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МатематикаРешебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 58
Задание 58
\[\boxed{\mathbf{58}.}\]
\[\textbf{а)}\ y - x^{2} = 0;\]
\[y = x^{2};\]
\[F(x) = \frac{x^{3}}{3}.\]
\[y^{2} - x = 0;\]
\[y = \sqrt{x} = x^{\frac{1}{2}}\]
\[F(x) = \frac{2}{3}x^{\frac{3}{2}}.\]
\[x^{2} = \pm \sqrt{x}\]
\[x^{4} = x\]
\[x = 0;\ \ x = 1.\]
\[\int_{0}^{1}{\sqrt{x}\text{dx}} = F(1) - F(0) =\]
\[= \frac{2}{3} \cdot 1 - \frac{2}{3} \cdot 0 = \frac{2}{3};\]
\[\int_{0}^{1}{x^{2}\text{dx}} = F(1) - F(0) =\]
\[= \frac{1}{3} - \frac{0}{3} = \frac{1}{3};\]
\[S = \int_{0}^{1}{\sqrt{x}\text{dx}} - \int_{0}^{1}{x^{2}\text{dx}} =\]
\[= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\ кв.\ ед.\]
\[\textbf{б)}\ y - x^{2} = 0;\]
\[y = x^{2};\]
\[F(x) = \frac{x^{3}}{3}.\]
\[y^{2} + x = 0;\]
\[y = \sqrt{- x} = - x^{\frac{1}{2}}\]
\[F(x) = - \frac{2}{3}x^{\frac{3}{2}}.\]
\[x^{2} = \pm \sqrt{- x}\]
\[x^{4} = x\]
\[x = 0;\ \ x = - 1.\]
\[\int_{- 1}^{0}{\sqrt{- x}\text{dx}} = F(0) - F( - 1) =\]
\[= 0 - \left( - \frac{2}{3} \cdot 1 \right) = \frac{2}{3};\]
\[\int_{- 1}^{0}{x^{2}\text{dx}} = F(0) - F( - 1) =\]
\[= 0 - \left( - \frac{1}{3} \right) = \frac{1}{3};\]
\[S = \int_{- 1}^{0}{\sqrt{- x}\text{dx}} - \int_{- 1}^{0}{x^{2}\text{dx}} =\]
\[= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\ кв.\ ед.\]
\[\textbf{в)}\ y = (1 - x)(x - 5);\ \ y = 4;\]
\[x = 1;\]
\[(1 - x)(x - 5) = 4\]
\[x - x^{2} + 5x - 5 - 4 = 0\]
\[- x^{2} + 6x - 9 = 0\]
\[x^{2} - 6x + 9 = 0\]
\[(x - 3)^{2} = 0\]
\[x = 3.\]
\[y = - x^{2} + 6x - 5;\]
\[F(x) = - \frac{x^{3}}{3} + \frac{6x^{2}}{2} - 5x =\]
\[= - \frac{x^{3}}{3} + 3x^{2} - 5x;\]
\[S_{прям} = 4 \cdot 2 = 8\ кв.\ ед.\]
\[S_{2} = \int_{1}^{3}{\left( - x^{2} + 6x - 5 \right)\text{dx}} =\]
\[= F(3) - F(1) =\]
\[= - 9 + 27 - 15 + 2\frac{1}{3} =\]
\[= 5\frac{1}{3}\ кв.\ ед.\]
\[S = S_{прям} - S_{2} = 8 - 5\frac{1}{3} =\]
\[= 2\frac{2}{3}\ кв.\ ед.\]
\[\textbf{г)}\ y = (x + 1)(3 - x);\ \ y = 4;\ \ \]
\[x = 3;\]
\[(x + 1)(3 - x) = 4\]
\[3x + 3 - x^{2} - x - 4 = 0\]
\[- x^{2} + 2x - 1 = 0\]
\[x^{2} - 2x + 1 = 0\]
\[(x - 1)^{2} = 0\]
\[x = 1.\]
\[S_{прям} = 4 \cdot 2 = 8\ кв.\ ед.\]
\[S_{2} = \int_{1}^{3}{\left( - x^{2} + 2x + 3 \right)\text{dx}} =\]
\[= F(3) - F(1) =\]
\[= 9 - 3\frac{2}{3} = 5\frac{1}{3}\ кв.\ ед.\]
\[S = S_{прям} - S_{2} = 8 - 5\frac{1}{3} =\]
\[= 2\frac{2}{3}\ кв.\ ед.\]
