16.5-bo'lim: Chiziqli integrallar uchun asosiy teorema

1. \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\)ni baholang, bu erda \(f\left( {x,y} \right) = {x^3}\left( {3 - {y^2}} \right) + 4y\) va C \(\vec r\left( t \right) = \left\langle {3 - {t^2},5 - t} \right\rangle \) tomonidan \( - 2 \le t \le 3\) bilan berilgan. Yechim
2. \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\)ni baholang, bu erda \(f\left( {x,y} \right) = y{{\bf{e}}^{{x^{\,2}} - 1}} + 4x\sqrt y \) va C \(\vec r\left( t \right) = \left\langle {1 - t,2{t^2} - 2t} \right\rangle \) tomonidan \(0 \le t \le 2\) bilan berilgan. Yechim
3. \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) yo'lini hisoblashdan mustaqil ekanligini hisobga olsak, bu erda C - \(\displaystyle \frac{{{{\left( {x - 5} \right)}^2}}}{4} + \frac{{{y^2}}}{9} = 1\) tomonidan soat miliga teskari aylanish bilan berilgan ellips. Yechim
4. \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\)-ni baholang, bu erda \(f\left( {x,y} \right) = {{\bf{e}}^{x\,y}} - {x^2} + {y^3}\) va C quyida ko'rsatilgan egri chiziqdir.
   
   Yechim