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El área bajo la curva de la función \( g(x)=(x-2)(4 x+1) \) en el intervalo \( [2,4] \) es igual a:

Ask by Vega Black. in Colombia
Mar 20,2025

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El área bajo la curva es \( \frac{86}{3} \).

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The Deep Dive

Para encontrar el área bajo la curva de la función \( g(x)=(x-2)(4x+1) \) en el intervalo \( [2,4] \), primero necesitamos integrar la función en ese intervalo. La integral definida se calcula como: \[ \int_{2}^{4} g(x) \, dx = \int_{2}^{4} (x-2)(4x+1) \, dx \] Expandemos la función: \[ g(x) = (x-2)(4x+1) = 4x^2 + x - 8 \] Ahora, integramos: \[ \int (4x^2 + x - 8) \, dx = \frac{4x^3}{3} + \frac{x^2}{2} - 8x \bigg|_{2}^{4} \] Evaluamos en los límites: \[ \left( \frac{4(4)^3}{3} + \frac{(4)^2}{2} - 8(4) \right) - \left( \frac{4(2)^3}{3} + \frac{(2)^2}{2} - 8(2) \right) \] Calculando ambos términos y restando, obtenemos el área bajo la curva. Al final, el resultado será el área buscada, que es igual a: \[ \text{Área} = \frac{56}{3} \] ¡Y ahí lo tienes! Una integral que resulta en el área bajo la curva de esta función polinómica tan colorida.

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