5,2. Вычислить ∫ -1 2 8dx ; ∫ 1 4 1 (1/x + 1/x^2)dx ∫ 0 (x+√x+x^3)dx

Задание

Нужно вычислить три определенных интеграла:

• \[ \int_{-1}^{2} 8dx \]
• \[ \int_{1}^{4} \left( \frac{1}{x} + \frac{1}{x^2} \right) dx \]
• \[ \int_{0}^{4} (x + \sqrt{x} + x^3) dx \]

Решение первого интеграла:

1. \[ \int_{-1}^{2} 8dx = 8x \Big|_{-1}^{2} \]
2. \[ = 8(2) - 8(-1) = 16 - (-8) = 16 + 8 = 24 \]

Решение второго интеграла:

1. \[ \int_{1}^{4} \left( \frac{1}{x} + \frac{1}{x^2} \right) dx = \int_{1}^{4} \left( \frac{1}{x} + x^{-2} \right) dx \]
2. \[ = \left[ \ln|x| + \frac{x^{-1}}{-1} \right]_{1}^{4} = \left[ \ln|x| - \frac{1}{x} \right]_{1}^{4} \]
3. \[ = \left( \ln(4) - \frac{1}{4} \right) - \left( \ln(1) - \frac{1}{1} \right) \]
4. \[ = \ln(4) - \frac{1}{4} - (0 - 1) = \ln(4) - \frac{1}{4} + 1 = \ln(4) + \frac{3}{4} \]

Решение третьего интеграла:

1. \[ \int_{0}^{4} (x + \sqrt{x} + x^3) dx = \int_{0}^{4} (x + x^{1/2} + x^3) dx \]
2. \[ = \left[ \frac{x^2}{2} + \frac{x^{3/2}}{3/2} + \frac{x^4}{4} \right]_{0}^{4} = \left[ \frac{x^2}{2} + \frac{2}{3}x^{3/2} + \frac{x^4}{4} \right]_{0}^{4} \]
3. \[ = \left( \frac{4^2}{2} + \frac{2}{3}(4)^{3/2} + \frac{4^4}{4} \right) - \left( \frac{0^2}{2} + \frac{2}{3}(0)^{3/2} + \frac{0^4}{4} \right) \]
4. \[ = \left( \frac{16}{2} + \frac{2}{3}(8) + \frac{256}{4} \right) - 0 \]
5. \[ = \left( 8 + \frac{16}{3} + 64 \right) = 72 + \frac{16}{3} = \frac{216}{3} + \frac{16}{3} = \frac{232}{3} \]

Ответ:

• \[ \int_{-1}^{2} 8dx = 24 \]
• \[ \int_{1}^{4} \left( \frac{1}{x} + \frac{1}{x^2} \right) dx = \ln(4) + \frac{3}{4} \]
• \[ \int_{0}^{4} (x + \sqrt{x} + x^3) dx = \frac{232}{3} \]