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El área del recinto limitado por \( y=5 x \), el eje \( x, y \) las rectas \( x=-1, x=2 \) vale

Ask by George Burns. in Spain
Mar 09,2025

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El área del recinto es 7.5 unidades cuadradas.

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Para encontrar el área del recinto limitado por la recta \( y = 5x \), el eje \( x \) y las rectas \( x = -1 \) y \( x = 2 \), primero debemos calcular las intersecciones de la recta con los límites verticales. En \( x = -1 \), \( y = 5(-1) = -5 \) (pero solo consideraremos la parte positiva), y en \( x = 2 \), \( y = 5(2) = 10 \). La forma del área se obtiene al integrar la función entre los límites de \( x \): \[ \text{Área} = \int_{-1}^{2} 5x \, dx \] Calculando la integral: \[ \int 5x \, dx = \frac{5x^2}{2} \] Evaluamos de \(-1\) a \(2\): \[ = \left[ \frac{5(2)^2}{2} - \frac{5(-1)^2}{2} \right] = \left[ \frac{20}{2} - \frac{5}{2} \right] = \left[ 10 - 2.5 \right] = 7.5 \] Por tanto, el área del recinto es \( 7.5 \) unidades cuadradas.

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