Imagine a thin vertical strip, going from $x$ to $x+dx$. The height of the strip is everywhere about $2-7x^2$.
The strip is almost a rectangle. So the area of the strip is about $(2-7x^2)\,dx$. The mass of the strip is therefore about $(1/4)(2-7x^2)\,dx$.
Every point in the strip is at about a perpendicular distance $x$ from the $y$-axis. So the moment of the strip about the $y$-axis is about $x(1/4)(2-7x^2)\,dx$.
"Add up" over all such strips (integrate). It is not hard to see that the curve $y=2-7x^2$ meets the $x$-axis at $x=\sqrt{2/7}$. So our moment is $\int_0^{\sqrt{2/7}}x(1/4)(2-7x^2)\,dx.$
Remark: The above is an informal way to find the appropriate integral. If you want to be more technical, divide the part of the $x$-axis from $x=0$ to $x=\sqrt{2/7}$ into a large number $n$ of equal subintervals. Let the division points be $0=x_0,x_1,\dots,x_n=\sqrt{2/7}$. Let $\delta x$ be the length of these subintervals. (Here $\Delta x=\sqrt{2/7}/n$.) The moment of the lamina can be approximated by $\sum_{i=0}^{n-1} x_i(1/4)(2-x_i^2)\Delta x.$ This is a Riemann sum. As $n\to\infty$, the approximation approaches the true moment. It also approaches the integral of the answer.