1000) Найти площадь криволинейной трапеции, ограниченной прямыми х = а, х = b, осью Ох и графиком функции у = f (x):

Решение:

Площадь криволинейной трапеции вычисляется по формуле определённого интеграла:

\( S = \int_{a}^{b} f(x) dx \)

Варианты решений:

1. \( a = 2, b = 4, f(x) = x^3 \)

\( S = \int_{2}^{4} x^3 dx = \left[ \frac{x^4}{4} \right]_{2}^{4} = \frac{4^4}{4} - \frac{2^4}{4} = \frac{256}{4} - \frac{16}{4} = 64 - 4 = 60 \)

2. \( a = 3, b = 4, f(x) = x^2 \)

\( S = \int_{3}^{4} x^2 dx = \left[ \frac{x^3}{3} \right]_{3}^{4} = \frac{4^3}{3} - \frac{3^3}{3} = \frac{64}{3} - \frac{27}{3} = \frac{37}{3} \approx 12.33 \)

3. \( a = -2, b = 1, f(x) = x^2 + 1 \)

\( S = \int_{-2}^{1} (x^2 + 1) dx = \left[ \frac{x^3}{3} + x \right]_{-2}^{1} = (\frac{1^3}{3} + 1) - (\frac{(-2)^3}{3} + (-2)) = (\frac{1}{3} + 1) - (\frac{-8}{3} - 2) = \frac{4}{3} - (\frac{-8 - 6}{3}) = \frac{4}{3} + \frac{14}{3} = \frac{18}{3} = 6 \)

4. \( a = 0, b = 2, f(x) = x^3 + 1 \)

\( S = \int_{0}^{2} (x^3 + 1) dx = \left[ \frac{x^4}{4} + x \right]_{0}^{2} = (\frac{2^4}{4} + 2) - (\frac{0^4}{4} + 0) = (\frac{16}{4} + 2) - 0 = 4 + 2 = 6 \)

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

\( S = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \sin x dx = \left[ -\cos x \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} = -\cos(\frac{2\pi}{3}) - (-\cos(\frac{\pi}{3})) = -(-\frac{1}{2}) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1 \)

6. \( a = -\frac{\pi}{6}, b = 0, f(x) = \cos x \)

\( S = \int_{-\frac{\pi}{6}}^{0} \cos x dx = \left[ \sin x \right]_{-\frac{\pi}{6}}^{0} = \sin(0) - \sin(-\frac{\pi}{6}) = 0 - (-\frac{1}{2}) = \frac{1}{2} \)