Centre of gravity of a planar region

Area.
$$A=\int_0^5(6x-x^2-x)\,dx=\int_0^5(5x-x^2)\,dx=\frac{125}{6}.$$

$\bar x$.
$$\iint_D x\,dx\,dy=\int_0^5 x(5x-x^2)\,dx=\frac{625}{12},$$
so $\bar x=\frac{625/12}{125/6}=\frac52$.

$\bar y$.
$$\iint_D y=\frac12\int_0^5\left[(6x-x^2)^2-x^2\right]dx=\frac{625}{3},$$
so $\bar y=\frac{625/3}{125/6}=10$... wait, careful: $\bar y=625/(3\cdot125/6)=10$? Check numeric: $625\cdot 6/(3\cdot125)=10$. But the geometric answer is $5$. Recompute carefully: $\iint_D y=\frac12\int_0^5(35x^2-12x^3+x^4)\,dx=\frac12\left(\frac{35\cdot125}{3}-3\cdot625+\frac{3125}{5}\right)=\frac12\left(\frac{4375}{3}-1875+625\right)$. Compute: $\frac{4375}{3}-1250=\frac{4375-3750}{3}=\frac{625}{3}$. So $\iint_D y=\frac{625}{6}$, and $\bar y=\frac{625/6}{125/6}=5$.

Conclusion. Centroid at $(5/2,5)$.