My task is to Find the volume of the largest cone can be inscribed in a sphere of radius R. But the problem is that I have to use the formula for the integral, which is: $$V (f, a, b) = \pi \int_{a}^{b}f^{2}(x)dx.$$ I managed to solve a problem without the integral, but it is explicitly written that the problem must be solved with the integral.
My solution is: $$V cone = \frac{1}{3}\pi R^{2}h=\frac{1}{3}\pi(\sqrt{r^{2}-x^{2}})^{2}(r+x)=\frac{1}{3}\pi({r^{2}-x^{2}})(r+x)=\frac{1}{3}\pi(r^{3}+r^{2}x-rx^{2}-x^{3})$$ $$\frac{dv}{dx}=\frac{1}{3}\pi(0+r^{2}-2r-2x^{2})$$ . $$\frac{1}{3}\pi(r^{2}-2r-2x^{2})=0$$ $$(r+x)(r-3x)=0$$ $r=-x$ and $r =\frac{1}{3}x$
$$Vcone=\pi \frac{1}{3}\left(r^{2}-\frac{1}{9}r^{2}\right)\left(r+\frac{1}{3}r\right)=\pi\left(\frac{8}{9}r^{2}\right)\left(\frac{4}{3}r\right)=\frac{32\pi}{81}r^{3}$$
If the sphere was centered at the origin and the cone aligned with the horizontal axis beginning at the circle's leftmost point, then as you found, $x=\dfrac{1}{3}R$. We then integrate the square of $f(x)=\dfrac{\sqrt{2}}{2}(x+R)$ (the equation of the line that governs the radius as $x$ varies) over $[-R, R/3]$
$$V = \pi\int_{-R}^{R/3} \left[\dfrac{\sqrt{2}}{2}(x+R)\right]^2 dx$$
The slope of the line I found with $r/(x+R) = \dfrac{\sqrt{R^2-x^2}}{x+R}=\dfrac{\sqrt{8}/3R}{4/3R} = \dfrac{\sqrt{2}}{2}$.
See this link for a visual: https://www.desmos.com/calculator/cajasutkfb