I-Riemann Sum: Enye Yezinsika Ze-Integral Calculus
Ezibalweni, ikakhulukazi ekubaleni okuhlanganisiwe, umqondo wesamba sikaRiemann udlala indima ebalulekile. Le ndlela, eyethulwa yisazi sezibalo esidumile saseJalimane uBernhard Riemann, iyindlela ebalulekile yokuchaza ukuhlanganiswa komsebenzi esikhathini esithile esinikeziwe. Ukuqonda isamba sikaRiemann kusenza sikwazi ukulinganisa indawo ngaphansi kwejika, okuwukusetshenziswa okubalulekile emikhakheni eminingi yesayensi nobuchwepheshe, kusukela ku-physics kuya kwezomnotho.
Ukuze siqonde umongo wesamba sikaRiemann, kumele sihlole izakhi zaso eziyisisekelo, okuhlanganisa ukwahlukanisa izikhawu, ukunquma amaphuzu okuhlola, ukwakha izamba, kanye nokusetshenziswa kwazo ekuhlanganisweni. Ake singene sijule kulesi sihloko.
Isingeniso Semiqondo Eyisisekelo
Isamba sikaRiemann siyindlela yokubala i-integral eqondile yomsebenzi phezu kwe-closed interval \([a, b]\). Le ndlela ihilela ukuhlukanisa i-interval ibe yi-subinterval encane, ukuhlola umsebenzi ezindaweni ezithile ku-subinterval ngayinye, bese kufingqwa imikhiqizo yamanani omsebenzi ngobude be-subintervals ehambisanayo.
Ukuhlukaniswa Kwesikhawu
Isinyathelo sokuqala ekuchazeni isamba sikaRiemann ukuhlukanisa isikhawu \([a, b]\) sibe yizikhawu ezincane zobude obunikeziwe. Ake sithi isikhawu \([a, b]\) sihlukaniswe ngezingxenye \(n\) ezilinganayo, bese:
\[ \Delta x = \frac{b – a}{n} \]
I-subinterval ngayinye inobude \(\Delta x\), futhi lawa maphuzu okuhlukanisa ngokuvamile yi-\((x_0, x_1, x_2, …, x_n)\), lapho \(x_0 = a\), \(x_1 = a + \Delta x\), \(x_2 = a + 2\Delta x\), njalo njalo kuze kufike ku-\(x_n = b\).
Ukunqunywa Kwephuzu Lokuhlola
Ku-subinterval ngayinye \([x_{i-1}, x_i]\), kudingeka iphuzu lokuhlola \(x_i \) elingaphakathi kwalelo subinterval. Leli phuzu linganqunywa kanje:
1. Iphuzu Lesobunxele: \(x_i^ = x_{i-1}\)
2. Iphuzu Lesokudla: \(x_i^ = x_i\)
3. Iphuzu eliphakathi: \(x_i^ = \frac{x_{i-1} + x_i}{2}\)
4. Amaphuzu Angahleliwe: Yonke i-\(x_i \) iyiphuzu elingahleliwe ku-\([x_{i-1}, x_i]\)
Ukukhethwa kwamaphuzu okuhlola kungathinta umphumela wesamba sikaRiemann, ikakhulukazi uma umsebenzi ungaqhubeki noma unokuguquguquka okusheshayo.
Ukwakheka kweSum
Uma ukunqunywa kwephuzu lokuhlukanisa izikhawu kanye nokuhlola sekuqediwe, isinyathelo esilandelayo ukubala inani lomsebenzi endaweni ngayinye yokuhlola \(f(x_i^ )\) bese uphinda lelo nani ngobude be-subinterval \(\Delta x\). Isamba sikaRiemann \(R\) sichazwa ngokuthi:
\[ R = \sum_{i=1}^nf(x_i^ ) \Delta x \]
Uma inani lama-subintervals \(n\) likhuliswa ngaphandle kokuboshwa (\(n \rightarrow \infty\)), ubude be-subinterval \(\Delta x\) buba buncane kakhulu futhi isamba sikaRiemann sisondela ekuhlanganisweni komsebenzi \(f\) esikhaleni \([a, b]\). Lo mkhawulo ubhalwe kanje:
\[ \int_a^bf(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i^ ) \Delta x \]
Isibonelo Sokusetshenziswa kweRiemann Sum
Njengomfanekiso, ake sisebenzise isamba sikaRiemann ukuze sinqume ukuhlanganiswa komsebenzi \(f(x) = x^2\) esikhaleni \([0, 1]\).
Isinyathelo 1: Ukuhlukaniswa Kwesikhawu
Ake sithi sihlukanisa isikhawu \([0, 1]\) sibe yizikhawu ezingaphansi \(n\) zobude obulinganayo, khona-ke ubude bezikhawu ezingaphansi buyi:
\[ \Delta x = \frac{1 – 0}{n} = \frac{1}{n} \]
Isinyathelo 2: Iphuzu Lokuhlola
Sebenzisa i-midpoint \(x_i \) ukuhlola umsebenzi ku-subinterval ngayinye \([x_{i-1}, x_i]\):
\[ x_i^ = \frac{x_{i-1} + x_i}{2} = \frac{\left(\frac{i-1}{n}\right) + \left(\frac{i}{n}\right)}{2} = \frac{2i – 1}{2n} \]
Isinyathelo 3: Bala Isamba
Inani lomsebenzi \(f(x_i^ ) = \left( \frac{2i – 1}{2n} \right)^2 = \frac{(2i-1)^2}{4n^2}\), bese kuba yi-Riemann sum:
\[ R = \sum_{i=1}^nf\left(\frac{2i – 1}{2n}\right) \Delta x = \sum_{i=1}^n \frac{(2i-1)^2}{4n^2} \cdot \frac{1}{n} = \frac{1}{4n^3} \sum_{i=1}^n (2i-1)^2 \]
Ngokuhlolwa okwengeziwe, isamba sezikwele zezinombolo ezingalingani sinikeza uphawu lwe-sigma olungenziwa lula kuze kube yilapho lufinyelela umkhawulo.
Ekugcineni, njengoba i-\(n\) iya ku-infinity, inani lesamba sikaRiemann lizosondela kumphumela we-integral eqondile:
\[ \lim_{n \to \infty} \frac{1}{4n^3} \sum_{i=1}^n (2i-1)^2 = \frac{1}{3} \]
Futhi emphumeleni wokuhlaziya we-integral sithola:
\[ \int_0^1 x^2 \, dx = \kwesobunxele[ \frac{x^3}{3} \kwesokudla]_0^1 = \frac{1}{3} \]
Izinhlobo kanye nokusetshenziswa kwe-Riemann Sums
Ngaphandle kokuhlanganiswa kwendabuko, isamba sikaRiemann sinezinye izinhlobo, okuhlanganisa isamba sikaRiemann-Kronecker kanye nesamba sikaRiemann-Stieltjes sezikhala ze-metric kanye nezicelo ezibanzi ekuhlaziyweni kokusebenza. Futhi sakha isisekelo sezindlela zezinombolo ezifana nezindlela zeTrapzoid kanye neSimpson ezisetshenziswa ekubaleni kwesayensi.
Iyavala
Izibalo zikaRiemann zinikeza indlela eqinile neguquguqukayo yokuchaza nokubala ama-integral ezimweni ezahlukahlukene zezibalo. Njengethuluzi lokufundisa lezinkinga eziyisisekelo zama-integral ekubaleni, ukuqonda okuphelele kwalomqondo kuvula ukuqonda ngokusetshenziswa okubanzi kwama-integral empilweni yangempela, kokubili kwezesayensi eziqondile kanye nasezindaweni zenhlalo-mnotho. UBernhard Riemann akagcinanga ngokucebisa ithiyori yezibalo ngalokhu kutholwa kodwa futhi wavula izindlela ezintsha ekuhlaziyeni kwesimanje kwama-integral.
