A-5. Find the maximum value of the following expression:
$$
\int_{0}^{y} \sqrt{x^{4}+\left(y-y^{2}\right)^{2}} d x \text {, }
$$

where $0 \leqslant y \leqslant \mathrm{i}$.
The maximum value is $\frac{1}{3}$, which is obtained when $y=1$. In fact, when $y=1$, $\int_{0}^{1} \sqrt{x^{4}} d x=\frac{1}{3}$. On the other hand, $\int_{0}^{y} \sqrt{x^{3}+\left(y-y^{2}\right)^{2}} d x \leqslant \int_{0}^{y}\left(x^{2}+\left(y-y^{2}\right)\right) d x$ $=y^{2}-\frac{2}{3} y^{3}$. Let $g(y)=y^{2}-\frac{2}{3} y^{3}$, then: $g^{\prime}(y)$ $=2 y-2 y^{2}=2 y(1-y)>0, y \in(0,1)$. Therefore, $g(y)$ is strictly increasing on $[0,1]$.

From this, we can derive $\int_{0}^{y} \sqrt{x^{4}+\left(y-y^{2}\right)^{2}} d x \leqslant g(y)$ $\leqslant g(1)=\frac{1}{3}$, with equality holding if and only if $y=1$.