Упр.49.5 ГДЗ Мордкович 10-11 класс (Алгебра)

а) $$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx$$; б) $$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx$$; в) $$\int_{\frac{2}{3}}^{11}5\sqrt[5]{3x-1}\,dx$$; г) $$\int_{2}^{3}(5x-7)^{-\frac{2}{3}}\,dx$$.

а) $$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=\int_{-1}^{0}(1-2x)^{\frac13}\,dx.$$

Сделаем замену: $$u=1-2x,\quad du=-2\,dx,\quad dx=-\frac12\,du.$$ Тогда

$$\int_{-1}^{0}(1-2x)^{\frac13}\,dx=-\frac12\int_{3}^{1}u^{\frac13}\,du=-\frac12\cdot \frac{u^{\frac43}}{\frac43}\Bigg|_{3}^{1}=-\frac38\,u^{\frac43}\Bigg|_{3}^{1}.$$

$$-\frac38\left(1^{\frac43}-3^{\frac43}\right)=-\frac38\left(1-3\sqrt[3]{3}\right).$$

Ответ: $$-\frac38\left(1-3\sqrt[3]{3}\right).$$

б) $$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=\int_{4}^{5}(x-3)^{-3}\,dx.$$

Первообразная:

$$\int (x-3)^{-3}\,dx=\frac{(x-3)^{-2}}{-2}=-\frac{1}{2(x-3)^2}.$$

Тогда

$$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=-\frac{1}{2(x-3)^2}\Bigg|_{4}^{5}=-\frac{1}{2\cdot 2^2}+\frac{1}{2\cdot 1^2}=\frac38.$$

Ответ: $$\frac38.$$

в) $$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\int_{\frac23}^{11}5(3x-1)^{\frac15}\,dx.$$

Сделаем замену: $$u=3x-1,\quad du=3\,dx,\quad dx=\frac13\,du.$$ Тогда

$$\int_{\frac23}^{11}5(3x-1)^{\frac15}\,dx=\frac53\int_{1}^{32}u^{\frac15}\,du=\frac53\cdot \frac{u^{\frac65}}{\frac65}\Bigg|_{1}^{32}=\frac{25}{18}u^{\frac65}\Bigg|_{1}^{32}.$$

$$\frac{25}{18}\left(32^{\frac65}-1\right)=\frac{25}{18}\left(32\sqrt[5]{32}-1\right)=87{,}5.$$

Ответ: $$87{,}5.$$

г) $$\int_{2}^{3}(5x-7)^{-\frac23}\,dx.$$

Положим $$u=5x-7,\quad du=5\,dx,\quad dx=\frac15\,du.$$ Тогда

$$\int_{2}^{3}(5x-7)^{-\frac23}\,dx=\frac15\int_{3}^{8}u^{-\frac23}\,du=\frac15\cdot \frac{u^{\frac13}}{\frac13}\Bigg|_{3}^{8}=\frac35\,u^{\frac13}\Bigg|_{3}^{8}.$$

$$\frac35\left(8^{\frac13}-3^{\frac13}\right)=\frac35\left(2-\sqrt[3]{3}\right).$$

Ответ: $$\frac35\left(2-\sqrt[3]{3}\right).$$