Christmas number 5

A nice problem! Geometrically it is the volume of a pyramid with base being a right-angled isosceles triangle that is half of a unit square. The pyramid is of unit height with the apex being directly above the right-angle vertex of the base.

Volume of pyramid = $\frac13 \times base$ $area \times height = \frac13 \times \frac12 \times 1= \frac16$ .

$\begin{aligned} I & = \int_0^1 \int_x^1 \int_0^{y-x} dz \ dy \ dx \\ & = \int_0^1 \int_x^1 z \ \bigg|_0^{y-x} dy \ dx \\ & = \int_0^1 \int_x^1 y-x \ dy \ dx \\ & = \int_0^1 \frac {y^2}2-xy \ \bigg|_x^1 \ dx \\ & = \int_0^1 \frac 12-x-\frac {x^2}2+x^2 \ dx \\ & = \int_0^1 \frac {({\color{#3D99F6}x-1})^2}2 \ dx & \small \color{#3D99F6} \text{Let }u = x-1, \ du = dx \\ & = \int_{-1}^0 \frac {{\color{#3D99F6}u}^2}2 \ du \\ & = \frac {u^3}6 \ \bigg|_{-1}^0 \\ & = \boxed{\dfrac 16} \end{aligned}$