ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 365

\[x = a;\ x = b;\ y = f(x).\]

\[1)\ a = 3,\ \ \ b = 4,\ \ \ f(x) = x^{2}:\]

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

\[= \frac{64 - 27}{3} = \frac{37}{3} = 12\frac{1}{3}.\]

\[Ответ:\ \ 12\frac{1}{3}.\]

\[2)\ a = 0,\ \ \ b = 2,\ \ \ f(x) = x^{3} + 1:\]

\[S = \int_{0}^{2}{\left( x^{3} + 1 \right)\text{dx}} = \left. \ \left( \frac{x^{4}}{4} + x \right) \right|_{0}^{2} =\]

\[= \frac{2^{4}}{4} + 2 = \frac{16}{4} + 2 = 6.\]

\[Ответ:\ \ 6.\]

\[3)\ a = 1,\ \ \ b = 8,\ \ \ f(x) = \sqrt[3]{x}:\]

\[S = \int_{1}^{8}{x^{\frac{1}{3}}\text{\ dx}} = \left. \ \frac{3}{4}x^{\frac{4}{3}} \right|_{1}^{8} =\]

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

\[= \frac{3}{4}(16 - 1) = \frac{3}{4} \bullet 15 = \frac{45}{4} =\]

\[= 11\frac{1}{4}.\]

\[Ответ:\ \ 11\frac{1}{4}.\]

\[4)\ a = 4,\ \ \ b = 9,\ \ \ f(x) = \sqrt{x}:\]

\[S = \int_{4}^{9}{x^{\frac{1}{2}}\text{\ dx}} = \left. \ \frac{2}{3}x^{\frac{3}{2}} \right|_{4}^{9} =\]

\[= \frac{2}{3} \bullet 9^{\frac{3}{2}} - \frac{2}{3} \bullet 4^{\frac{3}{2}} = \frac{2}{3}\left( 3^{3} - 2^{3} \right) =\]

\[= \frac{2}{3}(27 - 8) = \frac{2}{3} \bullet 19 =\]

\[= \frac{38}{3} = 12\frac{2}{3}.\]

\[Ответ:\ \ 12\frac{2}{3}.\]

\[5)\ a = \frac{\pi}{3},\ \ \ b = \frac{2\pi}{3},\ \ \ f(x) = \sin x:\]

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

\[= - \cos\frac{2\pi}{3} + \cos\frac{\pi}{3} =\]

\[= - \left( - \frac{1}{2} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1.\]

\[Ответ:\ \ 1.\]

\[6)\ a = - \frac{\pi}{6};\ b = 0;f(x) = \cos x:\]

\[S = \int_{- \frac{\pi}{6}}^{0}{\cos x\text{dx}} = \left. \ \sin x \right|_{- \frac{\pi}{6}}^{0} =\]

\[= \sin 0 - \sin\left( - \frac{\pi}{6} \right) = \frac{1}{2}.\]

\[Ответ:\ \ \frac{1}{2}.\]