Find the function $f(x)$ knowing that the volumen of the solid of revolution of $y=f(x)$ around the x axis, for $a \gt 1$ is $V=a^2 -a$ , $f(x)$ continuos and $ f(x) \gt 0$
Basically what I tried to do was use the mean value theorem for integrals by doing:
$$V=\pi\int_{0}^{a} (f(x))^2 dx = a^2 -a = a(a-1)$$
$$V=\int_{0}^{a} (f(x))^2 dx = \frac{a(a-1)}{\pi}$$
Then, by the Mean Value Theorem for Integral
$$\int_{0}^{a} (f(x))^2 dx = f(c)(a-0) = f(c)a = \frac{a(a-1)}{\pi}$$
and so we have that
$$(f(c))^2 = \frac{(a-1)}{\pi}$$ and
$$(f(c)) = \sqrt{\frac{(a-1)}{\pi}}$$
Is this the way to do it? Anyways, I'm stuck here. Any Hints? Thanks.
Differentiate both sides with respect to $a$ (using the FTC): $$\pi(f(a))^2=\frac{d}{da}(a^2-a)=2a-1$$ so, since $f(x)>0$ for all $x$, $$f(x)=\sqrt{\frac{2x-1}{\pi}}.$$