Question

Помогите с алгеброй, пожалуйста

Answer 1

$1)\; \; f(x)=\sqrt[4]{x}-x\, \sqrt{x}\\\\F(x)=\int \Big(\sqrt[4]{x}-x\, \sqrt{x}\Big)\, dx=\int \Big(x^{1/4}-x^{3/2}\Big)\, dx=\frac{x^{5/4}}{5/4}-\frac{x^{5/2}}{5/2}+C=\\\\=\frac{4\sqrt[4]{x^5}}{5}-\frac{2\sqrt{x^5}}{5}+C\; ;\\\\M(16;21):\; \; 21=\frac{4\sqrt[4]{16^5}}{5}-\frac{2\sqrt{16^5}}{5}+C\; \; ,\; \; 21=\frac{4\cdot 2^5}{5}-\frac{2\cdot 4^5}{5}+C\; \; ,\\\\21=25,6-409,6+C\; \; \; \; \Rightarrow \; \; C=405\\\\F(x)\Big|_{M}=\frac{4\sqrt[4]{16^5}}{5}-\frac{2\sqrt{16^5}}{5}+405\; .$

$2)\; \; y=\sqrt{x-2}\; \; ,\; \; x=11\; ,\; y=2\\\\\sqrt{x-2}=2\; \; \to \; \; \; x-2=4\; ,\; \; x=6\\\\\int \limits _6^{11}\, \sqrt{x-2}\, dx=\frac{(x-2)^{3/2}}{3/2}\Big|_6^{11}=\frac{2}{3}\cdot \Big(9^{3/2}-4^{3/2}\Big)=\frac{2}{3}\cdot (3^3-2^3)=\\\\=\frac{2}{3}\cdot (27-8)=\frac{38}{3}=12\frac{2}{3}$