The area moment of inertia about the y-axis of a linearly tapered section shown in the figure is _____________ m⁴ . (Answer in integer)

To find the area moment of inertia of the linearly tapered section about the y-axis, we can integrate the contributions of infinitesimal vertical strips of width $dx$ located at a distance $x$ from the y-axis.

From the given image, we can identify the following dimensions of the tapered section:
- The length of the section along the x-axis is $L=12\text{m}$.
- At $x=0$ (left end on the y-axis), the height is symmetric about the x-axis with a top distance of $1.5\text{m}$ and a bottom distance of $1.5\text{m}$, giving a total initial height of:
$h_1=1.5+1.5=3\text{m}$
- At $x=12\text{m}$ (right end), the height is also symmetric with a top distance of $3\text{m}$ and a bottom distance of $3\text{m}$, giving a total final height of:
$h_2=3+3=6\text{m}$

Since the section tapers linearly, the height $h\left(x\right)$ at any position $x$ is given by the linear relation:
$h\left(x\right)=h_1+\frac{h_2-h_1}{L}x$
Substituting the values:
$h\left(x\right)=3+\frac{6-3}{12}x=3+\frac{x}{4}$

The area of an infinitesimal vertical strip of width $dx$ is:
$dA=h\left(x\right)dx=(3+\frac{x}{4})dx$

The area moment of inertia about the y-axis $I_y$ is defined as:
$I_y={\int }_0^L{x}^{2}dA$
Substituting $dA$ and the limits of integration from $0$ to $12$:
$I_y={\int }_0^{12}{x}^2(3+\frac{x}{4})dx$
$I_y={\int }_0^{12}(3x^2+\frac{x^3}{4})dx$

Now, perform the integration:
$I_y={[x^3+\frac{x^4}{16}]}_0^{12}$
Evaluate at the upper limit $x=12$:
$I_y={12}^3+\frac{{12}^4}{16}$
$I_y=1728+\frac{20736}{16}$
$I_y=1728+1296=3024{\text{m}}^4$

Thus, the area moment of inertia about the y-axis is 3024.