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

\[\mathbf{1.\ }\]

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

\[f(x) = 2e^{2x} + 3x^{2} + \sin x;\]

\[F^{'}(x) =\]

\[= 2 \bullet e^{2x} + 3x^{2} - \left( - \sin x \right) =\]

\[= 2e^{2x} + 3x^{2} + \sin x.\]

\[Что\ и\ требовалось\ доказать.\]

\[\mathbf{2}\text{.\ }\]

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

\[F(x) = 3 \bullet \frac{x^{3}}{3} + 2 \bullet \frac{x^{2}}{2} - 3x + C =\]

\[= x^{3} + x^{2} - 3x + C;\]

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

\[C = - 1.\]

\[Ответ:\ \ \]

\[F(x) = x^{3} + x^{2} - 3x - 1.\]

\[\mathbf{3}\text{.\ }\]

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

\[= \frac{2}{3} \bullet 8 - \frac{2}{3} \bullet 1 = \frac{2}{3} \bullet 7 = \frac{14}{3};\]

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

\[= - \frac{1}{18} + \frac{1}{8} = \frac{5}{72};\]

\[3)\ \int_{0}^{\frac{\pi}{2}}{\cos{2x}\text{dx}} = \left. \ \frac{1}{2}\sin{2x} \right|_{0}^{\frac{\pi}{2}} =\]

\[= \frac{1}{2}\sin\pi - \frac{1}{2}\sin 0 = 0;\]

\[4)\ \int_{\frac{\pi}{2}}^{\pi}{\sin x\text{dx}} = \left. \ - \cos x \right|_{\frac{\pi}{2}}^{\pi} =\]

\[= - \cos\pi + \cos\frac{\pi}{2} = 1.\]

\[\mathbf{4}\text{.\ }\]

\[y = x^{2} + x - 6;\text{\ \ \ y} = 0:\]

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

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - 3;\]

\[x_{2} = \frac{- 1 + 5}{2} = 2;\]

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

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

\[= \left( \frac{8}{3} + \frac{4}{2} - 12 \right) - \left( - \frac{27}{3} + \frac{9}{2} + 18 \right) =\]

\[= \frac{8}{3} + 2 - 12 + 9 - \frac{9}{2} - 18 =\]

\[= - 20\frac{5}{6}.\]

\[Ответ:\ \ 20\frac{5}{6}.\]