16.3-bo'lim : Chiziqli integrallar - II qism
1. \( \displaystyle \int\limits_{C}{{\sqrt {1 + y} \,dy}}\) ni baholang, bu erda \(C\) \(y = {{\bf{e}}^{2x}}\) ning \(x = 0\) dan \(x = 2\) gacha bo'lgan qismidir.
Yechim
Bu erda ko'rsatilgan muammo bayonotida ko'rsatilgan yo'nalish bilan \(C\) ning qisqacha eskizi.
2-qadam
Keyinchalik, egri chiziqni parametrlashimiz kerak.
\[\vec r\left( t \right) = \left\langle {t,{{\bf{e}}^{2t}}} \right\rangle \hspace{0.25in}0 \le t \le 2\]3-qadam
Endi biz chiziqli integralni baholashimiz kerak. Ushbu turdagi chiziqli integral bilan ehtiyot bo'ling. E'tibor bering, bu holda differentsial \(dy\) bo'lib, avvalgi bo'limda bo'lgani kabi \(ds\) emas.
Chiziqli integralni “standart” integralga aylantirganda, biz faqat \(dy = y'\,dt\)ni eslashimiz kerak.
Shunday qilib, chiziq integralini baholaylik. Faqat parametrlarni integralga "ulash" ni unutmang (i.e. integraldagi \(x\) va \(y\) ni parametrlashning \(x\) va \(y\) komponentlari bilan almashtiring) va differentsialni to'g'ri aylantirish uchun.
Mana chiziqli integrali.
\[\begin{align*}\int\limits_{C}{{\sqrt {1 + y} \,dy}} &= \int_{0}^{2}{{\sqrt {1 + {{\bf{e}}^{2t}}} \,\,\left( {2{{\bf{e}}^{2t}}} \right)\,dt}}\\ & = \left. {\left[ {\frac{2}{3}{{\left( {1 + {{\bf{e}}^{2t}}} \right)}^{\frac{3}{2}}}} \right]} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{3}\left[ {{{\left( {1 + {{\bf{e}}^4}} \right)}^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right] = 274.4897}}\end{align*}\]E'tibor bering, bu holda integral oddiy almashtirish bilan yakunlanadi.