Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 69

\[\boxed{\mathbf{69}.}\]

\[\textbf{а)}\ y = x^{2} - 5;\]

\[y = - 0,5x^{2} + 1;\]

\[x^{2} - 5 = - 0,5x^{2} + 1\]

\[1,5x^{2} = 6\]

\[x^{2} = 4\]

\[x = \pm 2;\]

\[y( - 2) = - 1;\]

\[y(2) = - 1.\]

\[= - \left. \ \frac{1,5}{3}x^{2} + 6x \right|_{- 2}^{3} =\]

\[= - 0,5 \cdot 8 + 12 - 0,5 \cdot 8 + 12 =\]

\[= - 4 + 24 - 4 = 16\ кв.\ ед.\]

\[Ответ:16\ кв.ед.\]

\[\textbf{б)}\ y = x^{2} - 4x + 1;\]

\[y = - 2x^{2} + 8x + 1;\]

\[x^{2} - 4x + 1 = - 2x^{2} + 8x + 1\]

\[3x^{2} - 12x = 0\]

\[3x(x - 4) = 0\]

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

\[y(0) = 1;\]

\[y(4) = 16 - 16 + 1 = 1.\]

\[= - \left. \ \frac{3}{3}x^{3} + \frac{12}{2}x^{2} \right|_{0}^{4} =\]

\[= ( - 1 \cdot 64 + 6 \cdot 16) - (1 \cdot 0 - 0) =\]

\[= - 64 + 96 = 32\ ед.\ кв.\]

\[Ответ:32\ ед.\ кв.\]