16.3-bo'lim : Chiziqli integrallar - II qism
7. Quyidagi egri chiziqlarning har biri uchun \( \displaystyle \int\limits_{C}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}}\) ni baholang.
- \(C\) - \(\left( {1,7} \right)\) dan \(\left( { - 2,4} \right)\) gacha bo'lgan chiziq segmenti.
- \(C\) - \(\left( { - 2,4} \right)\) dan \(\left( {1,7} \right)\) gacha bo'lgan chiziq segmenti.
Yechim
Keling, muammoning ushbu qismi uchun egri chiziqning qisqacha chizmasidan boshlaylik.
Ushbu muammoning ikkinchi qismiga kelganimizdan keyin aniq bo'ladigan sabablarga ko'ra, keling, bu egri chiziqni \(C\) o'rniga \({C_1}\) deb ataymiz. Mana, \({C_1}\) parametrizatsiyasi.
\[{C_1}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1,7} \right\rangle + t\left\langle { - 2,4} \right\rangle = \left\langle {1 - 3t,7 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1\]2-qadam
Mana bu egri chiziq uchun integrali.
\[\begin{align*}\int\limits_{{{C_1}}}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}} & = \int\limits_{{{C_1}}}{{{x^3}\,dy}} - \int\limits_{{{C_1}}}{{y + 1\,dx}}\\ & = \int_{0}^{1}{{{{\left( {1 - 3t} \right)}^3}\left( { - 3} \right)\,dt}} - \int_{0}^{1}{{\left( {7 - 3t + 1} \right)\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{ - 3{{\left( {1 - 3t} \right)}^3} + 3\left( {8 - 3t} \right)\,dt}}\\ & = \left. {\left[ {\frac{1}{4}{{\left( {1 - 3t} \right)}^4} + 24t - \frac{9}{2}{t^2}} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{93}}{4}}}\end{align*}\]Chiziqli integralni "standart" integralga o'tkazishda differensiallar bilan to'g'ri ishlashni unutmang.
b \(C\) - \(\left( { - 2,4} \right)\) dan \(\left( {1,7} \right)\) gacha bo'lgan chiziq segmenti.
Yechim
Endi, oldingi qismda qilganimizdek, keling, bu egri chiziqni \(C\) o‘rniga \({C_2}\) deb “qayta nomlaymiz”.
Keyinchalik, bu egri chiziq qarama-qarshi yo'nalishdagi birinchi qadamdagi egri chiziq ekanligini unutmang. Boshqacha qilib aytganda, bizda bu \({C_2} = - {C_1}\) mavjud. Mana, to'liqlik uchun \({C_2}\) ning qisqacha eskizi.
2-qadam
Endi, bu nuqtada biz integralni baholash uchun ikki xil usuldan foydalanishimiz mumkin.
Birinchi usul, agar biz egri chiziq yo'nalishini o'zgartirsak, bu turdagi chiziqli integralning qiymati faqat belgilarni o'zgartirishi haqidagi faktdan foydalaning. Ushbu faktni ushbu qismdagi egri chiziq va birinchi qismdagi egri chiziq o'rtasidagi munosabat bilan birgalikda ishlatib, i.e. \({C_2} = - {C_1}\) , chiziqli integrali shunchaki,
\[\int\limits_{{{C_2}}}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}} = \int\limits_{{ - {C_1}}}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}} = - \int\limits_{{{C_1}}}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{93}}{4}}}\]E'tibor bering, yuqoridagi birinchi tenglik belgisi faqat ikkita egri chiziq o'rtasidagi munosabatni tan oldi. Ikkinchi tenglik belgisi - biz qaydlardagi faktdan foydalanganmiz.
Bu muammoni hal qilishning "oson" usuli. Shu bilan bir qatorda, biz egri chiziqni parametrlashimiz va chiziqli integralni to'g'ridan-to'g'ri hisoblashimiz mumkin. Biz buni muammoning qolgan qismida qanday qilishimizni ko'rsatish uchun qilamiz.
3-qadam
Mana bu egri chiziq uchun parametrlash.
\[{C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle { - 2,4} \right\rangle + t\left\langle {1,7} \right\rangle = \left\langle { - 2 + 3t,4 + 3t} \right\rangle \hspace{0.25in} 0 \le t \le 1\]
4-qadam
Endi biz faqat chiziq integralini hisoblashimiz kerak.
\[\begin{align*}\int\limits_{{{C_2}}}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}} & = \int\limits_{{{C_2}}}{{{x^3}\,dy}} - \int\limits_{{{C_2}}}{{y + 1\,dx}}\\ & = \int_{0}^{1}{{{{\left( { - 2 + 3t} \right)}^3}\left( 3 \right)\,dt}} - \int_{0}^{1}{{\left( {4 + 3t + 1} \right)\left( 3 \right)\,dt}}\\ & = \int_{0}^{1}{{3{{\left( { - 2 + 3t} \right)}^3} - 3\left( {5 + 3t} \right)\,dt}}\\ & = \left. {\left[ {\frac{1}{4}{{\left( { - 2 + 3t} \right)}^4} - 15t - \frac{9}{2}{t^2}} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{93}}{4}}}\end{align*}\]Demak, bu qismning chiziqli integrali biz kutganimizdek, belgidan tashqari birinchi qismdagi chiziqli integrali bilan bir xil qiymatga ega edi.