I'm pretty confused by how we should determine Integration limits.
$f_p(x)=x^3-px^2$
$V$ is the surface area enclosed by the graph of $f_2$ and the x-axis.
$V$ is revolved around the x-axis. Analytically calculate the volume of the resulting solid of revolution
I know the formula for figuring out the volume of revolutions around the $x$-achse is to just integrate $\pi(f(x)^2$, but I would still need to figure out the integration limits and I'm really lost on how to do that.
Thanks in advance for any help!
To find the integration limits, you must find the places where $f_2 (x)$ and the $x$-axis cross. You do this by setting $$x^3 - 2x^2 = 0$$ $$x^2 (x-2) = 0$$ $$x = 0, 2$$ Because $f_2 (x) \le 0$ for $x \in [0, 2]$, $V$ would be equal to the opposite of the integral of $f_2 (x)$. $$V = -\int_{0}^{2} f_2 (x)~ {\rm d}x = -\int_{0}^{2} \left(x^3 - 2x^2\right) {\rm d}x$$ I hope this is of help!