Страница 424 номер 1539, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Найти площадь фигуры, ограниченной линиями (1538—1542). 1) \(y = 4x - x^2\), \(y = 5\), \(x = 0\), \(x = 3\); 2) \(y = x^2 - 2x + 8\), \(y = 6\), \(x = -1\), \(x = 3\); 3) \(y = \sin x\), \(y = 0\), \(x = \frac{2\pi}{3}\), \(x = \pi\); 4) \(y = \cos x\), \(y = 0\), \(x = -\frac{\pi}{6}\), \(x = \frac{\pi}{6}\).
1) \(y = 4x - x^{2}\) – синий график; \(y = 5\) – красный график;
\(x = 0, x = 3;\)
\(S = 5 \cdot 3 - \int_{0}^{3}{\left( 4x - x^{2} \right)dx} =\)
\(= 15 - \left. \ \left( 2x^{2} - \dfrac{x^{3}}{3} \right) \right|_{0}^{3} =\)
\(= 15 - 18 + 9 = 6;\)
2) \(y = x^{2} - 2x + 8\) – синий график;
\(y = 6\) – красный график;
\(x = - 1, x = 3;\)
\(S = \int_{-1}^{3}{\left( x^{2} - 2x + 8 \right)dx} - 24 =\)
\(= \left. \ \left( \dfrac{x^{3}}{3} - x^{2} + 8x \right) \right|_{-1}^{3} - 24 =\)
\(= (9 - 9 + 24) -\)
\(- \left(-\dfrac{1}{3} - 1 - 8 \right) - 24 = 9\dfrac{1}{3};\)
3) \(y = \sin x\) – синий график;
\(y = 0\) – красный график;
\(x = \dfrac{2\pi}{3},x = \pi;\)
\(S = \int_{\frac{2\pi}{3}}^{\pi}{\sin x\,dx} = \left. \ - \cos x \right|_{\frac{2\pi}{3}}^{\pi} =\)
\(= 1 - \dfrac{1}{2} = \dfrac{1}{2};\)
4) \(y = \cos x\) – синий график;
\(y = 0\) – красный график;
\(x = - \dfrac{\pi}{6}, x = \dfrac{\pi}{6};\)
\(S = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}{\cos xdx} =\)
\(= \left. \ \sin x \right|_{-\frac{\pi}{6}}^{\frac{\pi}{6}} = \dfrac{1}{2} + \dfrac{1}{2} = 1.\)
