Covariance under a triangular density c(1 minus y)

Random variables $X_1,X_2$ have joint density $c(1-x_2)$ on the triangle $0\le x_1\le x_2\le1$ and zero elsewhere, where $c$ is the normalising constant. Find $\operatorname{Cov}(X_1,X_2)$.

$1/40$. Normalising gives $c=6$. Integrating out $x_1$ yields the weight $6x_2(1-x_2)$, from which $\mathbb{E}X_2=\tfrac12$, $\mathbb{E}X_1=\int_0^1 3x_2^2(1-x_2)\,dx_2=\tfrac14$ and $\mathbb{E}X_1X_2=\int_0^1 3x_2^3(1-x_2)\,dx_2=\tfrac{3}{20}$. Hence $\tfrac{3}{20}-\tfrac18=\tfrac{1}{40}$.