# Find maximum of integral

I am trying to find the maximum value of

$$I=\int_0^y \sqrt{x^4+(y-y^2)^2}\,dx$$

for $0 \leq y \leq 1$.

At $y=1$ the value $I=1/3$ which I think is the answer. How can you prove this?

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Hint: Use differentiation under the integral sign to find the derivative of $I(y)$.