Answer
$$\dfrac{7}{6}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^1_0 \int^{2-x}_0 \int^{2-x-y}_0 dz \space dy \space dx \\=\int^1_0 \int^{2-x}_0 (2-x-y) \space dy \space dx \\= \int^1_0 [(2-x)^2-\frac{1}{2}(2-x)^2]\space dx\\=\dfrac{1}{2} \times \int^1_0 (2-x)^2 dx\\= -\dfrac{1}{6} \times [(2-x)^3]^1_0 \\=-\dfrac{1}{6} +\dfrac{8}{6}\\=\dfrac{7}{6}$$
