Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 59

\[\boxed{\mathbf{59}.}\]

\[\textbf{а)}\ f(x) = 2x + 4;\]

\[y = 6x + 3;\ \ y = 0;\]

\[f(x) = 2x + 4:\]

\[F(x) = x^{2} + 4x + c;\]

\[c - некоторое\ число.\]

\[x^{2} + 4x + c = 6x + 3\]

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

\[D_{1} = 1 - (c - 3) = 0\]

\[1 - c + 3 = 0\]

\[c = 4;\]

\[Тогда:\]

\[F(x) = x^{2} + 4x + 4 = (x + 2)^{2}.\]

\[x^{2} + 4x + 4 = 6x + 3\]

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

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

\[x = 1.\]

\[6x + 3 = 0\]

\[x = - \frac{1}{2}.\]

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

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

\[= \left( \frac{1}{3} + 2 + 4 \right) - \left( - \frac{8}{3} + 8 - 8 \right) =\]

\[= 6\frac{1}{3} + 2\frac{2}{3} = 9\ кв.\ ед.\]

\[S_{2} = \frac{\frac{1}{2} + 1}{2} \cdot 9 = 6,75\ кв.\ ед.\]

\[S = S_{1} - S_{2} = 9 - 6,75 =\]

\[= 2,25\ кв.\ ед.\]

\[Ответ:2,25\ кв.\ ед.\]

\[\textbf{б)}\ f(x) = 2x - 2;\]

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

\[f(x) = 2x - 2:\]

\[F(x) = x^{2} - 2x + c;\]

\[c - некоторое\ число.\]

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

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

\[D_{1} = 1 - c = 0\]

\[1 - c = 0\]

\[c = 1.\]

\[Тогда:\]

\[F(x) = x^{2} - 2x + 1 = (x - 1)^{2};\]

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

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

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

\[x = - 1.\]

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

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

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

\[Ответ:\frac{2}{3}\ кв.\ ед.\]