ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1013

\[\boxed{\mathbf{1013}\mathbf{.}}\]

\[\textbf{а)}\ a = - 1,\ \ \ b = 1:\text{\ \ }\]

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

\[\int_{- 1}^{1}{\left( x^{2} + 4 \right)\text{\ dx}} =\]

\[= \left. \ \left( \frac{x^{3}}{3} + 4 \bullet \frac{x^{1}}{1} \right) \right|_{- 1}^{1} =\]

\[= \left. \ \left( \frac{x^{3}}{3} + 4x \right) \right|_{- 1}^{1} =\]

\[= \frac{1^{3}}{3} + 4 \bullet 1 - \frac{( - 1)^{3}}{3} - 4 \bullet ( - 1) =\]

\[= \frac{1}{3} + 4 + \frac{1}{3} + 4 = 8\frac{2}{3}\]

\[Ответ:\ \ 8\frac{2}{3}.\]

\[\textbf{б)}\ a = 0,\ \ \ b = 1:\text{\ \ }\]

\[y = \sqrt{x} + 1\]

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

\[= \left. \ \left( x^{\frac{3}{2}}\ :\frac{3}{2} + 1 \bullet \frac{x^{1}}{1} \right) \right|_{0}^{1} =\]

\[= \left. \ \left( \frac{2}{3}x\sqrt{x} + x \right) \right|_{0}^{1} =\]

\[= \frac{2}{3} \bullet 1\sqrt{1} + 1 - \frac{2}{3} \bullet 0\sqrt{0} - 0 =\]

\[= \frac{2}{3} + 1 = 1\frac{2}{3}\]

\[Ответ:\ \ 1\frac{2}{3}.\]

\[\textbf{в)}\ a = 1,\ \ \ b = 4:\ \ \]

\[y = \frac{2}{x}\]

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

\[= 2\ln 4 - 2\ln 1 = 2\ln 4 - 2 \bullet 0 =\]

\[= 2\ln 4\]

\[Ответ:\ \ 2\ln 4.\]