , P ) − s ( f , P ) = ∑ k = 1 n ( f ( x k ) − f ( x k − 1 ) ) ( x k − x k − 1 ) < ϵ f ( b ) − f ( a ) ∑ k = 1 n ( f ( x k ) − f ( x k − 1 ) ) = ϵ . {\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))(x_{k}-x_{k-1})<{\frac {\epsilon }{f(b)-f(a)}}\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))=\epsilon .} Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.
Proprijetà 2 :Il-kontinwità hi suffiċjenti għall-integrabbiltà
Prova: La l-funzjoni f : [ a , b ] → R {\displaystyle \ f:[a,b]\to \mathbb {R} } ϵ > 0 {\displaystyle \epsilon >0} δ > 0 {\displaystyle \delta >0}
| f ( x ) − f ( y ) | < ϵ b − a {\displaystyle |f(x)-f(y)|<{\frac {\epsilon }{b-a}}} kull meta| x − y | < δ {\displaystyle |x-y|<\delta } P {\displaystyle P} [ a , b ] {\displaystyle \ [a,b]} [ x k − 1 , x k ] {\displaystyle \ [x_{k-1},x_{k}]} δ {\displaystyle \delta }
0 < M k − m k < ϵ b − a {\displaystyle 0<M_{k}-m_{k}<{\frac {\epsilon }{b-a}}} u mela
0 < S ( f , P ) − s ( f , P ) = ∑ k = 1 n ( M k − m k ) ( x k − x k − 1 ) < ϵ b − a ∑ k = 1 n ( x k − x k − 1 ) = ϵ . {\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(M_{k}-m_{k})(x_{k}-x_{k-1})<{\frac {\epsilon }{b-a}}\sum _{k=1}^{n}(x_{k}-x_{k-1})=\epsilon .} Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.
Proprijetà 3 :Il-proprijetà tal-linjarità
Prova: Jekkα ≥ 0 {\displaystyle \displaystyle {\alpha \geq 0}} S ( α f ) = α S ( f ) {\displaystyle \displaystyle {S(\alpha f)=\alpha S(f)}} s ( α f ) = α s ( f ) {\displaystyle \displaystyle {s(\alpha f)=\alpha s(f)}}
S ( f ) = s ( f ) = ∫ a b f ( x ) d x , {\displaystyle \displaystyle {S(f)=s(f)=\int _{a}^{b}f(x){\rm {d}}x,}} għandna
∫ a b α f ( x ) d x = S ( α f ) = s ( α f ) = α ∫ a b f ( x ) d x . {\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=S(\alpha f)=s(\alpha f)=\alpha \int _{a}^{b}f(x){\rm {d}}x.}} B'mod simili jekkα < 0 {\displaystyle \displaystyle {\alpha <0}} S ( α f ) = α s ( f ) {\displaystyle \displaystyle {S(\alpha f)=\alpha s(f)}} s ( α f ) = α S ( f ) {\displaystyle \displaystyle {s(\alpha f)=\alpha S(f)}}
∫ a b α f ( x ) d x = α ∫ a b f ( x ) d x . {\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=\alpha \int _{a}^{b}f(x){\rm {d}}x.}} Mela issa biżżejjed li nipprovaw li
∫ a b ( f ( x ) + g ( x ) ) d x = ∫ a b f ( x ) d x + ∫ a b g ( x ) d x . {\displaystyle \ \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x){\rm {d}}x+\int _{a}^{b}g(x){\rm {d}}x.} Niftakru li
sup x ∈ D [ f ( x ) + g ( x ) ] ≤ sup x ∈ D f ( x ) + sup x ∈ D g ( x ) u inf x ∈ D ( [ f ( x ) + g ( x ) ] ) ≥ inf x ∈ D f ( x ) + inf x ∈ D g ( x ) {\displaystyle \ \sup _{x\in D}[f(x)+g(x)]\leq \sup _{x\in D}f(x)+\sup _{x\in D}g(x)\ \ \ {\rm {u}}\ \ \ \inf _{x\in D}([f(x)+g(x)])\geq \inf _{x\in D}f(x)+\inf _{x\in D}g(x)} u għalhekk għal kull partizzjoniP {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
S ( f + g , P ) ≤ S ( f , P ) + S ( g , P ) u s ( f + g , P ) ≥ s ( f , P ) + s ( g , P ) . {\displaystyle \ S(f+g,P)\leq S(f,P)+S(g,P)\ \ \ {\rm {u}}\ \ \ s(f+g,P)\geq s(f,P)+s(g,P).} Mit-Teorema 1 nafu li għall kullϵ > 0 {\displaystyle \displaystyle {\epsilon >0}} P 1 {\displaystyle \displaystyle {P_{1}}} P 2 {\displaystyle \displaystyle {P_{2}}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
S ( f , P 1 ) < s ( f , P 1 ) + ϵ 2 u S ( g , P 2 ) < s ( g , P 2 ) + ϵ 2 . {\displaystyle \displaystyle {S(f,P_{1})<s(f,P_{1})+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P_{2})<s(g,P_{2})+{\frac {\epsilon }{2}}.}} ĦalliP = P 1 ∪ P 2 {\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}
S ( f , P ) < s ( f , P ) + ϵ 2 u S ( g , P ) < s ( g , P ) + ϵ 2 . {\displaystyle \displaystyle {S(f,P)<s(f,P)+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P)<s(g,P)+{\frac {\epsilon }{2}}.}} Jekk nikkumbinaw id-diżugwaljanzi niksbu
S ( f + g , P ) < S ( f , P ) + S ( g , P ) < s ( f , P ) + s ( g , P ) + ϵ < s ( f + g , P ) + ϵ {\displaystyle \displaystyle {S(f+g,P)<S(f,P)+S(g,P)<s(f,P)+s(g,P)+\epsilon <s(f+g,P)+\epsilon }} u allura l-funzjonif + g {\displaystyle \displaystyle {f+g}}
Nin-naħa l-oħra, la
∫ a b ( f ( x ) + g ( x ) ) d x = S ( f + g ) ≤ S ( f + g , P ) < s ( f , P ) + s ( g , P ) + ϵ ≤ s ( f ) + s ( g ) + ϵ = ∫ a b f ( x ) d x + ∫ a b g ( x ) d x + ϵ {\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=S(f+g)\leq S(f+g,P)\\&{}<s(f,P)+s(g,P)+\epsilon \\&{}\leq s(f)+s(g)+\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x+\epsilon \end{aligned}}} u
∫ a b ( f ( x ) + g ( x ) ) d x = s ( f + g ) ≥ s ( f + g , P ) > S ( f , P ) + S ( g , P ) − ϵ ≥ S ( f ) + S ( g ) − ϵ = ∫ a b f ( x ) d x + ∫ a b g ( x ) d x − ϵ {\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=s(f+g)\geq s(f+g,P)\\&{}>S(f,P)+S(g,P)-\epsilon \\&{}\geq S(f)+S(g)-\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x-\epsilon \end{aligned}}} nikkonkludu li
∫ a b ( f ( x ) + g ( x ) ) d x = ∫ a b f ( x ) d x + ∫ a b g ( x ) d x . {\displaystyle \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x.} Proprijetà 4 :Il-proprijetà tal-additività
Prova:
Mit-Teorema 1 nafu li għall kullϵ > 0 {\displaystyle \displaystyle {\epsilon >0}} P 1 {\displaystyle \displaystyle {P_{1}}} [ a , c ] {\displaystyle \displaystyle {[a,c]}} P 2 {\displaystyle \displaystyle {P_{2}}} [ c , b ] {\displaystyle \displaystyle {[c,b]}}
S ( f , P 1 ) − s ( f , P 1 ) < ϵ 2 u S ( f , P 2 ) − s ( f , P 2 ) < ϵ 2 . {\displaystyle \displaystyle {S(f,P_{1})-s(f,P_{1})<{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(f,P_{2})-s(f,P_{2})<{\frac {\epsilon }{2}}.}} ĦalliP = P 1 ∪ P 2 {\displaystyle \displaystyle {P=P_{1}\cup P_{2}}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
S ( f , P ) − s ( f , P ) = S ( f , P 1 ) + S ( f , P 2 ) − s ( f , P 1 ) − s ( f , P 2 ) = ( S ( f , P 1 ) − s ( f , P 1 ) ) + ( S ( f , P 2 ) − s ( f , P 2 ) ) < ϵ 2 + ϵ 2 = ϵ . {\displaystyle \displaystyle {S(f,P)-s(f,P)=S(f,P_{1})+S(f,P_{2})-s(f,P_{1})-s(f,P_{2})=(S(f,P_{1})-s(f,P_{1}))+(S(f,P_{2})-s(f,P_{2}))<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon .}} Melaf {\displaystyle \displaystyle {f}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
Nin-naħa l-oħra, la
∫ a b f ( x ) d x ≤ S ( f , P ) = S ( f , P 1 ) + S ( f , P 2 ) < s ( f , P 1 ) + s ( f , P 2 ) + ϵ ≤ ∫ a c f ( x ) d x + ∫ c b f ( x ) d x + ϵ {\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\leq S(f,P)=S(f,P_{1})+S(f,P_{2})\\&{}<s(f,P_{1})+s(f,P_{2})+\epsilon \\&{}\leq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x+\epsilon \end{aligned}}} u
∫ a b f ( x ) d x ≥ s ( f , P ) = s ( f , P 1 ) + s ( f , P 2 ) > S ( f , P 1 ) + S ( f , P 2 ) − ϵ ≥ ∫ a c f ( x ) d x + ∫ c b f ( x ) d x − ϵ {\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\geq s(f,P)=s(f,P_{1})+s(f,P_{2})\\&{}>S(f,P_{1})+S(f,P_{2})-\epsilon \\&{}\geq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x-\epsilon \end{aligned}}} nikkonkludu li
∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x {\displaystyle \int _{a}^{b}f(x){\rm {d}}x=\int _{a}^{c}f(x){\rm {d}}x+\int _{c}^{b}f(x){\rm {d}}x} kif nixiequ.
Proprijetà 5 :Il-propijetà tal-monotonija
Prova : Jekkf ( x ) ≤ g ( x ) {\displaystyle f(x)\leq g(x)} x ∈ [ a , b ] {\displaystyle \ x\in [a,b]} P {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
S ( f , P ) ≤ S ( g , P ) u s ( f , P ) ≤ s ( g , P ) {\displaystyle \ S(f,P)\leq S(g,P)\ \ \ {\rm {u}}\ \ \ s(f,P)\leq s(g,P)} Minn dawn id-diżugwaljanzi nikkonkludu l-monotonija tal-integral.
Teorema :Teorema tal-valur assolut
Prova :ĦalliP {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}} n {\displaystyle \displaystyle {n}}
P = { a = x 0 < x 1 < x 2 < … < x n − 1 < x n = b } {\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}} u
m ~ k = inf { | f ( x ) | : x ∈ [ x k − 1 , x k ] } , M ~ k = sup { | f ( x ) | : x ∈ [ x k − 1 , x k ] } . {\displaystyle {\tilde {m}}_{k}=\inf\{|f(x)|:x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{|f(x)|:x\in [x_{k-1},x_{k}]\}.} Mid-diżugwaljanza
| f ( x ) | − | f ( y ) | ≤ | f ( x ) − f ( y ) | ≤ M k − m k {\displaystyle |f(x)|-|f(y)|\leq |f(x)-f(y)|\leq M_{k}-m_{k}} għal kullx , y ∈ [ x k − 1 , x k ] {\displaystyle \displaystyle {x,y\in [x_{k-1},x_{k}]}}
M ~ k − m ~ k ≤ M k − m k {\displaystyle {\tilde {M}}_{k}-{\tilde {m}}_{k}\leq M_{k}-m_{k}} u allura
S ( | f | , P ) − s ( | f | , P ) ≤ S ( f , P ) − s ( f , P ) . {\displaystyle S(|f|,P)-s(|f|,P)\leq S(f,P)-s(f,P).} Mela laf {\displaystyle \displaystyle {f}} | f | {\displaystyle \displaystyle {|f|}}
Id-diżugwaljanza bejn l-integrali, niksbuha mir-relazzjoni± f ( x ) ≤ | f ( x ) | {\displaystyle \pm f(x)\leq |f(x)|} x ∈ [ a , b ] {\displaystyle x\in [a,b]}
Teorema :Teorema integrali tal-medja
Prova: Laf {\displaystyle f\!} [ a , b ] {\displaystyle [a,b]\!} teorema ta' Weierstrass għandhamassimu M {\displaystyle M\!} minimu m {\displaystyle m\!} [ a , b ] {\displaystyle [a,b]\!}
sup x ∈ [ a , b ] f ( x ) = M u inf x ∈ [ a , b ] f ( x ) = m . {\displaystyle \sup _{x\in [a,b]}f(x)=M{\mbox{ u }}\inf _{x\in [a,b]}f(x)=m.\!} Mela
m ≤ f ( x ) ≤ M . {\displaystyle m\leq f(x)\leq M\!.} Mill-proprijetà tal-monotonija tal-integral jirriżulta li
m ( b − a ) = ∫ a b m d x ≤ ∫ a b f ( x ) d x ≤ ∫ a b M d x = M ( b − a ) {\displaystyle m(b-a)=\int _{a}^{b}m\,{\rm {d}}x\leq \int _{a}^{b}f(x)\,{\rm {d}}x\leq \int _{a}^{b}M\,{\rm {d}}x=M(b-a)\!} u allura
m ≤ 1 b − a ∫ a b f ( x ) d x ≤ M . {\displaystyle m\leq {{1} \over {b-a}}\int _{a}^{b}f(x)\,{\rm {d}}x\leq M.\!} Issa mill-proprijetajiet tal-funzjonijiet kontinwi nafu lif {\displaystyle f\!} [ a , b ] {\displaystyle [a,b]\!} kollha f'[ m , M ] {\displaystyle [m,M]\!} c ∈ [ a , b ] {\displaystyle c\in [a,b]\!} f ( c ) = 1 b − a ∫ a b f ( x ) d x {\displaystyle f(c)={{1} \over {b-a}}\int _{a}^{b}f(x){\rm {d}}x}
F'din is-sezzjoni nagħtu ż-żewġ teoremi fundamentali tal-kalkulu integrali li jistabillixxu il-konnessjoni intima li teżisti bejn il-kalkulu differenzjali u l-kalkulu integrali. Dawn it-teoremi huma s-sissien tal-analisi integrali fis-sens li huma l-ħolqa li tgħaqqad il-kalkulu differenzjali mal-kalkulu integrali.
Teorema :Teorema fundamentali tal-kalkulu integrali I
Prova: Nieħduc ∈ [ a , b ] {\displaystyle c\in [a,b]} ϵ > 0 {\displaystyle \displaystyle {\epsilon >0}} δ > 0 {\displaystyle \displaystyle {\delta >0}}
| f ( t ) − f ( c ) | < ϵ , {\displaystyle \displaystyle {|f(t)-f(c)|<\epsilon ,}} jekk| t − c | ≤ δ . {\displaystyle |t-c|\leq \delta .}
f ( c ) = 1 δ ∫ c c + δ f ( c ) d t {\displaystyle f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }f(c){\rm {d}}t} u li
F ( c + δ ) − F ( c ) δ = 1 δ ∫ c c + δ f ( t ) d t . {\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}={\frac {1}{\delta }}\int _{c}^{c+\delta }f(t){\rm {d}}t.} Allura għandna
F ( c + δ ) − F ( c ) δ − f ( c ) = 1 δ ∫ c c + δ ( f ( t ) − f ( c ) ) d t {\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}-f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }(f(t)-f(c)){\rm {d}}t} u għalhekk
| F ( c + δ ) − F ( c ) δ − f ( c ) | ≤ 1 δ ∫ c c + δ | f ( t ) − f ( c ) | d t . {\displaystyle \left|{\frac {F(c+\delta )-F(c)}{\delta }}-f(c)\right|\leq {\frac {1}{\delta }}\int _{c}^{c+\delta }|f(t)-f(c)|{\rm {d}}t.} MelaF ( c + δ ) − F ( c ) δ {\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}} f ( c ) {\displaystyle \displaystyle {f(c)}} δ {\displaystyle \displaystyle {\delta }}
F ′ ( c ) := lim δ → 0 F ( c + δ ) − F ( c ) δ = f ( c ) . {\displaystyle F'(c):=\lim _{\delta \to 0}{\frac {F(c+\delta )-F(c)}{\delta }}=f(c).}
Nota: Fil-kalkulu differenzjali hemm dan il-kunċett tal-primittiva :
Funzjoni F {\displaystyle \ F} [ a , b ] {\displaystyle \ [a,b]} primittiva ta' f {\displaystyle \ f} [ a , b ] {\displaystyle \ [a,b]}
F ′ ( x ) = f ( x ) {\displaystyle \ F'(x)=f(x)} għal kullx ∈ [ a , b ] {\displaystyle x\in [a,b]}
Mela dan it-teorema jiggarantixxi l-eżistenza ta' primittiva.
Teorema :Teorema fundamentali tal-kalkulu integrali II
Prova: ĦalliP {\displaystyle \displaystyle {P}} [ a , b ] {\displaystyle \displaystyle {[a,b]}} n {\displaystyle \displaystyle {n}}
P = { a = x 0 < x 1 < x 2 < … < x n − 1 < x n = b } , {\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\},} u
m ~ k = inf { f ′ ( x ) : x ∈ [ x k − 1 , x k ] } , M ~ k = sup { f ′ ( x ) : x ∈ [ x k − 1 , x k ] } . {\displaystyle {\tilde {m}}_{k}=\inf\{f'(x):x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{f'(x):x\in [x_{k-1},x_{k}]\}.} Billi napplikaw it-teorema tal-valur medju għall kull intervall[ x k − 1 , x k ] {\displaystyle \displaystyle {[x_{k-1},x_{k}]}} t k ∈ ( x k − 1 , x k ) {\displaystyle \displaystyle {t_{k}\in (x_{k-1},x_{k})}}
f ( x k ) − f ( x k − 1 = f ′ ( t k ) ( x k − x k − 1 ) . {\displaystyle \displaystyle {f(x_{k})-f(x_{k-1}=f'(t_{k})(x_{k}-x_{k-1}).}} Mela għandna
f ( b ) − f ( a ) = ∑ k = 1 n [ f ( x k ) − f ( x k − 1 ) ] = ∑ k = 1 n f ′ ( t k ) ( x k − x k − 1 ) . {\displaystyle f(b)-f(a)=\sum _{k=1}^{n}[f(x_{k})-f(x_{k-1})]=\sum _{k=1}^{n}f'(t_{k})(x_{k}-x_{k-1}).} Lam ~ k ≤ f ( t k ) ≤ M ~ k {\displaystyle \displaystyle {{\tilde {m}}_{k}}\leq f(t_{k})\leq {\tilde {M}}_{k}} k {\displaystyle \displaystyle {k}}
s ( f ′ , P ) ≤ f ( b ) − f ( a ) ≤ S ( f ′ , P ) . {\displaystyle s(f',P)\leq f(b)-f(a)\leq S(f',P).} La din hi valida għal kull partizzjoniP {\displaystyle \displaystyle {P}}
s ( f ′ ) ≤ f ( b ) − f ( a ) ≤ S ( f ′ ) . {\displaystyle s(f')\leq f(b)-f(a)\leq S(f').} Imma qegħdin nassumu lif ′ {\displaystyle \displaystyle {f'}} [ a , b ] {\displaystyle \displaystyle {[a,b]}}
s ( f ′ ) = S ( f ′ ) = ∫ a b f ′ ( t ) d t . {\displaystyle s(f')=S(f')=\int _{a}^{b}f'(t){\rm {d}}t.} Allura
∫ a b f ′ ( t ) d t = f ( b ) − f ( a ) . {\displaystyle \int _{a}^{b}f'(t){\rm {d}}t=f(b)-f(a).} Ngħidu li l-funzjonif {\displaystyle f} [ a , ∞ ) {\displaystyle [a,\infty )} f {\displaystyle f}
Hemm ukoll teorema li tiggarantixxi li funzjoni li hi assolutament integrabbli hi integrabbli, fuq l-intervall tat-tip[ a , ∞ ] {\displaystyle [a,\infty ]}
Teorema :Teorema tal-integrabbiltà assoluta
Jekkf {\displaystyle f}
Prova: Bit-teorema fuq l-eżistenza ta'l-integrali nafu li l-kondizzjoni neċessarja u suffiċjenti biex∫ a ∞ f ( x ) d x {\displaystyle \int _{a}^{\infty }f(x){\rm {d}}x}
∀ ϵ > 0 ∃ γ > 0 : ∀ x 1 , x 2 < γ | ∫ x 1 x 2 f ( x ) d x | < ϵ . {\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \left|\int _{x_{1}}^{x_{2}}f(x){\rm {d}}x\right|<\ \epsilon .}
| f | {\displaystyle \ |f|} | f ( x ) | {\displaystyle \ |f(x)|} f ( x ) {\displaystyle \ f(x)}
∀ ϵ > 0 ∃ γ > 0 : ∀ x 1 , x 2 < γ | ∫ x 1 x 2 | f ( x ) | d x | < ϵ . {\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \left|\int _{x_{1}}^{x_{2}}\left|f(x)\right|{\rm {d}}x\right|<\ \epsilon .}
Imma mill-proprietà tal-valur assolut għall-integrali għandna
| ∫ a b f ( x ) d x | ≤ ∫ a b | f ( x ) | d x . {\displaystyle \left|\int _{a}^{b}f(x){\rm {d}}x\right|\leq \int _{a}^{b}\left|f(x)\right|{\rm {d}}x.}
U mela nistgħu niktbu
∀ ϵ > 0 ∃ γ > 0 : ∀ x 1 , x 2 < γ ∫ x 1 x 2 | f ( x ) d x | < ϵ . {\displaystyle \forall \epsilon >\ 0\quad \exists \gamma >\ 0:\quad \forall x_{1},x_{2}<\ \gamma \quad \int _{x_{1}}^{x_{2}}\left|f(x){\rm {d}}x\right|<\ \epsilon .} Mill-liema niksbu li f {\displaystyle \ f}
Hemm bżonn noqgħodu attenti li ma nħalltux dan it-teorema mal-kuntrarju tiegħu, li hu falz għax mhux il-funzjonijiet integrabbli kollha huma assolutament integrabbli. Eżempju ta' dan hi funzjoni ta' dan it-tip
sin x x . {\displaystyle {{\sin x} \over {x}}.} L-integral skont Riemann li tħaditna fuqu hawn fuq, għandu motivazzjoni tajba, hu sempliċi biex tiddeskrivih u hu biżżejjed għal ħtieġijiet tal-kalulu elmentari. Però, dan l-integral ma jissodisfax il-ħtieġijiet kollha tal-analisi avvanzata. L-integral skont Lebesgue jippermetti l-integrazzjoni ta' funzjonijiet iżjed ġenerali, jittratta l-funzjonijiet limitati u mhux limitati fl-istess ħin, u jħallina nbidlu lintervall[ a , b ] {\displaystyle \displaystyle {[a,b]}}
