Вычислить определенные интегралы: 1. $\int_{1}^{2} x dx.$

Привет! Давай решим эти интегралы. Будем использовать формулу Ньютона-Лейбница: $\int_{a}^{b} f(x) dx = F(b) - F(a)$. 1. $\int_{1}^{2} x dx = \left. \frac{x^2}{2} \right|_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} = 1.5$ 2. $\int_{0}^{3} x^2 dx = \left. \frac{x^3}{3} \right|_{0}^{3} = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9$ 3. $\int_{1/2}^{1} x^3 dx = \left. \frac{x^4}{4} \right|_{1/2}^{1} = \frac{1^4}{4} - \frac{(1/2)^4}{4} = \frac{1}{4} - \frac{1/16}{4} = \frac{1}{4} - \frac{1}{64} = \frac{16}{64} - \frac{1}{64} = \frac{15}{64}$ 4. $\int_{0}^{1} (2x + 1) dx = \left. (x^2 + x) \right|_{0}^{1} = (1^2 + 1) - (0^2 + 0) = 2 - 0 = 2$ 5. $\int_{-1}^{0} (3x^2 + 1) dx = \left. (x^3 + x) \right|_{-1}^{0} = (0^3 + 0) - ((-1)^3 + (-1)) = 0 - (-1 - 1) = 0 - (-2) = 2$ 6. $\int_{-1}^{1} (\frac{1}{2}t + 4t^2) dt = \left. (\frac{1}{2} \cdot \frac{t^2}{2} + 4 \cdot \frac{t^3}{3}) \right|_{-1}^{1} = \left. (\frac{t^2}{4} + \frac{4t^3}{3}) \right|_{-1}^{1} = (\frac{1^2}{4} + \frac{4 \cdot 1^3}{3}) - (\frac{(-1)^2}{4} + \frac{4 \cdot (-1)^3}{3}) = (\frac{1}{4} + \frac{4}{3}) - (\frac{1}{4} - \frac{4}{3}) = \frac{1}{4} + \frac{4}{3} - \frac{1}{4} + \frac{4}{3} = \frac{8}{3} \approx 2.67$ 7. $\int_{-1/2}^{1/2} 3(u^2 + 1) du = 3 \int_{-1/2}^{1/2} (u^2 + 1) du = 3 \left. (\frac{u^3}{3} + u) \right|_{-1/2}^{1/2} = 3 [(\frac{(1/2)^3}{3} + \frac{1}{2}) - (\frac{(-1/2)^3}{3} + (- \frac{1}{2}))] = 3 [(\frac{1/8}{3} + \frac{1}{2}) - (\frac{-1/8}{3} - \frac{1}{2})] = 3 [(\frac{1}{24} + \frac{1}{2}) - (-\frac{1}{24} - \frac{1}{2})] = 3 [\frac{1}{24} + \frac{12}{24} + \frac{1}{24} + \frac{12}{24}] = 3 \cdot \frac{26}{24} = 3 \cdot \frac{13}{12} = \frac{13}{4} = 3.25$ Дополнительный (снизу): $\int_{1}^{2} x^3 dx = \left. \frac{x^4}{4} \right|_{1}^{2} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} = 3.75$