ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 58

\[\boxed{\mathbf{58}.}\]

\[\textbf{а)}\ y - x^{2} = 0;\]

\[y = x^{2};\]

\[F(x) = \frac{x^{3}}{3}.\]

\[y^{2} - x = 0;\]

\[y = \sqrt{x} = x^{\frac{1}{2}}\]

\[F(x) = \frac{2}{3}x^{\frac{3}{2}}.\]

\[x^{2} = \pm \sqrt{x}\]

\[x^{4} = x\]

\[x = 0;\ \ x = 1.\]

\[\int_{0}^{1}{\sqrt{x}\text{dx}} = F(1) - F(0) =\]

\[= \frac{2}{3} \cdot 1 - \frac{2}{3} \cdot 0 = \frac{2}{3};\]

\[\int_{0}^{1}{x^{2}\text{dx}} = F(1) - F(0) =\]

\[= \frac{1}{3} - \frac{0}{3} = \frac{1}{3};\]

\[S = \int_{0}^{1}{\sqrt{x}\text{dx}} - \int_{0}^{1}{x^{2}\text{dx}} =\]

\[= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\ кв.\ ед.\]

\[\textbf{б)}\ y - x^{2} = 0;\]

\[y = x^{2};\]

\[F(x) = \frac{x^{3}}{3}.\]

\[y^{2} + x = 0;\]

\[y = \sqrt{- x} = - x^{\frac{1}{2}}\]

\[F(x) = - \frac{2}{3}x^{\frac{3}{2}}.\]

\[x^{2} = \pm \sqrt{- x}\]

\[x^{4} = x\]

\[x = 0;\ \ x = - 1.\]

\[\int_{- 1}^{0}{\sqrt{- x}\text{dx}} = F(0) - F( - 1) =\]

\[= 0 - \left( - \frac{2}{3} \cdot 1 \right) = \frac{2}{3};\]

\[\int_{- 1}^{0}{x^{2}\text{dx}} = F(0) - F( - 1) =\]

\[= 0 - \left( - \frac{1}{3} \right) = \frac{1}{3};\]

\[S = \int_{- 1}^{0}{\sqrt{- x}\text{dx}} - \int_{- 1}^{0}{x^{2}\text{dx}} =\]

\[= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\ кв.\ ед.\]

\[\textbf{в)}\ y = (1 - x)(x - 5);\ \ y = 4;\]

\[x = 1;\]

\[(1 - x)(x - 5) = 4\]

\[x - x^{2} + 5x - 5 - 4 = 0\]

\[- x^{2} + 6x - 9 = 0\]

\[x^{2} - 6x + 9 = 0\]

\[(x - 3)^{2} = 0\]

\[x = 3.\]

\[y = - x^{2} + 6x - 5;\]

\[F(x) = - \frac{x^{3}}{3} + \frac{6x^{2}}{2} - 5x =\]

\[= - \frac{x^{3}}{3} + 3x^{2} - 5x;\]

\[S_{прям} = 4 \cdot 2 = 8\ кв.\ ед.\]

\[S_{2} = \int_{1}^{3}{\left( - x^{2} + 6x - 5 \right)\text{dx}} =\]

\[= F(3) - F(1) =\]

\[= - 9 + 27 - 15 + 2\frac{1}{3} =\]

\[= 5\frac{1}{3}\ кв.\ ед.\]

\[S = S_{прям} - S_{2} = 8 - 5\frac{1}{3} =\]

\[= 2\frac{2}{3}\ кв.\ ед.\]

\[\textbf{г)}\ y = (x + 1)(3 - x);\ \ y = 4;\ \ \]

\[x = 3;\]

\[(x + 1)(3 - x) = 4\]

\[3x + 3 - x^{2} - x - 4 = 0\]

\[- x^{2} + 2x - 1 = 0\]

\[x^{2} - 2x + 1 = 0\]

\[(x - 1)^{2} = 0\]

\[x = 1.\]

\[S_{прям} = 4 \cdot 2 = 8\ кв.\ ед.\]

\[S_{2} = \int_{1}^{3}{\left( - x^{2} + 2x + 3 \right)\text{dx}} =\]

\[= F(3) - F(1) =\]

\[= 9 - 3\frac{2}{3} = 5\frac{1}{3}\ кв.\ ед.\]

\[S = S_{прям} - S_{2} = 8 - 5\frac{1}{3} =\]

\[= 2\frac{2}{3}\ кв.\ ед.\]