Вычислите интегралы:

Решение:

1. \[ \int_1^4 x^3 dx = \left[ \frac{x^4}{4} \right]_1^4 = \frac{4^4}{4} - \frac{1^4}{4} = \frac{256}{4} - \frac{1}{4} = 64 - \frac{1}{4} = 63.75 \]
2. \[ \int_0^1 (x^2 - 2x + 1) dx = \left[ \frac{x^3}{3} - x^2 + x \right]_0^1 = \left( \frac{1^3}{3} - 1^2 + 1 \right) - \left( \frac{0^3}{3} - 0^2 + 0 \right) = \frac{1}{3} - 1 + 1 - 0 = \frac{1}{3} \]
3. \[ \int_{-\frac{4}{3}}^0 (3x^2 + 4x) dx = \left[ x^3 + 2x^2 \right]_{-\frac{4}{3}}^0 = (0^3 + 2(0^2)) - \left( (-\frac{4}{3})^3 + 2(-\frac{4}{3})^2 \right) = 0 - \left( -\frac{64}{27} + 2(\frac{16}{9}) \right) = -\left( -\frac{64}{27} + \frac{32}{9} \right) = -\left( -\frac{64}{27} + \frac{96}{27} \right) = -\frac{32}{27} \]
4. \[ \int_0^{\frac{\pi}{6}} \cos x dx = \left[ \sin x \right]_0^{\frac{\pi}{6}} = \sin(\frac{\pi}{6}) - \sin(0) = \frac{1}{2} - 0 = \frac{1}{2} \]

Ответ: 63.75, 1/3, -32/27, 1/2