Question

Даю 50 баллов. Решите пожалуйста срочно

Answer 1

Ответ:

$\displaystyle a)\ \ \int x^2\sqrt{2x^3+5}\, dx=\Big[\ t=2x^3+5\ ,\ dt=6x^2\, dx\ \Big]=\frac{1}{6}\int \sqrt{t}\, dt=\\\\\\=\frac{1}{6}\cdot \frac{t^{\frac{3}{2}}}{3/2}+C=\frac{2}{18}\sqrt{t^3} +C=\frac{1}{9}\, \sqrt{(2x^3+5)^3}+C\ ;\\\\\\\\b)\ \ S=\int\limits^{a}_{b}\, y(x)\, dx=\int \limits _{-1}^1x^2\sqrt{2x^3+5}\, dx=\frac{1}{9}\, \sqrt{(2x^3+5)^3}\Big|_{-1}^1=\\\\\\=\frac{1}{9}\cdot \Big(\sqrt{7^3}-\sqrt{3^3}\Big)=\frac{1}{9}\cdot (7\sqrt7-3\sqrt3)\ ;$

$\displaystyle c)\ \ V_{ox}=\pi \int \limits^{a}_{b}\, y^2(x)\, dx=\pi \int \limits _{-1}^1\, x^4\cdot (2x^3+5)\, dx=\pi \int \limits _{-1}^1\, (2x^7+5x^4)\, dx=\\\\\\=\pi \cdot \Big(\frac{2x^8}{8}+\frac{5x^5}{5}\Big)\Big|_{-1}^1=\pi \cdot \Big(\frac{x^8}{4}+x^5\Big)\Big|_{-1}^1=\pi \cdot \Big(\frac{1}{4}+1-\frac{1}{4}+1\Big)=2\pi$