1. **Planteamiento del problema:** Evaluar la integral definida para cada caso dado. 2. **Fórmulas y reglas importantes:** - La integral definida de una función $f(x)$ de $a$ a $b$ es $$\int_a^b f(x) dx = F(b) - F(a)$$ donde $F(x)$ es una antiderivada de $f(x)$. - Para potencias, $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ si $n \neq -1$. - Para productos, expandir antes de integrar. - Simplificar expresiones antes de integrar facilita el cálculo. 3. **Resolución paso a paso:** (a) $$\int_0^2 (6x^2 - 4x + 5) dx$$ - Antiderivada: $$F(x) = 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 5x = 2x^3 - 2x^2 + 5x$$ - Evaluar en 2 y 0: $$F(2) = 2(2)^3 - 2(2)^2 + 5(2) = 16 - 8 + 10 = 18$$ $$F(0) = 0$$ - Resultado: $$18 - 0 = 18$$ (b) $$\int_0^4 (2v + 5)(3v - 1) dv$$ - Expandir: $$(2v)(3v) + (2v)(-1) + 5(3v) + 5(-1) = 6v^2 - 2v + 15v - 5 = 6v^2 + 13v - 5$$ - Antiderivada: $$F(v) = 6 \cdot \frac{v^3}{3} + 13 \cdot \frac{v^2}{2} - 5v = 2v^3 + \frac{13}{2} v^2 - 5v$$ - Evaluar en 4 y 0: $$F(4) = 2(64) + \frac{13}{2}(16) - 20 = 128 + 104 - 20 = 212$$ $$F(0) = 0$$ - Resultado: $$212 - 0 = 212$$ (c) $$\int_1^2 \frac{y + 5y^7}{y^3} dy = \int_1^2 (y^{1-3} + 5y^{7-3}) dy = \int_1^2 (y^{-2} + 5y^4) dy$$ - Antiderivada: $$F(y) = \int y^{-2} dy + 5 \int y^4 dy = -y^{-1} + 5 \cdot \frac{y^5}{5} = -\frac{1}{y} + y^5$$ - Evaluar en 2 y 1: $$F(2) = -\frac{1}{2} + 32 = \frac{-1}{2} + 32 = \frac{63}{2}$$ $$F(1) = -1 + 1 = 0$$ - Resultado: $$\frac{63}{2} - 0 = \frac{63}{2} = 31.5$$ (d) $$\int_{-2}^{-1} \left(4x^3 + \frac{2}{x^3}\right) dx = \int_{-2}^{-1} (4x^3 + 2x^{-3}) dx$$ - Antiderivada: $$F(x) = 4 \cdot \frac{x^4}{4} + 2 \cdot \frac{x^{-2}}{-2} = x^4 - x^{-2} = x^4 - \frac{1}{x^2}$$ - Evaluar en -1 y -2: $$F(-1) = (-1)^4 - \frac{1}{(-1)^2} = 1 - 1 = 0$$ $$F(-2) = (-2)^4 - \frac{1}{(-2)^2} = 16 - \frac{1}{4} = \frac{64}{4} - \frac{1}{4} = \frac{63}{4}$$ - Resultado: $$0 - \frac{63}{4} = -\frac{63}{4} = -15.75$$ (e) $$\int_1^4 \sqrt{z} (1 + z) dz = \int_1^4 z^{1/2} + z^{3/2} dz$$ - Antiderivada: $$F(z) = \int z^{1/2} dz + \int z^{3/2} dz = \frac{z^{3/2}}{3/2} + \frac{z^{5/2}}{5/2} = \frac{2}{3} z^{3/2} + \frac{2}{5} z^{5/2}$$ - Evaluar en 4 y 1: $$F(4) = \frac{2}{3} (4)^{3/2} + \frac{2}{5} (4)^{5/2} = \frac{2}{3} (8) + \frac{2}{5} (32) = \frac{16}{3} + \frac{64}{5} = \frac{80}{15} + \frac{192}{15} = \frac{272}{15} \approx 18.13$$ $$F(1) = \frac{2}{3} (1) + \frac{2}{5} (1) = \frac{2}{3} + \frac{2}{5} = \frac{10}{15} + \frac{6}{15} = \frac{16}{15}$$ - Resultado: $$\frac{272}{15} - \frac{16}{15} = \frac{256}{15} \approx 17.07$$ **Respuestas finales:** - (a) 18 - (b) 212 - (c) 31.5 - (d) -15.75 - (e) 17.07