Find the maximum value of an integral

Question

If $y \in [0,1]$, then find the maximum value of $$\int_0^y \sqrt{x^4+(y-y^2)^2}dx$$ I am unable to start with it. I know integrating isn't the way to go. Hints please. Thanks.

Answer 1

Use the bound $$\sqrt{a^2+b^2} \le a + b$$ which is valid for $a,b \ge 0$. From this we see $$\int^y_0 \sqrt{x^4 + (y-y^2)^2} dx \le \int^y_0 x^2 + (y-y^2) dx = y^2 - \frac{2y^3}{3}.$$ You can maximize the latter function and then show that all inequalities become equalities for a particular value of $y$; i.e., the maximum is achieved.