16.3-bo'lim : Chiziqli integrallar - II qism
4. \( \displaystyle \int\limits_{C}{{1 + {x^3}\,dx}}\) ni baholang, bunda \(C\) 2 radiusli aylananing o'ng yarmi soat miliga teskari aylanib, keyin \(\left( {0,2} \right)\) dan \(\left( { - 3, - 4} \right)\) gacha bo'lgan chiziq segmenti. Yo'nalish uchun quyidagi eskizga qarang.
Yechim
Muammoni hal qilish uchun har bir egri chiziqni quyidagicha belgilaymiz:
Endi bu egri chiziqlarning har birini parametrlashtiramiz.
\({C_1}:\,\,\vec r\left( t \right) = \left\langle {2\cos \left( t \right),2\sin \left( t \right)} \right\rangle \hspace{0.25in} - \frac{1}{2}\pi \le t \le \frac{1}{2}\pi \)\({C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle { - 3, - 4} \right\rangle = \left\langle { - 3t,2 - 6t} \right\rangle \hspace{0.25in}0 \le t \le 1\)
2-qadam
Endi har bir egri chiziq uchun chiziqli integralini hisoblashimiz kerak.
\[\begin{align*}\int\limits_{{{C_1}}}{{1 + {x^3}\,dx}} & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{\left[ {1 + {{\left( {2\cos \left( t \right)} \right)}^3}} \right]\left( { - 2\sin \left( t \right)} \right)\,dt}}\\ & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{ - 2\sin \left( t \right) - 16{{\cos }^3}\left( t \right)\sin \left( t \right)\,dt}}\\ & = \left. {\left( {2\cos \left( t \right) + 4{{\cos }^4}\left( t \right)} \right)} \right|_{ - \,\frac{1}{2}\pi }^{\frac{1}{2}\pi } = \underline 0 \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{1 + {x^3}\,dx}} & = \int_{0}^{1}{{\left[ {1 + {{\left( { - 3t} \right)}^3}} \right]\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{ - 3 + 81{t^3}\,dt}}\\ & = \left. {\left( { - 3t + \frac{{81}}{4}{t^4}} \right)} \right|_0^1 = \underline {\frac{{69}}{4}} \end{align*}\]Chiziqli integralni "standart" integralga o'tkazishda differensiallar bilan to'g'ri ishlashni unutmang.
3-qadam
Yaxshi, bu muammoni hal qilish uchun biz qilishimiz kerak bo'lgan narsa to'liq chiziqli integralni olish uchun ushbu egri chiziqlar ustiga chiziqli integrallarni qo'shishdir.
\[ \displaystyle \int\limits_{C}{{1 + {x^3}\,dx}} = \left( 0 \right) + \left( {\frac{{69}}{4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{69}}{4}}}\]Esda tutingki, biz har bir alohida chiziqli integral natijasiga uning qayerdan kelganini ko'rsatish uchun qavs qo'yamiz va ular umuman kerak emas.