L Integral

Fl-, l-integral ta' hu li fil-każ ta' funzjoni ta' varjabbli waħda jassoċja mal-funzjoni l- taħt il-funzjoni sal-axissa.

Ħjiel storiku

L-idea bażika tal-kunċett tal-integral kienet digà dehret fix-xogħol ta' Arkimedi ta' Siracusa, li għex bejn il- u il- Q.K, l-ewwel parzjalment, fil-metodu li uża biex jikkalkula l- ta' jew ta' segment ta' magħruf bħala l- u wara iżjed preċiżament fil-kalkulazzjoni tal-arja tas-superfiċi magħluqa mill-ewwel dawra tal-ispiral (li Arkimedi 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:

${\displaystyle x^{\alpha }(\alpha >-1)\;}$ ( ),

${\displaystyle \displaystyle {1/x}}$ (, ).

Imma dan kien qabel li Newton u skoprew indipendentement it- li tefgħet id-direzzjoni tal-problema fuq it-tfittix ta' primittiva jew antiderivata tal-funzjoni.

Il-kalkulu kiseb sisien iżjed sodi b'iżvilupp tal- u x-xogħol ta' Cauchy fl-ewwel nofs tas-seklu 19. L-integral kien formalizzat rigorużament għall-ewwel darba, bl-użu tal-limiti minn 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-. Oħrajn ipproponew definizzjonijiet oħra li jestendu l-approċċ ta' Riemann u Lebesgue.

Introduzzjoni ewristika

Il-problema oriġinali tal-kalkulu integrali hu dik tad-definizzjoni u l-kalkulazzjoni tal-arja (bis-sinjal) tal-figura li għandha bħala truf, intervall ${\displaystyle \ I}$ fuq l-assi tal-axissi, limitat u magħluq (l-intervall tal-integrazzjoni), il-funzjoni mogħtija ${\displaystyle \ f}$ (il-funzjoni integrata) definita fuq ${\displaystyle \ I}$ u limitata, u s-segmenti vertikali mit-truf tal-intervall ${\displaystyle \ I}$ għall-grafiku tal-funzjoni ${\displaystyle \ f}$. In-numru reali li jagħti dik l-arja nsejħulu l-integral tal-funzjoni ${\displaystyle \ f}$ fuq l-intervall ${\displaystyle \ I}$.

Jekk il-grafiku tal-funzjoni ${\displaystyle \ f}$ hu magħmul minn segmenti, il-problema nistgħu nirriżolvuh faċilment, sakemm il-figura tista' tinqasam f'rettangli jew trapeżi li nafu niddefinixxu u nikkalkulaw l-arji tagħhom: is-somma alġebrija ta' dawk l-arji hi – għad-definizzjoni – l-integral imfittex.

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 ${\displaystyle \ f}$ hi integrabbli fuq l-intervall ${\displaystyle \ I}$. Fil-każ tal-kuntrarju, ngħidu li l-funzjoni ${\displaystyle \ f}$ m'hijiex integrabbli fuq l-intervall ${\displaystyle \ I}$.

F'termini iżjed formali, naqsmu l-intervall ${\displaystyle \ [a,b]}$ f'${\displaystyle \ n}$ sottointervalli tat-tip ${\displaystyle \ [x_{s-1},x_{s}]}$ fejn ${\displaystyle \ s=1,2,...,n}$ u ${\displaystyle \ x_{0}=a;x_{n}=b}$. Għal kull sottointervall nagħżlu punt ${\displaystyle \ t_{s}}$, li l-immaġni tiegħu hi ${\displaystyle \ f(t_{s})}$, u nħażżu r-rettanglu ċkejken li għandu bażi l-intervall ${\displaystyle \ [x_{s-1},x_{s}]}$ u għoli ${\displaystyle \ f(t_{s})}$; l-arja tal-figura magħmula mir-rettangli ċkejkna kollha mibnija hekk tagħtina s-somma (msejħa ta' Cauchy-Riemann)

${\displaystyle \sum _{s=1}^{n}f(t_{s})\delta x_{s}:=\sum _{s=1}^{n}f(t_{s})(x_{s}-x_{s-1})}$ .

Jekk meta nċekknu l-wisa' tal-intervalli ${\displaystyle \ \delta x_{s}=x_{s}-x_{s-1}}$, il-valuri miksuba jinġemgħu fi dejjem iċken ta' numru ${\displaystyle \ i}$, il-funzjoni ${\displaystyle \ f}$ hi integrabbli fuq l-intervall ${\displaystyle \ [a,b]}$, u ${\displaystyle \ i}$ hu l-integral tagħha.

L-analisi sħieħa tiddependi mill-fatt li kemm il-mod ta' taqsim f'intervalli, u kemm l-għażla tal-punti ġewwa dawk l-intervalli iridu jispiċċaw 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.

Integral ta' Riemann

Ejjew naqsmu l-intervall kompatt ${\displaystyle \ [a,b]}$ permezz ta' ${\displaystyle \displaystyle {P}}$ f'${\displaystyle \displaystyle {n}}$ sottointervalli :

${\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}}$,

Ħalli jkunu

• ${\displaystyle m_{k}=\inf\{f(x):x\in [x_{k-1},x_{k}]\}}$

• ${\displaystyle M_{k}=\sup\{f(x):x\in [x_{k-1},x_{k}]\}.}$

Niddefinixxu s-somma integrali inferjuri (relattiva għall-partizzjoni ${\displaystyle \displaystyle {P}}$):

${\displaystyle s(f,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1}).}$

Jekk nammettu li ${\displaystyle \displaystyle {f}}$ tieħu valuri pożittivi fl-intervall, ${\displaystyle \displaystyle {s(f,P)}}$ hija s-somma tar-rettangli inskritti fir-reġjun tal-pjan ${\displaystyle \mathbb {R} }$, taħt il-grafiku ta' ${\displaystyle \displaystyle {f}}$.

Niddefinixxu s-somma integrali superjuri (relattiva għall-partizzjoni ${\displaystyle \displaystyle {P}}$):

${\displaystyle S(f,P)=\sum _{k=1}^{n}M_{k}(x_{k}-x_{k-1})}$

Analogament, ${\displaystyle \displaystyle {S(f,P)}}$ hi s-somma tal-arji tar-rettangli ċirkoskritti fir-regjun ${\displaystyle \mathbb {R} }$.

Jidher ċar li jekk ${\displaystyle m\leq f(x)\leq M,\ \forall x\in [a,b]}$ imbagħad għal kull partizzjoni ${\displaystyle \displaystyle {P}}$ ta' ${\displaystyle \displaystyle {[a,b]}\ }$:

${\displaystyle m(b-a)\leq s(f,P)\leq S(f,P)\leq M(b-a)}$.

Dawn iż-żewġ lemmata wieħed jista' jipprovahom faċilment:

Lemma 1.:

Jekk ${\displaystyle \displaystyle {P}}$ u ${\displaystyle \displaystyle {Q}}$ huma partizzjonijiet ta' ${\displaystyle \displaystyle {[a,b]}}$ u ${\displaystyle \displaystyle {Q}}$ hija rfinament ta' ${\displaystyle \displaystyle {P}\ }$:

${\displaystyle s(f,P)\leq s(f,Q)\leq S(f,Q)\leq S(f,P)}$.

Lemma 2.:

Għal kull żewġ partizzjonijiet ${\displaystyle \displaystyle {P,Q}}$ ta' ${\displaystyle \displaystyle {[a,b]}\ }$:

${\displaystyle s(f,P)\leq S(f,Q)}$.

Ħalli jkunu

${\displaystyle s(f)\displaystyle {=}\sup\{s(f,P):P}$ partizzjoni ta' ${\displaystyle \displaystyle {[a,b]}\}}$, ${\displaystyle S(f)\displaystyle {=}\inf\{S(f,P):P}$ partizzjoni ta' ${\displaystyle \displaystyle {[a,b]}\}}$.

${\displaystyle \displaystyle {s(f)}}$ ngħidulu l-integral inferjuri u ${\displaystyle \displaystyle {S(f)}}$ l-integral superjuri. Mill-lemma preċidenti nistgħu niddeduċu li dawn jissodisfaw

${\displaystyle s(f)\leq S(f).}$

Definizzjoni

Definizzjoni: Integral skont Riemann

Jekk ${\displaystyle \displaystyle {s(f)=S(f)}}$ ngħidu li l-funzjoni ${\displaystyle \displaystyle {f}}$ hi integrabbli skont Riemann fuq l-intervall magħluq limitat ${\displaystyle \ [a,b]}$ u l-valur kommuni ngħidulu l-integral ta' ${\displaystyle \displaystyle {f}}$ fuq ${\displaystyle \ [a,b]}$ u nuruh bis-simbolu:

${\displaystyle \int _{a}^{b}\!\!f(x){\rm {d}}x.}$

In-numri ${\displaystyle \displaystyle {a}}$, ${\displaystyle \displaystyle {b}}$ ngħidulhom it-truf tal-integrazzjoni u ${\displaystyle \displaystyle {f}\ }$ l-integrand (${\displaystyle \displaystyle {a}}$ l-ewwel tarf, ${\displaystyle \displaystyle {b}}$ it-tieni tarf). Il-varjabbli ta' integrazzjoni hi jiġifieri ${\displaystyle \int \!\!\!f(x){\rm {d}}x}$ tfisser l-istess bħal ${\displaystyle \int \!\!\!f(t){\rm {d}}t}$. Id-${\displaystyle \displaystyle {{\rm {d}}x}}$ insibuha bħala d-differenzjali tal-varjabbli tal-integrazzjoni.

Jekk il-funzjoni integrabbli ${\displaystyle \displaystyle {f}}$ hi posittiva l-integral hu daqs l-arja tar-reġjun:

${\displaystyle \{(x,y)\ |\ 0\leq y\leq f(x),\ x\in [a,b]\}.}$

Jekk il-funzjoni ${\displaystyle \displaystyle {f}}$ tibdel is-sinjal fuq ${\displaystyle \displaystyle {[a,b]}}$ l-integral jirrappreżenta is-somma tal-arji bis-sinjal differenti, posittiv jekk l-arja tkun fuq l-assi tal-axissa, negattiv jekk tkun taħt.

Ħalli tkun il-partizzjoni li taqsam l-intervall ${\displaystyle \displaystyle {\ [a,b]}}$ f'sottointervalli ugwali ta' tul ${\displaystyle \displaystyle {(b-a)/n}}$. Jekk il-limiti ta' ${\displaystyle \displaystyle {s(f,P_{n})}}$ u ta' ${\displaystyle \displaystyle {S(f,P_{n})}}$ meta ${\displaystyle \displaystyle {n}}$ tersaq lejn l-infinit huma l-istess, imbagħad ikollna

${\displaystyle s(f)\geq \lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})\geq S(f)}$

u allura, la ${\displaystyle \displaystyle {s(f)\leq S(f)}}$, ikollna wkoll

${\displaystyle \displaystyle {s(f)=S(f).}}$

Eżempju 1.

Ħalli ${\displaystyle \displaystyle {f(x)=x^{2}}}$ u l-intervall ikun ${\displaystyle \displaystyle {\ [0,1]}}$. Imbagħad

${\displaystyle \displaystyle {M_{k}=\sup\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=(k/n)^{2}}}$

u

${\displaystyle \displaystyle {m_{k}=\inf\{x^{2}\,|\,x\in [(k-1)/n,k/n]\}=((k-1)/n)^{2}}.}$

Mela

${\displaystyle \displaystyle {S(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n}k^{2}={\frac {1}{3}}\left({\frac {n+1}{n}}\right)\left({\frac {2n+1}{2n}}\right),}}$

fejn użajna l-formula ${\displaystyle \displaystyle {1^{2}+2^{2}+\ldots +n^{2}=n(n+1)(2n+1)/6}}$.

Bl-istess mod

${\displaystyle \displaystyle {s(f,P_{n})={\frac {1}{n}}\sum _{k=1}^{n}\left({\frac {k-1}{n}}\right)^{2}={\frac {1}{n^{3}}}\sum _{k=1}^{n-1}k^{2}={\frac {1}{3}}\left({\frac {n-1}{n}}\right)\left({\frac {2n-1}{2n}}\right).}}$

Allura

${\displaystyle \int _{0}^{1}x^{2}{\rm {d}}x=\lim _{n\to \infty }s(f,P_{n})=\lim _{n\to \infty }S(f,P_{n})={\frac {1}{3}}.}$

Eżempju 2.

B'kuntrast mal-eżempju ta' qabel, ejjew nikkunsidraw il-funzjoni ${\displaystyle \displaystyle {g:[0,1]\mapsto \mathbb {R} }}$ definita hekk

${\displaystyle {g(x)={\begin{cases}1,\ \mathrm {jekk} \ x\ \mathrm {hija\ razzjonali} \\0,\ \mathrm {jekk} \ x\ \mathrm {hija\ rrazzjonali} .\end{cases}}}}$

Għal kull partizzjoni tal-intervall ${\displaystyle \displaystyle {[0,1]}}$, f'kull sottointervall ${\displaystyle \displaystyle {[x_{k-1},x_{k}]}}$ hemm numri razzjonali u irrazzjonali u mela ${\displaystyle \displaystyle {M_{k}=1}}$ u ${\displaystyle \displaystyle {m_{k}=0}}$. Għalhekk

${\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=0}$ ${\displaystyle s(g,P)=\sum _{k=1}^{n}m_{k}(x_{k}-x_{k-1})=\sum _{k=1}^{n}(x_{k}-x_{k-1})=1.}$

Mela ${\displaystyle \displaystyle {s(g)=0}}$ u ${\displaystyle \displaystyle {S(g)=1}}$. La dawn mhux imdaqs nistgħu nikkonkludu li l-funzjoni ${\displaystyle \displaystyle {g}}$ mhux integrabbli.

La mhux kull funzjoni hi integrabbli, hemm bżonn li niddeċiedu meta l-integral ta' funzjoni jeżisti 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

Halli ${\displaystyle \displaystyle {f}}$ tkun funzjoni limitata fuq ${\displaystyle \displaystyle {[a,b]}}$. Imbagħad ${\displaystyle \displaystyle {f}}$ hi integrabbli jekkk (jekk u biss jekk), għal kull ${\displaystyle \displaystyle {\epsilon \,>\,0}}$, teżisti partizzjoni ${\displaystyle \displaystyle {P}}$ ta' ${\displaystyle \displaystyle {[a,b]}}$ li għaliha

${\displaystyle \displaystyle {S(f,P)-s(f,p)<\epsilon .}}$

Prova : Nissoponu li ${\displaystyle \displaystyle {f}}$ hi integrabbli u hekk ${\displaystyle \displaystyle {S(f)=s(f)}}$. Għall kull ${\displaystyle \displaystyle {\epsilon >0}}$ mogħtija, teżisti partizzjoni ${\displaystyle \displaystyle {P_{1}}}$ ta' ${\displaystyle \displaystyle {[a.b]}}$ li tissodisfa

${\displaystyle \displaystyle {s(f,P_{1})>s(f)-{\frac {\epsilon }{2}}.}}$

(Din issegwi mid-definizzjoni tas-supremum). Bl-istess mod teżisti partizzjoni ${\displaystyle \displaystyle {P_{2}}}$ ta' ${\displaystyle \displaystyle {[a.b]}}$ li tissodisfa

${\displaystyle \displaystyle {S(f,P_{2})<S(f)+{\frac {\epsilon }{2}}.}}$

Ħalli ${\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}$. Imbagħad minn Lemma 1, għandna

${\displaystyle \displaystyle {S(f,P)-s(f,P)\leq S(f,P_{2})-s(f,P_{1})<\left(S(f)+{\frac {\epsilon }{2}}\right)-\left(s(f)-{\frac {\epsilon }{2}}\right)=S(f)-s(f)+\epsilon =\epsilon .}}$

Min naħa l-oħra, nissoponu li għall kull ${\displaystyle \displaystyle {\epsilon >0}}$ mogħtija, teżisti partizzjoni ${\displaystyle \displaystyle {P}}$ ta' ${\displaystyle \displaystyle {[a.b]}}$ li tissodisfa ${\displaystyle \displaystyle {S(f,P)<s(f,P)+\epsilon }}$. Allura

${\displaystyle \displaystyle {S(f)\leq S(f,P)<s(f,P)+\epsilon \leq s(f)+\epsilon .}}$

La ${\displaystyle \displaystyle {\epsilon }}$ hi arbitrarja, bilfors li ${\displaystyle \displaystyle {S(f)\leq s(f)}}$ u allura ${\displaystyle \displaystyle {S(f)=s(f)}}$ u ${\displaystyle \displaystyle {f}}$ hi integrabbli.

Proprjetajiet tal-integral skont Riemann

Integrabbiltà

Proprijetà 1: Il-monotonija hi biżżejjed għall-integrabbiltà

Jekk il-funzjoni ${\displaystyle \ f:[a,b]\to \mathbb {R} }$ hi monotona, allura hi integrabbli.

Prova: Nissoponu li l-funzjoni ${\displaystyle \ f}$ tiżdied fuq ${\displaystyle \ [a,b]}$. Il-każ fejn tonqos hu simili. Jekk ningħataw ${\displaystyle \epsilon >0}$, nistgħu nagħzlu ${\displaystyle \delta >0}$ li tissodisfa

${\displaystyle \delta <{\frac {\epsilon }{f(b)-f(a)}}.}$

Ħalli ${\displaystyle P}$ tkun partizzjoni tal-intervall ${\displaystyle \ [a,b]}$ f'sottointervalli ${\displaystyle \ [x_{k-1},x_{k}]}$ ta' wisa' inqas minn ${\displaystyle \delta }$. Mill-monontonija għandna li ${\displaystyle \ M_{k}=f(x_{k})}$ u ${\displaystyle \ m_{k}=f(x_{k-1})}$. Mela

${\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))(x_{k}-x_{k-1})<{\frac {\epsilon }{f(b)-f(a)}}\sum _{k=1}^{n}(f(x_{k})-f(x_{k-1}))=\epsilon .}$

Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.

Proprijetà 2: Il-kontinwità hi suffiċjenti għall-integrabbiltà

Jekk il-funzjoni ${\displaystyle \ f:[a,b]\to \mathbb {R} }$ hi kontinwa, allura hi integrabbli.

Prova: La l-funzjoni ${\displaystyle \ f:[a,b]\to \mathbb {R} }$ hi kontinwa allura hi kontinwa uniformement. Jekk ningħataw ${\displaystyle \epsilon >0}$, teżisti ${\displaystyle \delta >0}$ li għaliha

${\displaystyle |f(x)-f(y)|<{\frac {\epsilon }{b-a}}}$

kull meta ${\displaystyle |x-y|<\delta }$. Jekk ${\displaystyle P}$ hi partizzjoni tal-intervall ${\displaystyle \ [a,b]}$ f'sottointervalli ${\displaystyle \ [x_{k-1},x_{k}]}$ ta' wisa' inqas minn ${\displaystyle \delta }$, imbagħad ikollna

${\displaystyle 0<M_{k}-m_{k}<{\frac {\epsilon }{b-a}}}$

u mela

${\displaystyle 0<S(f,P)-s(f,P)=\sum _{k=1}^{n}(M_{k}-m_{k})(x_{k}-x_{k-1})<{\frac {\epsilon }{b-a}}\sum _{k=1}^{n}(x_{k}-x_{k-1})=\epsilon .}$

Allura mit-Teorema 1 nistgħu nikkonkludu li l-funzjoni hi integrabbli.

Linjarità

Proprijetà 3: Il-proprijetà tal-linjarità

Ħalli ${\displaystyle f}$ u ${\displaystyle g}$ jkunu żewġ funzjonijiet kontinwi definiti f'intervall ${\displaystyle [a,b]}$ u ħalli jkunu ${\displaystyle \alpha ,\beta \in \mathbb {R} }$. Imbagħad:

${\displaystyle \int _{a}^{b}(\alpha f(x)+\beta g(x))\,dx=\alpha \int _{a}^{b}f(x)dx+\beta \int _{a}^{b}g(x){\rm {d}}x.}$

Prova: Jekk ${\displaystyle \displaystyle {\alpha \geq 0}}$ jidher ċar li ${\displaystyle \displaystyle {S(\alpha f)=\alpha S(f)}}$ u ${\displaystyle \displaystyle {s(\alpha f)=\alpha s(f)}}$. Mela la

${\displaystyle \displaystyle {S(f)=s(f)=\int _{a}^{b}f(x){\rm {d}}x,}}$

għandna

${\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=S(\alpha f)=s(\alpha f)=\alpha \int _{a}^{b}f(x){\rm {d}}x.}}$

B'mod simili jekk ${\displaystyle \displaystyle {\alpha <0}}$ għandna ${\displaystyle \displaystyle {S(\alpha f)=\alpha s(f)}}$ u ${\displaystyle \displaystyle {s(\alpha f)=\alpha S(f)}}$ u allura

${\displaystyle \displaystyle {\int _{a}^{b}\alpha f(x){\rm {d}}x=\alpha \int _{a}^{b}f(x){\rm {d}}x.}}$

Mela issa biżżejjed li nipprovaw li

${\displaystyle \ \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x){\rm {d}}x+\int _{a}^{b}g(x){\rm {d}}x.}$

Niftakru li

${\displaystyle \ \sup _{x\in D}[f(x)+g(x)]\leq \sup _{x\in D}f(x)+\sup _{x\in D}g(x)\ \ \ {\rm {u}}\ \ \ \inf _{x\in D}([f(x)+g(x)])\geq \inf _{x\in D}f(x)+\inf _{x\in D}g(x)}$

u għalhekk għal kull partizzjoni ${\displaystyle \displaystyle {P}}$ ta' ${\displaystyle \displaystyle {[a,b]}}$

${\displaystyle \ S(f+g,P)\leq S(f,P)+S(g,P)\ \ \ {\rm {u}}\ \ \ s(f+g,P)\geq s(f,P)+s(g,P).}$

Mit-Teorema 1 nafu li għall kull ${\displaystyle \displaystyle {\epsilon >0}}$ mogħtija, jeżistu partizzjonijiet ${\displaystyle \displaystyle {P_{1}}}$ u ${\displaystyle \displaystyle {P_{2}}}$ ta' ${\displaystyle \displaystyle {[a,b]}}$ li jissodisfaw

${\displaystyle \displaystyle {S(f,P_{1})<s(f,P_{1})+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P_{2})<s(g,P_{2})+{\frac {\epsilon }{2}}.}}$

Ħalli ${\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}$. Imbagħad minn Lemma 1, għandna

${\displaystyle \displaystyle {S(f,P)<s(f,P)+{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(g,P)<s(g,P)+{\frac {\epsilon }{2}}.}}$

Jekk nikkumbinaw id-diżugwaljanzi niksbu

${\displaystyle \displaystyle {S(f+g,P)<S(f,P)+S(g,P)<s(f,P)+s(g,P)+\epsilon <s(f+g,P)+\epsilon }}$

u allura l-funzjoni ${\displaystyle \displaystyle {f+g}}$ hi integrabbli.

Nin-naħa l-oħra, la

${\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=S(f+g)\leq S(f+g,P)\\&{}<s(f,P)+s(g,P)+\epsilon \\&{}\leq s(f)+s(g)+\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x+\epsilon \end{aligned}}}$

u

${\displaystyle {\begin{aligned}\int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x&{}=s(f+g)\geq s(f+g,P)\\&{}>S(f,P)+S(g,P)-\epsilon \\&{}\geq S(f)+S(g)-\epsilon =\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x-\epsilon \end{aligned}}}$

nikkonkludu li

${\displaystyle \int _{a}^{b}(f(x)+g(x))\,{\rm {d}}x=\int _{a}^{b}f(x)\,{\rm {d}}x+\int _{a}^{b}g(x)\,{\rm {d}}x.}$

Additività

Proprijetà 4: Il-proprijetà tal-additività

Jekk ${\displaystyle f}$ tkun integrabbli fuq l-intervalli ${\displaystyle \displaystyle {[a,c]}}$ u ${\displaystyle \displaystyle {[c,b]}}$, imbagħad tkun integrabbli fuq l-intervall ${\displaystyle \displaystyle {[a,b]}}$ u

${\displaystyle \int _{a}^{b}f(x){\rm {d}}x=\int _{a}^{c}f(x){\rm {d}}x+\int _{c}^{b}f(x){\rm {d}}x.}$

Prova:

Mit-Teorema 1 nafu li għall kull ${\displaystyle \displaystyle {\epsilon >0}}$ mogħtija, jeżistu partizzjonijiet ${\displaystyle \displaystyle {P_{1}}}$ ta' ${\displaystyle \displaystyle {[a,c]}}$ u ${\displaystyle \displaystyle {P_{2}}}$ ta' ${\displaystyle \displaystyle {[c,b]}}$ li jissodisfaw

${\displaystyle \displaystyle {S(f,P_{1})-s(f,P_{1})<{\frac {\epsilon }{2}}\ \ \ {\rm {u}}\ \ \ S(f,P_{2})-s(f,P_{2})<{\frac {\epsilon }{2}}.}}$

Ħalli ${\displaystyle \displaystyle {P=P_{1}\cup P_{2}}}$. Din partizzjoni ta' ${\displaystyle \displaystyle {[a,b]}}$ u għandna

${\displaystyle \displaystyle {S(f,P)-s(f,P)=S(f,P_{1})+S(f,P_{2})-s(f,P_{1})-s(f,P_{2})=(S(f,P_{1})-s(f,P_{1}))+(S(f,P_{2})-s(f,P_{2}))<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon .}}$

Mela ${\displaystyle \displaystyle {f}}$ hi integrabbli fuq ${\displaystyle \displaystyle {[a,b]}}$.

Nin-naħa l-oħra, la

${\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\leq S(f,P)=S(f,P_{1})+S(f,P_{2})\\&{}<s(f,P_{1})+s(f,P_{2})+\epsilon \\&{}\leq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x+\epsilon \end{aligned}}}$

u

${\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,{\rm {d}}x&{}\geq s(f,P)=s(f,P_{1})+s(f,P_{2})\\&{}>S(f,P_{1})+S(f,P_{2})-\epsilon \\&{}\geq \int _{a}^{c}f(x)\,{\rm {d}}x+\int _{c}^{b}f(x)\,{\rm {d}}x-\epsilon \end{aligned}}}$

nikkonkludu li

${\displaystyle \int _{a}^{b}f(x){\rm {d}}x=\int _{a}^{c}f(x){\rm {d}}x+\int _{c}^{b}f(x){\rm {d}}x}$

kif nixiequ.

Monotonija

Proprijetà 5: Il-propijetà tal-monotonija

Jekk ${\displaystyle {\displaystyle f}}$ u ${\displaystyle \ {\displaystyle g}}$ ikunu żewġ funzjonijiet integrabbli fuq l-intervall ${\displaystyle \ [a,b]}$ u ${\displaystyle f(x)\leq g(x)}$ għal kull ${\displaystyle \ x\in [a,b]}$, imbagħad ${\displaystyle \int _{a}^{b}f(x){\rm {d}}x\leq \int _{a}^{b}g(x){\rm {d}}x.}$

Prova : Jekk ${\displaystyle f(x)\leq g(x)}$ għal kull ${\displaystyle \ x\in [a,b]}$, għal kull partizzjoni ${\displaystyle \displaystyle {P}}$ ta' ${\displaystyle \displaystyle {[a,b]}}$ ikollna

${\displaystyle \ S(f,P)\leq S(g,P)\ \ \ {\rm {u}}\ \ \ s(f,P)\leq s(g,P)}$

Minn dawn id-diżugwaljanzi nikkonkludu l-monotonija tal-integral.

Valur assolut

Teorema: Teorema tal-valur assolut

Jekk ${\displaystyle f}$ tkun integrabbli fl-intervall ${\displaystyle \ [a,b]}$, imbagħad ${\displaystyle |f|}$ hi wkoll integrabbli u ${\displaystyle \left|\int _{a}^{b}f(x){\rm {d}}x\right|\leq \int _{a}^{b}\left|f(x)\right|{\rm {d}}x.}$

Prova: Ħalli ${\displaystyle \displaystyle {P}}$ tkun partizzjoni ta' ${\displaystyle \displaystyle {[a,b]}}$ f'${\displaystyle \displaystyle {n}}$ sottointervalli

${\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\}}$,

u

${\displaystyle {\tilde {m}}_{k}=\inf\{|f(x)|:x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{|f(x)|:x\in [x_{k-1},x_{k}]\}.}$

Mid-diżugwaljanza

${\displaystyle |f(x)|-|f(y)|\leq |f(x)-f(y)|\leq M_{k}-m_{k}}$

għal kull ${\displaystyle \displaystyle {x,y\in [x_{k-1},x_{k}]}}$, nikkonkludu li

${\displaystyle {\tilde {M}}_{k}-{\tilde {m}}_{k}\leq M_{k}-m_{k}}$

u allura

${\displaystyle S(|f|,P)-s(|f|,P)\leq S(f,P)-s(f,P).}$

Mela la ${\displaystyle \displaystyle {f}}$ hi integrabbli, ${\displaystyle \displaystyle {|f|}}$ hi integrabbli wkoll.

Id-diżugwaljanza bejn l-integrali, niksbuha mir-relazzjoni ${\displaystyle \pm f(x)\leq |f(x)|}$ valida għal kull ${\displaystyle x\in [a,b]}$.

Teorema tal-medja

Teorema: Teorema integrali tal-medja

Jekk ${\displaystyle f:[a,b]\to \mathbb {R} \!}$ tkun imbagħad teżisti ${\displaystyle c\in [a,b]\!}$ li għaliha

${\displaystyle {{1} \over {b-a}}\int _{a}^{b}f(x)\,{\rm {d}}x=f(c).\!}$

Prova: La ${\displaystyle f\!}$ hi kontinwa f'${\displaystyle [a,b]\!}$, bit- għandha ${\displaystyle M\!}$ u ${\displaystyle m\!}$ f'${\displaystyle [a,b]\!}$:

${\displaystyle \sup _{x\in [a,b]}f(x)=M{\mbox{ u }}\inf _{x\in [a,b]}f(x)=m.\!}$

Mela

${\displaystyle m\leq f(x)\leq M\!.}$

Mill-proprijetà tal-monotonija tal-integral jirriżulta li

${\displaystyle m(b-a)=\int _{a}^{b}m\,{\rm {d}}x\leq \int _{a}^{b}f(x)\,{\rm {d}}x\leq \int _{a}^{b}M\,{\rm {d}}x=M(b-a)\!}$

u allura

${\displaystyle m\leq {{1} \over {b-a}}\int _{a}^{b}f(x)\,{\rm {d}}x\leq M.\!}$

Issa mill-proprijetajiet tal- nafu li ${\displaystyle f\!}$ f' ${\displaystyle [a,b]\!}$ trid tieħu il-valuri kollha f'${\displaystyle [m,M]\!}$. Allura, in partikulari teżisti ${\displaystyle c\in [a,b]\!}$ li tissodisfa ${\displaystyle f(c)={{1} \over {b-a}}\int _{a}^{b}f(x){\rm {d}}x}$.

Kalkulu differenzjali u kalkulu integrali

F'din is-sezzjoni nagħtu ż-żewġ teoremi fundamentali tal-kalkulu integrali li jistabillixxu il-konnessjoni intima li teżisti bejn il- 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 fundamentali tal-kalkulu integrali I

Teorema: Teorema fundamentali tal-kalkulu integrali I

Jekk ${\displaystyle f:[a,b]\to \mathbb {R} }$ tkun , imbagħad il-funzjoni integrali ${\displaystyle F:[a,b]\to \mathbb {R} }$ definita bħala

${\displaystyle F(x):=\int _{a}^{x}f(t){\rm {d}}t}$

tkun f' ${\displaystyle \displaystyle {[a,b]}}$ u jkollha derivata ${\displaystyle F^{\prime }(c)=f(c)}$ għal kull ${\displaystyle c\in [a,b]}$.

Prova: Nieħdu ${\displaystyle c\in [a,b]}$. Imbagħad għal kull ${\displaystyle \displaystyle {\epsilon >0}}$ mogħtija, teżisti ${\displaystyle \displaystyle {\delta >0}}$ li għaliha

${\displaystyle \displaystyle {|f(t)-f(c)|<\epsilon ,}}$

jekk ${\displaystyle |t-c|\leq \delta .}$ Jidher ċar li

${\displaystyle f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }f(c){\rm {d}}t}$

u li

${\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}={\frac {1}{\delta }}\int _{c}^{c+\delta }f(t){\rm {d}}t.}$

Allura għandna

${\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}-f(c)={\frac {1}{\delta }}\int _{c}^{c+\delta }(f(t)-f(c)){\rm {d}}t}$

u għalhekk

${\displaystyle \left|{\frac {F(c+\delta )-F(c)}{\delta }}-f(c)\right|\leq {\frac {1}{\delta }}\int _{c}^{c+\delta }|f(t)-f(c)|{\rm {d}}t.}$

Mela ${\displaystyle {\frac {F(c+\delta )-F(c)}{\delta }}}$ tikkonverġi lejn ${\displaystyle \displaystyle {f(c)}}$ meta ${\displaystyle \displaystyle {\delta }}$ tersaq lejn 0, u allura

${\displaystyle F'(c):=\lim _{\delta \to 0}{\frac {F(c+\delta )-F(c)}{\delta }}=f(c).}$

Nota: Fil-kalkulu differenzjali hemm dan il-kunċett tal-primittiva:

Funzjoni ${\displaystyle \ F}$ derivabbli f'intervall ${\displaystyle \ [a,b]}$ ngħidulha l-primittiva ta' ${\displaystyle \ f}$ f' ${\displaystyle \ [a,b]}$ jekk:

${\displaystyle \ F'(x)=f(x)}$

għal kull ${\displaystyle x\in [a,b]}$.

Mela dan it-teorema jiggarantixxi l-eżistenza ta' primittiva.

Teorema fundamentali tal-kalkulu integrali II

Teorema: Teorema fundamentali tal-kalkulu integrali II

Jekk ${\displaystyle f:[a,b]\to \mathbb {R} }$ tkun , u d-derivata ${\displaystyle f'}$ tkun , imbagħad

${\displaystyle \int _{a}^{b}f'(t){\rm {d}}t=f(b)-f(a).}$

Prova: Ħalli ${\displaystyle \displaystyle {P}}$ tkun partizzjoni ta' ${\displaystyle \displaystyle {[a,b]}}$ f'${\displaystyle \displaystyle {n}}$ sottointervalli

${\displaystyle P=\{a=x_{0}<x_{1}<x_{2}<\ldots <x_{n-1}<x_{n}=b\},}$

u

${\displaystyle {\tilde {m}}_{k}=\inf\{f'(x):x\in [x_{k-1},x_{k}]\},\ \ {\tilde {M}}_{k}=\sup\{f'(x):x\in [x_{k-1},x_{k}]\}.}$

Billi napplikaw it-teorema tal-valur medju għall kull intervall ${\displaystyle \displaystyle {[x_{k-1},x_{k}]}}$, niksbu punti ${\displaystyle \displaystyle {t_{k}\in (x_{k-1},x_{k})}}$