Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 54

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\[\textbf{а)}\ y = x^{2} - 4x + 6;y = 6\]

\[x^{2} - 4x + 6 = 6\]

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

\[x(x - 4) = 0\]

\[x = 0;\ \ x = 4.\]

\[S_{1} = 6 \cdot 4 = 24\ кв.\ ед.\]

\[S_{2} = \int_{0}^{4}{\left( x^{2} - 4x + 6 \right)\text{dx}} =\]

\[= \left. \ \frac{x^{3}}{3} - 2x^{2} + 6x \right|_{0}^{4} =\]

\[= \frac{4^{3}}{3} - 32 + 24 = 13\frac{1}{3}\ кв.\ ед.\]

\[S = S_{1} - S_{2} = 24 - 13\frac{1}{3} =\]

\[= 10\frac{2}{3}\ кв.\ ед.\]

\[\textbf{б)}\ y = - x^{2} - 4x + 5;\ \ y = 5;\]

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

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

\[- x(x + 4) = 0\]

\[x = 0;\ \ x = - 4;\]

\[S_{1} = 5 \cdot 4 = 20\ ед.\ кв.\]

\[= 0 - \left( - \frac{( - 4)^{3}}{3} - 32 - 20 \right) =\]

\[= - 21\frac{1}{3} + 52 = 30\frac{2}{3}\ кв.ед.\]

\[S = S_{2} - S_{1} = 30\frac{2}{3} - 20 =\]

\[= 10\frac{2}{3}\ кв.ед.\]

\[\textbf{в)}\ y = x^{2} + 1;\ \ y = 3 - x;\]

\[x^{2} + 1 = 3 - x\]

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 2;\ \ x_{2} = 1;\]

\[S_{1} = \frac{2 + 5}{2} \cdot 3 = 10,5\ кв.\ ед.\]

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

\[= \left. \ \frac{x^{3}}{3} + x \right|_{- 2}^{1} =\]

\[= \frac{1}{3} + 1 - \left( \frac{( - 2)^{3}}{3} + ( - 2) \right) =\]

\[= 1 + \frac{1}{3} + \frac{8}{3} + 2 = 6\ кв.\ ед.\]

\[S = S_{1} - S_{2} = 10,5 - 6 =\]

\[= 4\ кв.\ ед.\]

\[\textbf{г)}\ y = 4 - x^{2};\ \ y = x + 2;\]

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

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]

\[x_{1} = - 2;\ \ x_{2} = 1;\]

\[S_{1} = \frac{1}{2} \cdot 3 \cdot 3 = 4,5\ кв.ед.\]

\[S_{2} = \int_{- 2}^{1}{\left( 4 - x^{2} \right)\text{dx}} =\]

\[= \left. \ 4x - \frac{x^{3}}{3} \right|_{- 2}^{1} =\]

\[= 4 - \frac{1}{3} - \left( - 8 + \frac{8}{3} \right) = 9\ кв.\ ед.\]

\[S = S_{2} - S_{1} = 9 - 4,5 =\]

\[= 4,5\ кв.\ ед.\]