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МатематикаРешебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 55
Задание 55
\[\boxed{\mathbf{55}.}\]
\[\textbf{а)}\ y = x^{2} - 2x + 3;\]
\[y = 3 + 2x;\]
\[x^{2} - 2x + 3 = 3 + 2x\]
\[x^{2} - 4x = 0\]
\[x(x - 4) = 0\]
\[x = 0;\ \ x = 4.\]
\[S_{1} = \frac{(3 + 11)}{2} \cdot 4 = 28\ кв.\ ед.\]
\[S_{2} = \int_{0}^{4}{\left( x^{2} - 2x + 3 \right)\text{dx}} =\]
\[= \left. \ \frac{x^{3}}{3} - x^{2} + 3x \right|_{0}^{4} =\]
\[= \frac{4^{3}}{3} - 4^{2} + 12 - 0 =\]
\[= 17\frac{1}{3}\ кв.\ ед.\]
\[S = S_{1} - S_{2} = 28 - 17\frac{1}{3} =\]
\[= 10\frac{2}{3}\ кв.\ ед.\]
\[Ответ:10\frac{2}{3}\ кв.\ ед.\]
\[\textbf{б)}\ y = x^{2} + 2x + 5;\]
\[y = 5 - 2x;\]
\[x^{2} + 2x + 5 = 5 - 2x\]
\[x^{2} + 4x = 0\]
\[x(x + 4) = 0\]
\[x = 0;\ \ x = - 4.\]
\[S_{1} = \frac{13 + 5}{2} \cdot 4 = 36\ кв.\ ед.\]
\[S_{2} = \int_{- 4}^{0}{\left( x^{2} + 2x + 5 \right)\text{dx}} =\]
\[= \left. \ \frac{x^{3}}{3} + x^{2} + 5x \right|_{- 4}^{0} =\]
\[= \frac{64}{3} - 16 + 20 = 25\frac{1}{3}\ кв.\ ед.\]
\[S = S_{1} - S_{2} = 36 - 25\frac{1}{3} =\]
\[= 10\frac{2}{3}\ кв.\ ед.\]
\[Ответ:10\frac{2}{3}\ кв.\ ед.\]
\[\textbf{в)}\ y = x^{2} + 2x + 4;\]
\[y = 4 - 2x;\]
\[x^{2} + 2x + 4 = 4 - 2x\]
\[x^{2} + 4x = 0\]
\[x(x + 4) = 0\]
\[x = 0;\ \ x = - 4.\]
\[S_{1} = \frac{12 + 4}{2} \cdot 4 = 32\ кв.\ ед.\]
\[S_{2} = \int_{- 4}^{0}{\left( x^{2} + 2x + 4 \right)\text{dx}} =\]
\[= \left. \ \frac{x^{3}}{3} + x^{2} + 4x \right|_{- 4}^{0} =\]
\[= \frac{64}{3} - 16 + 16 = 21\frac{1}{3}\ кв.\ ед.\]
\[S = S_{1} - S_{2} = 32 - 21\frac{1}{3} =\]
\[= 10\frac{2}{3}\ кв.\ ед.\]
\[Ответ:10\frac{2}{3}\ кв.\ ед.\]
\[\textbf{г)}\ y = x^{2} - 2x + 6;\]
\[y = 2x + 6;\]
\[x^{2} - 2x + 6 = 2x + 6\]
\[x^{2} - 4x = 0\]
\[x(x - 4) = 0\]
\[x = 0;\ \ x = 4.\]
\[S_{1} = \frac{14 + 6}{2} \cdot 4 = 40\ кв.\ ед.\]
\[S_{2} = \int_{0}^{4}{\left( x^{2} - 2x + 6 \right)\text{dx}} =\]
\[= \left. \ \frac{x^{3}}{3} - x^{2} + 6x \right|_{0}^{4} =\]
\[= \left( \frac{4^{3}}{3} - 4^{2} + 6 \cdot 4 \right) - 0 =\]
\[= \frac{64}{3} - 16 + 24 = 29\frac{1}{3}\ кв.\ ед.\]
\[S = S_{1} - S_{2} = 40 - 29\frac{1}{3} =\]
\[= 10\frac{2}{3}\ кв.\ ед.\]
\[Ответ:10\frac{2}{3}\ кв.\ ед.\]
