ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 68

\[\boxed{\mathbf{68}.}\]

\[\textbf{а)}\ y = \frac{x^{2}}{4};\ \ y = 3 - \frac{x^{2}}{2};\]

\[x \in \lbrack - 2;2\rbrack.\]

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

\[S_{2} = \int_{- 2}^{2}{\frac{x^{2}}{4}\text{dx}};\]

\[S =\]

\[= \int_{- 2}^{2}{\left( 3 - \frac{x^{2}}{2} \right)\text{dx}} - \int_{- 2}^{2}{\frac{x^{2}}{4}\text{dx}} =\]

\[= \int_{- 2}^{2}{\left( 3 - \frac{x^{2}}{2} - \frac{x^{2}}{4} \right)\text{dx}} =\]

\[= \int_{- 2}^{2}{\left( 3 - \frac{3x^{2}}{4} \right)\text{dx}} =\]

\[= 2\int_{0}^{2}{\left( 3 - \frac{3x^{2}}{4} \right)\text{dx}} =\]

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

\[= 2 \cdot 4 = 8\ кв.\ ед.\]

\[Ответ:8\ кв.\ ед.\]

\[\textbf{б)}\ y = x^{2} - 6x + 10;\]

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

\[x \in \lbrack 1;5\rbrack.\]

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

\[S_{2} = \int_{1}^{5}{\left( x^{2} - 6x + 10 \right)\text{dx}};\]

\[= \int_{1}^{5}{\left( - 2x^{2} + 12x - 10 \right)\text{dx}} =\]

\[= {(\left. \ - \frac{2x^{3}}{3} + 6x^{2} - 10x) \right|}_{1}^{5} =\]

\[16\frac{2}{3} + 4\frac{2}{3} = 21\frac{1}{3}\ кв.\ ед.\]

\[Ответ:21\frac{1}{3}\ кв.\ ед.\]