Haqiqatan ham bu muammo bilan bu erda ko'p narsa qilish mumkin emas. Biz qilishimiz kerak bo'lgan narsa - vektorning barcha komponentlarini birlashtirish.

\[\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \left\langle {\int{{6\,dt}},\int{{6{t^2} - 4t\,dt}},\int{{t{{\bf{e}}^{2t}}\,dt}}} \right\rangle \\ & = \left\langle {\int{{6\,dt}},\int{{6{t^2} - 4t\,dt}},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{2}\int{{{{\bf{e}}^{2t}}\,dt}}} \right\rangle = \left\langle {6t,2{t^3} - 2{t^2},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{4}{{\bf{e}}^{2t}}} \right\rangle \end{align*}\]

Asosiy integratsiya qoidalarini unutmang, masalan, qismlar bo'yicha integratsiya (uchinchi muddat). Shuni ham esda tutingki, aniq integralni bajarishning bir usuli avval noaniq integralni bajarish va keyin baholashdir.

Bu muammoning javobi shundan iboratki,

\[\begin{align*}\int_{{ - 1}}^{2}{{\vec r\left( t \right)\,dt}} & = \left. {\left\langle {6t,2{t^3} - 2{t^2},\frac{1}{2}t{{\bf{e}}^{2t}} - \frac{1}{4}{{\bf{e}}^{2t}}} \right\rangle } \right|_{ - 1}^2\\ & = \left\langle {12,8,\frac{3}{4}{{\bf{e}}^4}} \right\rangle - \left\langle { - 6, - 4, - \frac{3}{4}{{\bf{e}}^{ - 2}}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {18,12,\frac{3}{4}\left( {{{\bf{e}}^4} + {{\bf{e}}^{ - 2}}} \right)} \right\rangle }}\end{align*}\]

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.