Find the maximum value of $\int_{0}^{y}\sqrt{x^{4}+(y-y^{2})^{2}}dx$

Find the maximum value of $$\displaystyle \int_{0}^{y}\sqrt{x^{4}+(y-y^{2})^{2}}dx$$ for $$0 \leq y\leq 1$$.

Try: Let $$I(y) = \int^{y}_{0}\sqrt{x^4+(y-y^2)^2}dx$$

Then $$I'(y)=\sqrt{y^{4}+y^{2}(1-y)^{2}}+y(1-y)(1-2y)\int_{0}^{y}\frac{dx}{\sqrt{x^{4}+y^{2}(1-y)^{2}}}$$

Now did not find any clue how I find $$\int_{0}^{y}\frac{dx}{\sqrt{x^{4}+y^{2}(1-y)^{2}}}$$

Could some help me find it? Thanks.

• How did you get that second term after differentiation? – MisterRiemann Oct 28 '18 at 14:30
• @Sobi See here – TheSimpliFire Oct 28 '18 at 14:32
• @TheSimpliFire Ahh indeed! Thanks. – MisterRiemann Oct 28 '18 at 14:33
• Do you want to maximize $\int_{0}^{y}\sqrt{x^4+(y-y^2)}\,dx$ or $\int_{0}^{y}\sqrt{x^4+(y-y^2)^{\color{red}{2}}}\,dx$ over $y\in[0,1]$? – Jack D'Aurizio Oct 28 '18 at 15:03
• HINT: you may use Leibnitz theorem to find the derivative and then equate it to 0. – PradyumanDixit Oct 28 '18 at 15:12

Actually you do not need differentiation. $$I(y) = y^3 \int_{0}^{1}\sqrt{x^4+(1-y)^2}\,dx\geq y^3\int_{0}^{1}x^2\,dx=\frac{y^3}{3}$$ hence the maximum value of $$I(y)$$ over $$[0,1]$$ is at least $$\frac{1}{3}=I(1)$$.
On the other hand $$I(y) \leq y^3 \int_{0}^{1} x^2+(1-y)\,dx = y^3\left(\frac{4}{3}-y\right)\leq \frac{1}{3}$$ hence $$I(1)=\frac{1}{3}$$ is precisely the maximum.
It is possible to show that the integrand $$f(x,y)$$ is increasing in $$y\in[0,1]$$ since the stationary point is at $$\frac{2x^3}{\sqrt{x^4+(y-y^2)^2}}=0\implies x=0$$ Since the integrand is continuous and differentiable on $$\mathbb{R}$$, with $$f(1,y)>f(0,y)$$ (taking the principal root), $$f$$ is increasing. Therefore, $$I$$ is at its maximum at $$y=1$$. $$\boxed{I(1) = \int^{1}_{0}\sqrt{x^4}\,dx=\left[\frac{x^3}3\right]_0^1=\frac13}$$