11.5. Найдите площадь криволинейной трапеции, ограниченной:

1. Параболой $$y = x^2 + 1$$ и прямыми $$y = 0$$, $$x = 0$$, $$x = 2$$. $$S = \int_{0}^{2} (x^2 + 1) dx = (\frac{x^3}{3} + x) \Big|_{0}^{2} = (\frac{2^3}{3} + 2) - (\frac{0^3}{3} + 0) = \frac{8}{3} + 2 = \frac{8 + 6}{3} = \frac{14}{3}$$
2. Косинусоидой $$y = cosx$$ и прямыми $$y = 0$$, $$x = -\frac{\pi}{6}$$, $$x = \frac{\pi}{2}$$. $$S = \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} cosx dx = sinx \Big|_{-\frac{\pi}{6}}^{\frac{\pi}{2}} = sin(\frac{\pi}{2}) - sin(-\frac{\pi}{6}) = 1 - (-\frac{1}{2}) = 1 + \frac{1}{2} = \frac{3}{2}$$
3. Графиком функции $$y = -x^3$$ и прямыми $$y = 0$$, $$x = -2$$. $$S = |\int_{-2}^{0} -x^3 dx| = |-\int_{-2}^{0} x^3 dx| = |-\frac{x^4}{4} \Big|_{-2}^{0}| = |-(\frac{0^4}{4} - \frac{(-2)^4}{4})| = |-(0 - \frac{16}{4})| = |-(-4)| = |4| = 4$$
4. Параболой $$y = 3 - 2x - x^2$$ и прямыми $$y = 0$$, $$x = -2$$, $$x = 0$$. $$S = \int_{-2}^{0} (3 - 2x - x^2) dx = (3x - x^2 - \frac{x^3}{3}) \Big|_{-2}^{0} = (3(0) - 0^2 - \frac{0^3}{3}) - (3(-2) - (-2)^2 - \frac{(-2)^3}{3}) = 0 - (-6 - 4 + \frac{8}{3}) = 10 - \frac{8}{3} = \frac{30 - 8}{3} = \frac{22}{3}$$
5. Гиперболой $$y = \frac{1}{2x}$$ и прямыми $$y = 0$$, $$x = \frac{1}{4}$$, $$x = 2$$. $$S = \int_{\frac{1}{4}}^{2} \frac{1}{2x} dx = \frac{1}{2} \int_{\frac{1}{4}}^{2} \frac{1}{x} dx = \frac{1}{2} ln|x| \Big|_{\frac{1}{4}}^{2} = \frac{1}{2} (ln(2) - ln(\frac{1}{4})) = \frac{1}{2} (ln(2) - ln(1) + ln(4)) = \frac{1}{2} (ln(2) + ln(4)) = \frac{1}{2} (ln(2) + ln(2^2)) = \frac{1}{2} (ln(2) + 2ln(2)) = \frac{1}{2} (3ln(2)) = \frac{3}{2} ln(2)$$
6. Параболой $$y = 2x - x^2$$ и осью абсцисс. Сначала найдём точки пересечения с осью абсцисс: $$2x - x^2 = 0$$, $$x(2 - x) = 0$$, $$x = 0$$ или $$x = 2$$. $$S = \int_{0}^{2} (2x - x^2) dx = (x^2 - \frac{x^3}{3}) \Big|_{0}^{2} = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3}) = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$$
7. Синусоидой $$y = sin 2x$$ и прямыми $$y = 0$$, $$x = \frac{\pi}{12}$$, $$x = \frac{\pi}{4}$$. $$S = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} sin 2x dx = -\frac{1}{2} cos 2x \Big|_{\frac{\pi}{12}}^{\frac{\pi}{4}} = -\frac{1}{2} (cos (2 \cdot \frac{\pi}{4}) - cos (2 \cdot \frac{\pi}{12})) = -\frac{1}{2} (cos (\frac{\pi}{2}) - cos (\frac{\pi}{6})) = -\frac{1}{2} (0 - \frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$$