Упр.21.14 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

Задача

21.14. Вычислите интеграл:

а) $$\int_{0}^{4}\sqrt{x(x+1)}\,dx$$;

в) $$\int_{\frac{2}{3}}^{11}5\sqrt[5]{3x-1}\,dx$$;

б) $$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx$$;

г) $$\int_{2}^{3}(5x-7)^{-\frac{2}{3}}\,dx$$.

Подробный ответ

а) $$\int_0^4 \sqrt{x}(x+1)\,dx=\int_0^4 \left(x^{3/2}+x^{1/2}\right)\,dx$$

$$=\left(\frac{2}{5}x^{5/2}+\frac{2}{3}x^{3/2}\right)\Bigg|_0^4$$

$$=\frac{2}{5}\cdot 4^{5/2}+\frac{2}{3}\cdot 4^{3/2}=\frac{2}{5}\cdot 32+\frac{2}{3}\cdot 8=\frac{64}{5}+\frac{16}{3}=\frac{272}{15}=18\frac{2}{15}.$$

б) $$\int_{-1}^0 \sqrt[3]{1-2x}\,dx=\int_{-1}^0 (1-2x)^{1/3}\,dx$$

Положим $$u=1-2x,\quad du=-2\,dx,\quad dx=-\frac{1}{2}du.$$ Тогда

$$\int_{-1}^0 (1-2x)^{1/3}\,dx=-\frac12\int_3^1 u^{1/3}\,du=\frac12\int_1^3 u^{1/3}\,du.$$

$$=\frac12\cdot \frac{3}{4}u^{4/3}\Bigg|_1^3=\frac{3}{8}\left(3^{4/3}-1\right)=\frac{3}{8}\left(3\sqrt[3]{3}-1\right).$$

в) $$\int_{2/3}^{11} 5\sqrt[5]{3x-1}\,dx=\int_{2/3}^{11} 5(3x-1)^{1/5}\,dx$$

Положим $$u=3x-1,\quad du=3\,dx,\quad dx=\frac{1}{3}du.$$ Тогда

$$\int_{2/3}^{11} 5(3x-1)^{1/5}\,dx=\frac{5}{3}\int_1^{32} u^{1/5}\,du.$$

$$=\frac{5}{3}\cdot \frac{5}{6}u^{6/5}\Bigg|_1^{32}=\frac{25}{18}\left(32^{6/5}-1\right).$$

Так как $$32^{6/5}=(2^5)^{6/5}=2^6=64,$$ получаем

$$\frac{25}{18}(64-1)=\frac{25}{18}\cdot 63=\frac{175}{2}=87{,}5.$$

г) $$\int_2^3 (5x-7)^{-2/3}\,dx$$

Положим $$u=5x-7,\quad du=5\,dx,\quad dx=\frac{1}{5}du.$$ Тогда