Решебник по алгебре и начала математического анализа 11 класс Колягин Проверь себя IV. Сложный вариант

\[\mathbf{1}\text{.\ }\]

\[f(x) = e^{x} - 3\sin x;\ A(0;\ 2):\]

\[F(x) = e^{x} + 3\cos x + C;\]

\[F(0) = e^{0} + 3\cos 0 + C = 2;\]

\[F(0) = 1 + 3 + C = 2;\]

\[C = - 2.\]

\[Ответ:\ \ F(x) = e^{x} + 3\cos x - 2.\]

\[\mathbf{2}.\ \]

\[1)\ \int_{1}^{4}{\sqrt{x}\text{dx}} = \left. \ \frac{2}{3}x^{\frac{3}{2}} \right|_{1}^{4} =\]

\[= \frac{2}{3} \bullet 8 - \frac{2}{3} = \frac{2}{3} \bullet 7 = \frac{14}{3};\]

\[2)\ \int_{0}^{1}{\frac{2}{3x + 1}\text{dx}} =\]

\[= \left. \ \frac{2}{3}\ln|3x + 1| \right|_{0}^{1} = \frac{2}{3}\ln 4 = \frac{4}{3}\ln 2.\]

\[\mathbf{3.}\ \]

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

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

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

\[= 3 - \frac{7}{3} = \frac{2}{3}.\]

\[Ответ:\ \ \frac{2}{3}.\]

\[\mathbf{4}.\ \]

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

\[y = x^{2} - 2x + 2:\]

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

\[2x^{2} - 6x = 0\]

\[2x(x - 3) = 0\]

\[x_{1} = 0;\text{\ \ \ }x_{2} = 3.\]

\[S =\]

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

\[= \int_{0}^{3}{\left( 2x^{2} - 6x \right)\text{dx}} =\]

\[= \left. \ \left( 2 \bullet \frac{x^{3}}{3} - 6 \bullet \frac{x^{2}}{2} \right) \right|_{0}^{3} =\]

\[= \frac{2}{3} \bullet 27 - 3 \bullet 9 = 18 - 27 = - 9.\]

\[Ответ:\ \ 9.\]

\[\mathbf{5}.\]

\[\ y = x^{2} + 1;\text{\ \ \ }(0;\ - 3):\]

\[y^{'}(x) = 2x;\]

\[y(a) = a^{2} + 1;\]

\[y^{'}(a) = 2a;\]

\[y = a^{2} + 1 + 2a(x - a) =\]

\[= 1 + 2ax - a^{2};\]

\[- 3 = 1 + 2a \bullet 0 - a^{2}\]

\[a^{2} = 4\]

\[a = \pm 2.\]

\[y_{1} = 1 - 4x - 4 = - 4x - 3;\]

\[y_{2} = 1 + 4x - 4 = 4x - 3.\]

\[= \frac{16}{3} + 16 - 16 = \frac{16}{3}.\]

\[Ответ:\ \ \frac{16}{3}.\]