Answer
$\dfrac{8}{35} $
Work Step by Step
Consider $I=\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{x+1} (2x-y-z) \ dz \ dy \ dx$
or, $=\int_{0}^{1} \int_{0}^{x^2} [\dfrac{3x^2}{2}-\dfrac{3y^2}{2}] \ dy \ dx$
or, $=\int_{0}^{1} [\dfrac{3x^4}{2}-\dfrac{x^6}{2}] \ dx$
or, $=\int_{0}^{1} \dfrac{3x^4}{2} \ dx -\int_{0}^{1} \dfrac{x^6}{2} \ dx$
or, $=(3/2) \int_{0}^{1} x^4 \ dx-(1/2) \times [\dfrac{x^5}{5}]_0^1$
or, $=\dfrac{3}{2} [\dfrac{x^5}{5}]_0^1 -\dfrac{1}{2} [\dfrac{x^5}{5}]_0^1$
or, $=\dfrac{3}{2} \times \dfrac{1}{5}-\dfrac{1}{2} \times \dfrac{1}{7} $
or, $=\dfrac{3}{10}-\dfrac{1}{14} $
or, $=\dfrac{8}{35} $
