Answer
$$
\begin{aligned}
& F_R=43.6 \mathrm{lb} \\
& \bar{x}=3.27 \mathrm{ft}
\end{aligned}
$$
Work Step by Step
$$
\begin{gathered}
d A=w d x \\
\begin{array}{c}
F_R=\int d A=\int_0^6(4+2 \sqrt{x}) d x \\
=\left[4 x+\frac{4}{3} x^{\frac{1}{2}}\right]_0^6 \\
F_R=43.6 \mathrm{lb} \\
\int \bar{x} d F=\int_0^6\left(4 x+2 x^{\frac{3}{2}}\right) d x \\
=\left[2 x^2+\frac{4}{5} x^{\frac{2}{2}}\right]_0^6 \\
=142.5 \mathrm{lb} \cdot \mathrm{ft} \\
\bar{x}=\frac{142.5}{43.6}=3.27 \mathrm{ft}
\end{array}
\end{gathered}
$$
