Show that when the density of the region is proportional to the distance from the $y$-axis, the first moment about the $x$-axis is

$I_x={\int }_1^2{\int }_{-x+2}^{2x-1}kx{y}^{2}dydx=\frac{28}{5}k$

The expression is

$I_x={\int }_1^2{\int }_{-x+2}^{2x-1}kx{y}^{2}dydx=\frac{28}{5}k$

Plot the vertices $(1,1),(2,0),$and $(2,3)$and join them.

Obtain the equation of $AB$by using the formula of coordinate geometry

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ y-1=\frac{0-1}{2-1}(x-1) \\ y=-x+2 \end{gathered}$$

Equation of $BC$

$$\begin{gathered} y-0=\frac{3-0}{2-2}(x-2) \\ y-0=\frac{3}{0}(x-2) \\ x-2=0\text{[Cross multiply]} \\ x=2 \end{gathered}$$

And equation of $CA$is

$y=2x-1$

The moment of inertia of the mass in $\Omega$about the $x$axis is

$I_x={\iint }_{\Omega }{y}^2\rho (x,y)dA$

Where $\rho (x,y)$is the density of the region $\Omega .$

Here $\rho (x,y)$is proportional to the distance from $y$- axis.

Assume $\rho (x,y)=kx$. Then

$I_x={\iint }_{\Omega }kx{y}^{2}dydx$

Impose the limits on integrals.

$I_x={\int }_1^2{\int }_{-x+2}^{2x-1}kx{y}^{2}dydx$

Integrate the inner integral with respect to $y$first.

$$\begin{gathered} I_x=k{\int }_1^{2}x{\left[\frac{y^3}{3}\right]}_{-x+2}^{2x-1}dx \\ {I}_x=\frac{k}{3}{\int }_1^{2}x\left[(2x-1{)}^3-(-x+2{)}^3\right]dx \\ {I}_x=\frac{9k}{3}{\int }_1^2\left[x^4-2x^3+2x^2-x\right]dx[\text{Simplify]} \end{gathered}$$

Integrate with respect to $x$.

$I_x=3k{\left[\frac{1}{5}{x}^5-\frac{2}{4}{x}^4+\frac{2}{3}{x}^3-\frac{1}{2}{x}^2\right]}_1^2$

Substitute the limits

$I_x=3k\left[\left(\frac{1}{5}(2{)}^5-\frac{2}{4}(2{)}^4+\frac{2}{3}(2{)}^3-\frac{1}{2}(2{)}^2\right)-\left(\frac{1}{5}-\frac{2}{4}+\frac{2}{3}-\frac{1}{2}\right)\right]$

$I_x=3k\left[\left(\frac{288}{5}-72+12-18\right)-\left(\frac{9}{5}-\frac{9}{2}+6-\frac{9}{2}\right)\right]$ [Simplify]

$I_x=\frac{28}{5}k$

Thus, the moment of inertia about the$x$ axis is

$I_x=\frac{28}{5}k$