First product of inertia of right-angled triangle

Take a right-angled triangle with the right-angle corner at the origin, base $b$ along the $x$-axis and height $h$ along the $y$-axis. The hypotenuse has equation $y = h\left(1 - \dfrac{x}{b}\right)$.

 y
 h|\
  | \
  |  \
  |   \
  O----- b   x

Derivation from first principles

Take a vertical strip of width $dx$ at distance $x$, of height $y_1 = h\left(1 - \dfrac{x}{b}\right)$. Its area is $dA = y_1\,dx$. The product of inertia of the strip about the $x$ and $y$ axes is obtained by applying the parallel-axis theorem to the strip (its own $I_{xy}$ is zero, as it is symmetrical about its centre line parallel to $y$):

$$dI_{xy} = (dA)\,x\cdot\frac{y_1}{2} = x\,\frac{y_1}{2}\,y_1\,dx = \frac{x\,y_1^2}{2}\,dx$$

Hence

$$I_{xy} = \int_0^b \frac{x\,h^2}{2}\left(1 - \frac{x}{b}\right)^2dx = \frac{h^2}{2}\int_0^b\left(x - \frac{2x^2}{b} + \frac{x^3}{b^2}\right)dx$$ $$= \frac{h^2}{2}\left[\frac{b^2}{2} - \frac{2b^2}{3} + \frac{b^2}{4}\right] = \frac{h^2}{2}\cdot\frac{b^2}{12}$$ $$\boxed{I_{xy} = \frac{b^2h^2}{24}}$$

(positive, since the triangle lies in the first quadrant). About the centroidal axes, $\bar I_{xy} = \dfrac{b^2h^2}{24} - \dfrac{bh}{2}\cdot\dfrac{b}{3}\cdot\dfrac{h}{3} = -\dfrac{b^2h^2}{72}$.