Compute volume of tetrahedron using a triple integral

The equation of the plane containing the points $(2,0,0),(0,2,0),(0,0,1)$ is given by $x+y+2z-2=0,$ that is, $z=1-\frac x2-\frac y2.$ Note that the basis of the tetrahedron is a triangle, one side on the $OX$-axis, one side on the $OY$-axis, and the equation of the third one is given by $x+y=2.$ Thus, the volume can be obtained by evaluating the integral

$$\int_0^2\int_0^{2-x} \left(1-\frac x2-\frac y2\right)dydx. $$

We have:

$$\int_0^2\int_0^{2-x} \left(1-\frac x2-\frac y2\right)dydx=\int_0^2 \left(\left(1-\frac x2\right)(2-x)-\frac{(2-x)^2}{4}\right)dx \\ =\int_0^2 \left(1-x+\frac{x^2}{4}\right)dx. $$