Calcula la integral triple \( \int_{0}^{1} \int_{0}^{2} \int_{0}^{3}(x+y+z) d z d y d x \)
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Para calcular la integral triple \( I = \int_{0}^{1} \int_{0}^{2} \int_{0}^{3}(x+y+z) \, dz \, dy \, dx \), empezamos integrando con respecto a \( z \): \[ I = \int_{0}^{1} \int_{0}^{2} \left[ (x+y+z) z \right]_{0}^{3} \, dy \, dx \] \[ = \int_{0}^{1} \int_{0}^{2} (x+y+3) \cdot 3 \, dy \, dx \] \[ = \int_{0}^{1} \int_{0}^{2} (3x + 3y + 9) \, dy \, dx \] Ahora integramos con respecto a \( y \): \[ I = \int_{0}^{1} \left[ (3x y + \frac{3}{2} y^2 + 9y) \big|_{0}^{2} \right] \, dx \] \[ = \int_{0}^{1} \left[ (3x \cdot 2 + \frac{3}{2} \cdot 4 + 9 \cdot 2) - (0) \right] \, dx \] \[ = \int_{0}^{1} (6x + 6 + 18) \, dx \] \[ = \int_{0}^{1} (6x + 24) \, dx \] Finalmente, integramos con respecto a \( x \): \[ = \left[ 3x^2 + 24x \right]_{0}^{1} \] \[ = (3 \cdot 1^2 + 24 \cdot 1) - (0) = 3 + 24 = 27 \] Por lo tanto, el resultado de la integral triple es \( \boxed{27} \).
