Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 34

\[\boxed{\mathbf{34}.}\]

\[\textbf{а)}\ y = \sqrt{1 - x^{2}};\ \ \lbrack - 1;1\rbrack;\]

\[\int_{- 1}^{1}{\sqrt{1 - x^{2}}\text{dx}} = \frac{1}{2}S_{круга} =\]

\[= \frac{1}{2}\pi R^{2} = \frac{1}{2}\pi \cdot 1 = \frac{\pi}{2}.\]

\[\textbf{б)}\ y = - \sqrt{1 - x^{2}};\ \ \lbrack - 1;1\rbrack;\]

\[\int_{- 1}^{1}{- \sqrt{1 - x^{2}}\text{dx}} = - \frac{1}{2}S_{круга} =\]

\[= - \frac{1}{2}\pi R^{2} = - \frac{1}{2}\pi \cdot 1 = - \frac{\pi}{2}.\]

\[\textbf{в)}\ y = \sqrt{9 - x^{2}};\ \ \lbrack - 3;0\rbrack;\]

\[\int_{- 3}^{0}{\sqrt{9 - x^{2}}\text{dx}} = \frac{1}{4}S_{круга} =\]

\[= \frac{1}{4}\pi R^{2} = \frac{1}{4}\pi \cdot 3^{2} = \frac{9\pi}{4}.\]

\[\textbf{г)}\ y = - \sqrt{16 - x^{2}};\ \ \lbrack 0;4\rbrack\]

\[\int_{0}^{4}{- \sqrt{16 - x^{2}}\text{dx}} = - \frac{1}{4}S_{круга} =\]

\[= - \frac{1}{4}\pi R^{2} = - \frac{1}{4}\pi \cdot 4^{2} = - 4\pi.\]