Isibonelo semibuzo yengxoxo yeRiemann Sum
I-Pendahuluan
Isamba sikaRiemann siwumqondo oyisisekelo ekubaleni osetshenziswa ukuchaza i-integral eqondile yomsebenzi. Le ndlela isebenzisa ukwahlukaniswa kwezikhawu kanye nesamba sezindawo zama-rectangles ukuze kulinganiswe i-integral. Lesi sihloko sizoxoxa kabanzi ngomqondo wesamba sikaRiemann, kufaka phakathi izibonelo nezingxoxo ukuze kube lula ukuqonda.
Umqondo Oyisisekelo We-Riemannian Sum
Ngaphambi kokuba sixoxe ngezibonelo, kubalulekile ukuqonda umqondo oyisisekelo wezibalo zeRiemannian. Izibalo zeRiemannian zingahlukaniswa zibe izinhlobo ezintathu eziyinhloko:
1. Isamba seRiemann esingakwesokunxele
2. Isamba sikaRiemann kwesokudla
3. Isilinganiso se-Midpoint Riemann
Le ndlela iphula isikhawu somsebenzi ukuze sihlanganiswe kuma-subinterval amancane anobude obulinganayo. Ngayinye yalezi zi-subinterval isetshenziswa ukwakha unxande okuphakama kwawo kunqunywa inani lomsebenzi endaweni ethile ngaphakathi kwe-subinterval (kwesobunxele, kwesokudla, noma phakathi nendawo).
Ifomula Ejwayelekile yeRiemann Sum
Ake sithi sifuna ukuhlanganisa umsebenzi \( f(x) \) kusukela ku-\( a \) kuya ku-\( b \). Sihlukanisa isikhawu \( [a, b] \) sibe yi-\( n \) ama-subinterval alinganayo obude \( \Delta x = \frac{ba}{n} \). Izibalo zikaRiemann zezinhlobo ezintathu ezishiwo ngenhla zingabhalwa kanje:
1. URiemann Okwesobunxele:
\[ L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x \]
2. Kwesokudla uRiemann:
\[ R_n = \sum_{i=1}^{n} f(x_i) \Delta x \]
3. URiemann Ophakathi:
\[ M_n = \sum_{i=0}^{n-1} f\left(\frac{x_i + x_{i+1}}{2}\right) \Delta x \]
Di mana:
– \( \Delta x \) ububanzi be-subinterval ngayinye.
– \( x_i \) yindawo yokuqala ye-i-th subinterval yesamba seRiemann sesobunxele.
– \( x_i \) yindawo yokugcina ye-i-th subinterval yesamba seRiemann esifanele.
– \( \frac{x_i + x_{i+1}}{2} \) yi-midpoint ye-i-th subinterval ye-middle Riemann sum.
Imibuzo Eyisibonelo Nengxoxo
Ake sixoxe ngezinkinga eziyisibonelo zohlobo ngalunye lwe-Riemann Sum ukuze sijulise ukuqonda kwethu.
Isibonelo 1: I-Riemann Sum Yesobunxele
Bala isamba seRiemann sesobunxele se- \( f(x) = x^2 \) esikhaleni \([0, 2]\) no- \( n = 4 \).
Ingxoxo:
1. Ububanzi be-Subival (Δx):
\[ \Delta x = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Indawo Yokuhlukanisa Isikhawu (kwesobunxele):
\[ x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5 \]
3. Inani Lomsebenzi Ephuzwini Lokuhlukanisa:
\[ f(x_0) = f(0) = 0^2 = 0 \]
\[ f(x_1) = f(0.5) = (0.5)^2 = 0.25 \]
\[ f(x_2) = f(1.0) = (1.0)^2 = 1 \]
\[ f(x_3) = f(1.5) = (1.5)^2 = 2.25 \]
4. I-Riemann Sum Yesobunxele (Ekugcineni):
\[ L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x = (0) \cdot 0.5 + (0.25) \cdot 0.5 + (1) \cdot 0.5 + (2.25) \cdot 0.5 \]
\[ L_n = 0 + 0.125 + 0.5 + 1.125 \]
\[L_n = 1.75 \]
Isibonelo 2: I-Riemann Sum Yesokudla
Bala isamba sikaRiemann esifanele se- \( f(x) = x^2 \) esikhaleni \([0, 2]\) no- \( n = 4 \).
Ingxoxo:
1. Ububanzi be-Subival (Δx):
\[ \Delta x = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Indawo Yokuhlukanisa Isikhawu (kwesokudla):
\[ x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, x_4 = 2.0 \]
3. Inani Lomsebenzi Ephuzwini Lokuhlukanisa:
\[ f(x_1) = f(0.5) = (0.5)^2 = 0.25 \]
\[ f(x_2) = f(1.0) = (1.0)^2 = 1 \]
\[ f(x_3) = f(1.5) = (1.5)^2 = 2.25 \]
\[ f(x_4) = f(2.0) = (2.0)^2 = 4 \]
4. I-Riemann Sum Yesokudla (Rn):
\[ R_n = \sum_{i=1}^{n} f(x_i) \Delta x = (0.25) \cdot 0.5 + (1) \cdot 0.5 + (2.25) \cdot 0.5 + (4) \cdot 0.5 \]
\[ R_n = 0.125 + 0.5 + 1.125 + 2 \]
\[ R_n = 3.75 \]
Isibonelo 3: I-Middle Riemann Sum
Bala isamba esiphakathi sikaRiemann sika- \( f(x) = x^2 \) esikhaleni \([0, 2]\) no- \( n = 4 \).
Ingxoxo:
1. Ububanzi be-Subival (Δx):
\[ \Delta x = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Iphuzu eliphakathi le-Subinterval:
\[ x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, \umbhalo{ kanye } x_{n-1}=2.0 \]
Iphuzu eliphakathi le-subinterval:
\[tm_0 = \kwesobunxele(\frac{0 + 0.5}{2}\kwesokudla)=0.25 \]
\[tm_1 = \kwesobunxele(\frac{0.5 + 1.0}{2}\kwesokudla)=0.75 \]
\[tm_2 = \kwesobunxele(\frac{1.0 + 1.5}{2}\kwesokudla)=1.25 \]
\[tm_3 = \kwesobunxele(\frac{1.5 + 2.0}{2}\kwesokudla)=1.75 \]
3. Inani Lomsebenzi ku-Midpoint:
\[ f(0.25) = (0.25)^2 = 0.0625 \]
\[ f(0.75) = (0.75)^2 = 0.5625 \]
\[ f(1.25) = (1.25)^2 = 1.5625 \]
\[ f(1.75) = (1.75)^2 = 3.0625 \]
4. I-Central Riemann Sum (Mn):
\[ M_n = \sum_{i=0}^{n-1} f(tm_i) \Delta x = (0.0625) \cdot 0.5 + (0.5625) \cdot 0.5 + (1.5625) \cdot 0.5 + (3.0625) \cdot 0.5 \]
\[ M_n = 0.03125 + 0.28125 + 0.78125 + 1.53125 \]
\[ M_n = 2.625 \]
Isiphetho
Lesi sihloko sixoxe ngendlela yokubala izamba zeRiemann zesobunxele, kwesokudla, kanye neziphakathi, kanye nezibonelo ezinemininingwane. Indlela yesamba seRiemann inikeza indlela ephumelelayo yokulinganisa ukuhlanganiswa komsebenzi ngokuhlukanisa isikhawu sawo sibe ama-subinterval amancane nokubala indawo iyonke ye-subinterval ngayinye. Ukuqonda okuhle kwesamba seRiemann kubalulekile kulabo abafunda i-calculus noma abasebenza ngemisebenzi eyinkimbinkimbi emikhakheni eyahlukene yesayensi.
