Imibuzo Yezibonelo kanye Nengxoxo Yokusetshenziswa Kwezinto Ezihlanganisiwe Ekubaleni Indawo Yendiza Eyisicaba
Ekufundiseni izibalo, ama-integral avame ukuhlangana nawo ekubaleni. Enye yezindlela ezaziwa kakhulu zokusebenzisa ama-integral ukubala indawo ngaphansi kwejika noma indiza. Lesi sihloko sizoxoxa ngezinkinga eziningana zezibonelo futhi sixoxe ngokusetshenziswa kwama-integral ekubaleni indawo yendiza.
Isingeniso kuThiyori
Ngaphambi kokuqhubekela enkingeni yesibonelo, ake sibukeze umqondo oyisisekelo wokubala indawo ngaphansi kwejika sisebenzisa ama-integral. Uma sinomsebenzi f(x) oqhubekayo esikhaleni [a, b], khona-ke indawo engaphansi kwejika y = f(x) kusukela ku-x = a kuya ku-x = b yile:
\[ L = \int_{a}^{b} f(x) \, dx \]
Ngokwejiyometri, lokhu kusho ukuthi sifingqa indawo yesikwele esincane kakhulu kusukela ku-x = a kuya ku-x = b.
Isibonelo Umbuzo 1
I-Soal
Bala indawo engaphansi kwejika u-y = x² esikhaleni [1, 3].
Ingxoxo
Ukuze sibale indawo, sisebenzisa i-integral:
\[ L = \int_{1}^{3} x^2 \, dx \]
Siqala ngokuthola i-antiderivative ye- \( x^2 \). I-antiderivative ye- \( x^2 \) ingu- \( \frac{x^3}{3} \). Bese i-integral iba:
\[ L = \kwesobunxele[ \frac{x^3}{3} \kwesokudla]_{1}^{3} \]
Khumbula ukuthi kufanele sihlole i-antiderivative ngemikhawulo ye-integral:
\[ L = \kwesobunxele( \frac{3^3}{3} \kwesokudla) – \kwesobunxele( \frac{1^3}{3} \kwesokudla) \]
\[ L = \kwesobunxele( \frac{27}{3} \kwesokudla) – \kwesobunxele( \frac{1}{3} \kwesokudla) \]
\[ L = 9 – \frac{1}{3} \]
\[ L = \frac{27}{3} – \frac{1}{3} \]
\[ L = \frac{26}{3} \]
Ngakho-ke, indawo engaphansi kwejika u-y = x² kusukela ku-x = 1 kuya ku-x = 3 yile:
\[ \frac{26}{3} \, \text{area unit} \]
Isibonelo Umbuzo 2
I-Soal
Nquma indawo yesifunda esilinganiselwe yijika u-y = x³ kanye nemigqa u-x = 1 no-x = 2.
Ingxoxo
Ukuze sibale indawo, sisebenzisa i-integral:
\[ L = \int_{1}^{2} x^3 \, dx \]
Njengokujwayelekile, siqala ngokuthola i-antiderivative ye- \( x^3 \). I-antiderivative ye- \( x^3 \) ingu- \( \frac{x^4}{4} \). I-integral iba:
\[ L = \kwesobunxele[ \frac{x^4}{4} \kwesokudla]_{1}^{2} \]
Hlola imikhawulo ye-integral:
\[ L = \kwesobunxele( \frac{2^4}{4} \kwesokudla) – \kwesobunxele( \frac{1^4}{4} \kwesokudla) \]
\[ L = \kwesobunxele( \frac{16}{4} \kwesokudla) – \kwesobunxele( \frac{1}{4} \kwesokudla) \]
\[ L = 4 – \frac{1}{4} \]
\[ L = \frac{16}{4} – \frac{1}{4} \]
\[ L = \frac{15}{4} \]
Ngakho-ke, indawo engaphansi kwejika u-y = x³ kusukela ku-x = 1 kuya ku-x = 2 yile:
\[ \frac{15}{4} \, \text{area unit} \]
Isibonelo Umbuzo 3
I-Soal
Nquma indawo yesifunda esilinganiselwe ngama-curve u-y = x² + 1 kanye no-y = 2x + 2 esikhaleni u-x = 0 kuya ku-x = 1.
Ingxoxo
Okokuqala, sidinga ukuthola izindawo zokuhlangana ukuze sinqume imikhawulo yokuhlanganiswa. Isixazululo ku-\( x^2 + 1 = 2x + 2 \):
\[ x^2 + 1 = 2x + 2 \]
\[ x^2 – 2x – 1 = 0 \]
Ukusebenzisa ifomula ye-quadratic:
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ x = \frac{2 \pm \sqrt{8}}{2} \]
\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ x = 1 \pm \sqrt{2} \]
Kodwa-ke, ngemikhawulo ephezulu nephansi phakathi kuka-0 no-1, akudingeki sisebenzise isisombululo se-quadratic, umkhawulo ojwayelekile ohlanganisiwe ukusuka ku-0 kuya ku-1. Okulandelayo, bala indawo yejika eliphezulu lika-y ususe ijika eliphansi lika-y ngokwale mikhawulo:
\[ L = \int_{0}^{1} [(2x + 2) – (x^2 + 1)] \, dx \]
Ukulula komsebenzi:
\[ L = \int_{0}^{1} (2x + 2 – x^2 – 1) \, dx \]
\[ L = \int_{0}^{1} (-x^2 + 2x + 1) \, dx \]
Okulandelayo, sithola i-antiderivative:
I-antiderivative ye- \( (-x^2) \) ingu- \( -\frac{x^3}{3} \),
I-antiderivative ye- \( (2x) \) ingu- \( x^2 \),
I-antiderivative ye-\( (1) \) ingu-\( x \).
Ukuze,
\[ L = \kwesobunxele. \kwesobunxele(-\frac{x^3}{3} + x^2 + x \kwesokudla) \kwesokudla|_0^1 \]
Ukuhlolwa okulandelayo:
\[ L = \kwesobunxele[ -\frac{1^3}{3} + 1^2 + 1 \kwesokudla] – \kwesobunxele[ -\frac{0^3}{3} + 0^2 + 0 \kwesokudla] \]
\[ L = \kwesobunxele[ -\frac{1}{3} + 1 + 1 \kwesokudla] – \kwesobunxele[ 0 \kwesokudla] \]
\[ L = -\frac{1}{3} + 2 \]
\[ L = \frac{6}{3} – \frac{1}{3} \]
\[ L = \frac{5}{3} \]
Ngakho-ke, indawo yesifunda elinganiselwe ngama-curve u-y = x² + 1 kanye no-y = 2x + 2 esikhaleni [0, 1] yile:
\[ \frac{5}{3} \, \text{area unit} \]
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Kusukela ezibonelweni ezingenhla, singabona ukuthi ama-integral angasetshenziswa kanjani ukubala indawo engaphansi kwe-curve noma phakathi kwama-curve amabili. Ngokuqonda kahle imiqondo eyisisekelo yama-integral kanye namasu okulwa nokususwa kwawo, ukubala lezi zindawo kuba yinto ehlelekile futhi esebenza kahle kakhulu. Ngethemba ukuthi lesi sihloko sikhulise ukuqonda kwethu ukusetshenziswa kwama-integral ezweni langempela, ikakhulukazi emkhakheni wokulinganisa indawo yezindawo zendiza.
