Упр.49.5 ГДЗ Мордкович 10-11 класс (Алгебра)
а) интеграл(-1 0) (3)корень(1- 2x) dx;
б) интеграл(4 5) 1 / (x — 3)^3 dx;
в) интеграл(2/3 11) 5(5)корень(Зх — 1) dx;
г) интеграл(2 3) (5х — 7)^-2/3 dx.
$$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=\int_{-1}^{0}(1-2x)^{\frac13}\,dx$$
Сделаем замену: $$u=1-2x,\quad du=-2\,dx,\quad dx=-\frac12\,du.$$ Тогда
$$\int (1-2x)^{\frac13}\,dx=-\frac12\int u^{\frac13}\,du=-\frac12\cdot \frac{u^{\frac43}}{\frac43}=-\frac38u^{\frac43}.$$
Подставим пределы:
$$\int_{-1}^{0}\sqrt[3]{1-2x}\,dx=-\frac38(1-2x)^{\frac43}\Big|_{-1}^{0}$$
$$=-\frac38\left(1^{\frac43}-3^{\frac43}\right)=-\frac38\left(1-3\sqrt[3]{3}\right).$$
$$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=\int_{4}^{5}(x-3)^{-3}\,dx$$
$$\int (x-3)^{-3}\,dx=\frac{(x-3)^{-2}}{-2}=-\frac{1}{2(x-3)^2}.$$
Тогда
$$\int_{4}^{5}\frac{1}{(x-3)^3}\,dx=-\frac{1}{2(x-3)^2}\Big|_{4}^{5}$$
$$=-\frac{1}{2\cdot 2^2}+\frac{1}{2\cdot 1^2}=\frac12-\frac18=\frac38.$$
$$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\int_{\frac23}^{11}5(3x-1)^{\frac15}\,dx$$
Сделаем замену: $$u=3x-1,\quad du=3\,dx,\quad dx=\frac13\,du.$$ Тогда
$$\int 5(3x-1)^{\frac15}\,dx=\frac53\int u^{\frac15}\,du=\frac53\cdot \frac{u^{\frac65}}{\frac65}=\frac{25}{18}u^{\frac65}.$$
Подставим пределы:
$$\int_{\frac23}^{11}5\sqrt[5]{3x-1}\,dx=\frac{25}{18}(3x-1)^{\frac65}\Big|_{\frac23}^{11}$$
$$=\frac{25}{18}\left(32^{\frac65}-1^{\frac65}\right)=\frac{25}{18}(32\sqrt[5]{32}-1)=87{,}5.$$
$$\int_{2}^{3}(5x-7)^{-\frac23}\,dx$$
Сделаем замену: $$u=5x-7,\quad du=5\,dx,\quad dx=\frac15\,du.$$ Тогда
$$\int (5x-7)^{-\frac23}\,dx=\frac15\int u^{-\frac23}\,du=\frac15\cdot \frac{u^{\frac13}}{\frac13}=\frac35u^{\frac13}.$$
Следовательно,
$$\int_{2}^{3}(5x-7)^{-\frac23}\,dx=\frac35(5x-7)^{\frac13}\Big|_{2}^{3}$$
$$=\frac35\left(\sqrt[3]{8}-\sqrt[3]{3}\right)=\frac35(2-\sqrt[3]{3}).$$
Ответ
а) $$-\frac38\left(1-3\sqrt[3]{3}\right);$$ б) $$\frac38;$$ в) $$87{,}5;$$ г) $$\frac35\left(2-\sqrt[3]{3}\right).$$
