0,001х⁶-1=(0,1х²-1)(0,01х²+0.1х+1)
Задача на уравнение касательной к графику функции. Решение см во вложении.
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Вычислить двойной интеграл
![\int\limits^2_0 {} \, dy \int\limits^1_0 {(x^2+2y)} \, dx](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E2_0+%7B%7D+%5C%2C+dy+%5Cint%5Climits%5E1_0+%7B%28x%5E2%2B2y%29%7D+%5C%2C+dx++)
Решение:
Найдем внутренний интеграл:
![\int\limits^1_0 {(x^2+2y)} \, dx =( \frac{x^3}{3}+2yx) \left. \right|_0^1=\frac{1^3}{3}+2y*1-\frac{0^3}{3}-2y*0=\frac{1}{3}+2y](https://tex.z-dn.net/?f=%5Cint%5Climits%5E1_0+%7B%28x%5E2%2B2y%29%7D+%5C%2C+dx+%3D%28+%5Cfrac%7Bx%5E3%7D%7B3%7D%2B2yx%29++%5Cleft.+%5Cright%7C_0%5E1%3D%5Cfrac%7B1%5E3%7D%7B3%7D%2B2y%2A1-%5Cfrac%7B0%5E3%7D%7B3%7D-2y%2A0%3D%5Cfrac%7B1%7D%7B3%7D%2B2y)
Результат подставим во внешний интеграл
![\int\limits^2_0 { (\frac{1}{3}+2y) } \, dy =( \frac{1}{3}y+y^2)\left. \right|_0^2= \frac{2}{3}+2^2-\frac{0}{3}-0^2= \frac{2}{3}+4=4 \frac{2}{3}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E2_0+%7B+%28%5Cfrac%7B1%7D%7B3%7D%2B2y%29+%7D+%5C%2C+dy+%3D%28+%5Cfrac%7B1%7D%7B3%7Dy%2By%5E2%29%5Cleft.+%5Cright%7C_0%5E2%3D+%5Cfrac%7B2%7D%7B3%7D%2B2%5E2-%5Cfrac%7B0%7D%7B3%7D-0%5E2%3D+%5Cfrac%7B2%7D%7B3%7D%2B4%3D4+%5Cfrac%7B2%7D%7B3%7D++)
Ответ:
![4 \frac{2}{3}](https://tex.z-dn.net/?f=4+%5Cfrac%7B2%7D%7B3%7D+)