12.7-bo'lim : Vektor funktsiyalari bilan hisoblash
9. \(\displaystyle \int{{\vec r\left( t \right)\,dt}}\) ni baholang, bu erda \(\vec r\left( t \right) = \left\langle {\left( {1 - t} \right)\cos \left( {{t^2} - 2t} \right),\cos \left( t \right)\sin \left( t \right),{{\sec }^2}\left( {4t} \right)} \right\rangle \).
Yechim
Haqiqatan ham bu muammo bilan bu erda ko'p narsa qilish mumkin emas. Biz qilishimiz kerak bo'lgan narsa - vektorning barcha komponentlarini birlashtirish.
\[\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \left\langle {\int{{\left( {1 - t} \right)\cos \left( {{t^2} - 2t} \right)\,dt}},\int{{\cos \left( t \right)\sin \left( t \right)\,dt}},\int{{{{\sec }^2}\left( {4t} \right)\,dt}}} \right\rangle \\ & = \left\langle {\int{{\left( {1 - t} \right)\cos \left( {{t^2} - 2t} \right)\,dt}},\int{{\frac{1}{2}\sin \left( {2t} \right)\,dt}},\int{{{{\sec }^2}\left( {4t} \right)\,dt}}} \right\rangle \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle { - \frac{1}{2}\sin \left( {{t^2} - 2t} \right), - \frac{1}{4}\cos \left( {2t} \right),\frac{1}{4}\tan \left( {4t} \right)} \right\rangle + \vec c}}\end{align*}\]O'zgartirish qoidasi (barcha shartlar) va ba'zi asosiy trig formulalari (yarim burchak va ikki burchak formulalari) kabi asosiy integratsiya qoidalarini unutmang, ba'zi integrallarni (ikkinchi muddat) bajarishingiz kerak.
Siz nima sodir bo'layotganini kuzatish uchun integratsiya qobiliyatingizni yaxshi bilasiz, degan taxminda biz ko'p integratsiya tafsilotlarini yechimga kiritmadik. Agar siz integratsiya qobiliyatingiz bilan biroz zanglagan bo'lsangiz, integratsiya qobiliyatingizni yangilash uchun Calculus I va Calculus II ning integratsiya materialiga qaytishingiz kerak bo'ladi.