Keyinchalik, egri chiziqni parametrlashimiz kerak.

\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {4, - 1,2} \right\rangle + t\left\langle {1,7, - 1} \right\rangle = \left\langle {4 - 3t, - 1 + 8t,2 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1\]

Endi biz chiziqli integralni baholashimiz kerak. Ushbu turdagi integral chiziqlar bilan ehtiyot bo'ling. Bunday holda bizda integranda \(dy\) va \(dz\) mavjud. Eslatib o'tamiz, bu shunchaki soddalashtirilgan belgi,

\[\int\limits_{C}{{{x^2}\,dy - yz\,dz}} = \int\limits_{C}{{{x^2}\,dy}} - \int\limits_{C}{{yz\,dz}}\]

Keyin biz qilishimiz kerak bo'lgan yagona narsa, biz chiziqli integralni "standart" integralga aylantirganimizda \(dy = y'\,dt\) va \(dz = z'\,dt\) ni eslashdir.

Shunday qilib, chiziq integralini baholaylik. Faqat parametrlarni integralga "ulash" ni unutmang (i.e. integraldagi \(x\), \(y\) va \(z\) parametrlarini \(x\), \(y\) va \(z\) komponentlari bilan almashtiring) va differentsiallarni to'g'ri konvertatsiya qilish.

Mana chiziqli integrali.

\[\begin{align*}\int\limits_{C}{{{x^2}\,dy - yz\,dz}} & = \int\limits_{C}{{{x^2}\,dy}} - \int\limits_{C}{{yz\,dz}}\\ & = \int_{0}^{1}{{{{\left( {4 - 3t} \right)}^2}\left( 8 \right)\,dt}} - \int_{0}^{1}{{\left( { - 1 + 8t} \right)\,\,\left( {2 - 3t} \right)\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{8{{\left( {4 - 3t} \right)}^2} - 3\left( {24{t^2} - 19t + 2\,dt} \right)\,dt}}\\ & = \left. {\left[ { - \frac{8}{9}{{\left( {4 - 3t} \right)}^3} - 3\left( {8{t^3} - \frac{{19}}{2}t^{2} + 2t} \right)} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{109}}{2}}}\end{align*}\]

E'tibor bering, bu holda biz integralni haqiqiy baholashdan oldin ikkita integralni bitta integralga birlashtirdik. Buni qilish shart emas, lekin vaqti-vaqti bilan integralni va shuning uchun integralni baholashni soddalashtirishi mumkin.