Imibuzo Yezibonelo kanye Nengxoxo Yezinhlelo Zokusebenza Ezihlanganisiwe
Ukuhlanganiswa kuwumqondo oyisisekelo ekubaleni onezinhlelo eziningi emikhakheni eyahlukene yesayensi, njengefiziksi, ezomnotho, i-biology, kanye nobunjiniyela. Ama-Integrals asetshenziselwa ukubala indawo engaphansi kwejika, umthamo wento eqinile, umsebenzi, ingcindezi, nokuningi. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zezinhlelo zokusebenza ezihlanganisiwe, kulandelwe izincazelo ezinemininingwane zendlela yokuzixazulula.
1. Ukunquma Indawo Engaphansi Kwejika
Enye yezindlela ezivame kakhulu zokusebenzisa ama-integral ukubala indawo engaphansi kwe-curve yomsebenzi esikhathini esithile. Ake sithi sifuna ukuthola indawo yesifunda elinganiselwe yi-curve \(y = x^2\) kanye ne-axis \(x\) kusukela ku-\(x = 0\) kuya ku-\(x = 2\).
Isibonelo sezinkinga:
Nquma indawo engaphansi kwejika \(y = x^2\) kusukela \(x = 0\) kuya \(x = 2\).
Ingxoxo:
Ukuze sithole indawo engaphansi kwejika \(y = x^2\) kusukela \(x = 0\) kuya ku \(x = 2\), sidinga ukubala i-integral eqondile yomsebenzi:
\[ \int_{0}^{2} x^2 \, dx \]
Isinyathelo 1: Thola i-integral ye-\(x^2\).
Qaphela ukuthi i-integral ye-\(x^2\) ithi:
\[ \int x^2 \, dx = \frac{x^3}{3} + C \]
Isinyathelo 2: Sebenzisa umkhawulo ohlanganisiwe \(0\) ku-\(2\).
\[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} \]
Isinyathelo 3: Bala inani lomkhawulo.
\[ \kwesobunxele. \frac{x^3}{3} \kwesokudla|_{0}^{2} = \frac{2^3}{3} – \frac{0^3}{3} = \frac{8}{3} – 0 = \frac{8}{3} \]
Ngakho-ke, indawo engaphansi kwejika \(y = x^2\) kusukela ku-\(x = 0\) kuya ku-\(x = 2\) iyi-\( \frac{8}{3} \) amayunithi endawo.
2. Ukubala Umthamo Wezinto Ezijikelezayo
Ama-Integrals asetshenziswa futhi ukubala ivolumu yezinto eziqinile zokuguquka. Uma isifunda sijikeleziswa nge-axis ye-\(x\), ivolumu yento ingatholakala kusetshenziswa indlela yediski noma indlela yendandatho.
Isibonelo sezinkinga:
Bala ivolumu yento ekhiqizwe lapho isifunda esilinganiselwe yi-curve \(y = \sqrt{x}\) kanye nomugqa \(x = 4\) kujikeleziswa ku-axis \(x\).
Ingxoxo:
Ukuze sithole ivolumu ye-solid of revolution, singasebenzisa indlela yediski. Ivolumu \(V\) ye-solid ephumayo ingachazwa kanje:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
Lapho \(f(x) = \sqrt{x}\), \(a = 0\), kanye \(b = 4\).
Isinyathelo 1: Yakha i-integral yevolumu.
\[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx \]
Isinyathelo 2: Yenza umsebenzi ube lula ku-integral.
\[ V = \pi \int_{0}^{4} x \, dx \]
Isinyathelo 3: Thola i-integral ye-\(x\).
\[ \int x \, dx = \frac{x^2}{2} + C \]
Isinyathelo 4: Sebenzisa imikhawulo \(0\) ku-\(4\).
\[ V = \pi \kwesobunxele[ \frac{x^2}{2} \kwesokudla]_{0}^{4} \]
Isinyathelo 5: Bala inani lomkhawulo.
\[ \kwesobunxele. \frac{x^2}{2} \kwesokudla|_{0}^{4} = \pi \kwesobunxele( \frac{4^2}{2} – \frac{0^2}{2} \kwesokudla) = \pi \kwesobunxele( \frac{16}{2} \kwesokudla) = 8\pi \]
Ngakho-ke, ivolumu yento ephumayo ingamayunithi evolumu angu-\(8\pi\).
3. Ukubala Umsebenzi Owenziwe Amandla Aguquguqukayo
Izinhlelo zokusebenza ezihlanganisiwe ziyatholakala naku-physics, enye yazo ukubala umsebenzi owenziwe amandla aguquguqukayo lapho into isuka kwelinye iphuzu iye kwelinye.
Isibonelo sezinkinga:
Amandla \(F(x) = 3x^2\) I-Newton isebenza kunhlayiya ehamba kusuka ku-\(x = 1\) imitha iye ku-\(x = 3\) amamitha. Bala umsebenzi owenziwe amandla.
Ingxoxo:
Umsebenzi \(W\) owenziwe ngamandla \(F(x)\) ungatholakala ngokubala i-integral ye \(F(x)\) phezu kokususwa kusuka ku \(a\) kuya ku \(b\):
\[ W = \int_{a}^{b} F(x) \, dx \]
Lapho \(a = 1\), \(b = 3\), kanye \(F(x) = 3x^2\).
Isinyathelo 1: Yakha ingxenye ebalulekile yomsebenzi.
\[ W = \int_{1}^{3} 3x^2 \, dx \]
Isinyathelo 2: Thola i-integral ye-\(3x^2\).
\[ \int 3x^2 \, dx = 3 \left( \frac{x^3}{3} \right) = x^3 + C \]
Isinyathelo 3: Sebenzisa imikhawulo \(1\) ku-\(3\).
\[ W = \kwesobunxele[ x^3 \kwesokudla]_{1}^{3} \]
Isinyathelo 4: Bala inani lomkhawulo.
\[ W = \kwesobunxele. x^3 \kwesokudla|_{1}^{3} = 3^3 – 1^3 = 27 – 1 = 26 \]
Ngakho-ke, umsebenzi owenziwa yi-force yi-\(26\) joules.
4. Ukunquma Ukucindezela Kwamanzi
Ku-physics, ama-integrals asetshenziswa futhi ukubala ingcindezi ye-hydrostatic ebusweni obucwiliswe oketshezini.
Isibonelo sezinkinga:
Ipuleti eliqondile elingamamitha ayi-6 ukuphakama kanye namamitha ama-4 ububanzi licwiliswa emanzini kanti ingaphezulu lalo lingaphezu kwamanzi. Bala amandla aphelele okucindezela kwamanzi epuletini.
Ingxoxo:
Ingcindezi ekujuleni \(h\) emanzini inikezwa yi-\(P = \rho gh\), lapho \(\rho\) kungubuningi bamanzi (cishe \(1000 \text{ kg/m}^3\)) kanye ne-\(g\) kungukusheshisa okubangelwa amandla adonsela phansi (cishe \(9.8 \text{ m/s}^2\)).
Ukuze sithole amandla okucindezela aphelele, kumelwe sihlanganise ingcindezi endaweni eqondile yepuleti.
Isinyathelo 1: Thola umsebenzi wokucindezela.
\[ P(y) = \rho gy \]
Isinyathelo 2: Amandla aphelele \(F\) ayingxenye ebalulekile yezikhathi zengcindezi endaweni eyisisekelo \(dA\) kusukela \(y = 0\) kuya ku-\(y = 6\).
\[ F = \int_{0}^{6} \rho gy \cdot 4 \, dy \]
Isinyathelo 3: Yenza kube lula ama-constant.
\[ F = 4 \rho g \int_{0}^{6} y \, dy \]
Isinyathelo 4: Thola i-integral ye-\(y\).
\[ \int y \, dy = \frac{y^2}{2} \]
Isinyathelo 5: Sebenzisa imikhawulo \(0\) ku-\(6\).
\[ F = 4 \cdot 1000 \cdot 9.8 \left[ \frac{y^2}{2} \right]_{0}^{6} \]
Isinyathelo 6: Bala inani lomkhawulo.
\[ F = 4 \cdot 1000 \cdot 9.8 \cdot \frac{6^2}{2} = 4 \cdot 1000 \cdot 9.8 \cdot 18 = 705600 \]
Ngakho-ke, amandla aphelele okucindezela kwamanzi epuletini ngu-\(705600\) Newton.
Isiphetho
Ukusetshenziswa kwama-integral ezinhlotsheni ezahlukene zokusebenza kunikeza amandla amakhulu okuhlaziya ekubaleni ubungako obuyinkimbinkimbi obungokwenyama. Kulesi sihloko, sixoxe ngendlela ama-integral asetshenziswa ngayo ekubaleni indawo engaphansi kwejika, umthamo we-solid of revolution, umsebenzi owenziwa amandla aguquguqukayo, kanye nokucindezela kwe-hydrostatic. Ngokuqonda okuhle kwamasu okuhlanganisa, singaxazulula izinkinga ezahlukahlukene ezisebenzayo ezivela kwisayensi nobunjiniyela.
