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.

  • $\begingroup$ How did you get that second term after differentiation? $\endgroup$ – MisterRiemann Oct 28 '18 at 14:30
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    $\begingroup$ @Sobi See here $\endgroup$ – TheSimpliFire Oct 28 '18 at 14:32
  • $\begingroup$ @TheSimpliFire Ahh indeed! Thanks. $\endgroup$ – MisterRiemann Oct 28 '18 at 14:33
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    $\begingroup$ 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 $y\in[0,1]$? $\endgroup$ – Jack D'Aurizio Oct 28 '18 at 15:03
  • $\begingroup$ HINT: you may use Leibnitz theorem to find the derivative and then equate it to 0. $\endgroup$ – 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 $y=1$. $$\boxed{I(1) = \int^{1}_{0}\sqrt{x^4}\,dx=\left[\frac{x^3}3\right]_0^1=\frac13}$$


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