![A=\int\limits_L \, Fdl=\int\limits_L \, 6xdx+6y^2dy](https://tex.z-dn.net/?f=+A%3D%5Cint%5Climits_L+%5C%2C+Fdl%3D%5Cint%5Climits_L+%5C%2C+6xdx%2B6y%5E2dy+)
По условию: x=sint; y=cost, тогда
![dx=costdt; \ \ dy=-sintdt](https://tex.z-dn.net/?f=+dx%3Dcostdt%3B+%5C+%5C+dy%3D-sintdt+)
Подставляем все известные данные в интеграл:
![A=\int\limits^{\frac{\pi}{2}}_0\, 6sint*costdt+6cos^2t*(-sint)dt= \\\\ =\int\limits^{\frac{\pi}{2}}_0 (6sint*cost-6cos^2t*sint)dt =\int\limits^{\frac{\pi}{2}}_0 6sint*cost \ dt \ - \\ \\ -\int\limits^{\frac{\pi}{2}}_0 6cos^2t*sint \ dt=6(\int\limits^{\frac{\pi}{2}}_0sint \ d(sint) \ +\int\limits^{\frac{\pi}{2}}_0cos^2t \ d(cost)= \\ \\ =6(\frac{sin^2t}{2} \ + \frac{cos^3t}{3} ) \ \ |^{\frac{\pi}{2}}_0=6(\frac{1}{2} -\frac{1}{3} )=3-2=1](https://tex.z-dn.net/?f=+A%3D%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0%5C%2C+6sint%2Acostdt%2B6cos%5E2t%2A%28-sint%29dt%3D++%5C%5C%5C%5C+%3D%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0+%286sint%2Acost-6cos%5E2t%2Asint%29dt+%3D%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0+6sint%2Acost+%5C+dt+%5C+-+%5C%5C+%5C%5C+-%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0+6cos%5E2t%2Asint+%5C+dt%3D6%28%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0sint+%5C+d%28sint%29+%5C+%2B%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0cos%5E2t+%5C+d%28cost%29%3D+%5C%5C+%5C%5C+%3D6%28%5Cfrac%7Bsin%5E2t%7D%7B2%7D+%5C+%2B+%5Cfrac%7Bcos%5E3t%7D%7B3%7D+%29+%5C+%5C+%7C%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0%3D6%28%5Cfrac%7B1%7D%7B2%7D+-%5Cfrac%7B1%7D%7B3%7D+%29%3D3-2%3D1+)
Ответ: 1
₊5678 _1972⊥58 Вот столбиком пойдёт?
9485 174 34
⁻⁻⁻⁻⁻⁻⁻ _232
15163 232
0
на 5 больше, значить нужно "+"
290*13-(29*50-17*20)= 1) 50*90=4500 2)20*17=340 3)4500-340=4160 4)290*13=3770 5) ???