Bunday holda, parametrlash uchun faqat egri chiziq tenglamasidan foydalanishimiz mumkin, chunki belgilangan yo'nalish \(x\) ni oshirish yo'nalishida ketmoqda, bu bizga kerak bo'lganda kichikroq qiymatdan kattaroq qiymatgacha integral chegaralarni beradi. Mana bu egri chiziq uchun parametrlash.

\[\vec r\left( t \right) = \left\langle {t,{t^2}} \right\rangle \,\,\,\,\,\,\,\, - 1 \le t \le 2\]

Biz buni quyidagi kabi parametr shakliga ham ajratishimiz mumkin.

\[\begin{array}{*{20}{l}}{x = t}\\{y = {t^2}}\end{array}\hspace{0.25in} - 1 \le t \le 2\]

Parametrlashtirishning har qanday shakli muammo uchun ishlaydi, ammo biz ushbu muammoning qolgan qismi uchun vektor shaklidan foydalanamiz.

Bizga parametrlash hosilasining kattaligi kerak bo'ladi, shuning uchun uni olamiz.

\[\vec r'\left( t \right) = \left\langle {1,2t} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2t} \right)}^2}} = \sqrt {1 + 4{t^2}} \]

Bizga parametrlashda "baholangan" integral ham kerak bo'ladi. Bularning barchasini eslang, biz integranddagi \(x\)/\(y\)ni parametrlashdan \(x\)/\(y\) bilan almashtiramiz. Bu erda parametrlashda baholangan integral.

\[6x = 6t\]

U holda chiziqli integrali,

\[\int\limits_{C}{{6x\,ds}} = \int_{{ - 1}}^{2}{{6t\sqrt {1 + 4{t^2}} \,dt}} = \left. {\frac{1}{2}{{\left( {1 + 4{t^2}} \right)}^{\frac{3}{2}}}} \right|_{ - 1}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left( {{{17}^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right)}}\]