Solution
A
We have
$$
\begin{aligned}
x^{3} y^{2}-x^{2} y^{2}-x y^{4}+x y^{3} & =x y^{2}\left(x^{2}-x-y^{2}+y\right) \\
& =x y^{2}\left(\left(x^{2}-y^{2}\right)-(x-y)\right) \\
& =x y^{2}((x-y)(x+y)-(x-y)) \\
& =x y^{2}(x-y)(x+y-1) .
\end{aligned}
$$

In the region where $0 \leq x \leq y$, we have $x y^{2} \geq 0$ and $x-y \leq 0$. Therefore in this region $x y^{2}(x-y)(x+y-1) \geq 0 \Leftrightarrow(x+y-1) \leq 0$.
Therefore the region of the $x y$-plane within which we have $x^{3} y^{2}-x^{2} y^{2}-x y^{4}+x y^{3} \geq 0$ and $0 \leq x \leq y$ is the region where $x+y-1 \leq 0$ and $0 \leq x \leq y$.

The line with equation $x=0$ (the $y$-axis) divides the plane into the region where $x>0$ and the region where $x0$ and - to indicate the side where $xx$ and the region where $y0$ and the region where $x+y-1<0$.
Therefore the region whose area we need to find is the triangle $O P Q$ which is bounded by the lines with equations $x=0, x=y$ and $y=1-x$, as shown in the diagram.
The point $O$ is the origin where the lines $x=0$ and $y=x$ meet. The point $P$ is the point where the lines $x=0$ and $y=1-x$ meet. The co-ordinates of $P$ are $(0,1)$. The point $Q$ is the point where the lines $y=x$ and $y=1-x$ meet. The co-ordinates of $Q$ are $\left(\frac{1}{2}, \frac{1}{2}\right)$.
The triangle $O P Q$ has a base $O P$ which has length 1. The height of the triangle with this base is the distance of $Q$ from the $y$-axis which is $\frac{1}{2}$. Therefore, the area of this triangle is
$$
\frac{1}{2}\left(1 \times \frac{1}{2}\right)=\frac{1}{4} .
$$
25. What is the area of the part of the $x y$-plane within which $x^{3} y^{2}-x^{2} y^{2}-x y^{4}+x y^{3} \geq 0$ and $0 \leq x \leq y$ ?
A $\frac{1}{4}$
B $\frac{1}{2}$
C 1
D 2
E 4