I'd like help with this exercise. (:
I have done a, but in b I'm stuck when I need to find the integral of (1/x)^2.
You could be failed for the entire semester for that!
$\displaystyle \int \left(\frac{1}{x}\right)^{2} dx = \int \frac{1}{x^{2}}\;dx = \int x^{-2}\;dx = -x^{-1} = C = -\frac{1}{x} + C$
I hope you are kicking yourself. Do NOT stumble on that one on an exam.
Please don't forget to do it both ways, at least for the practice.
$\displaystyle 2\pi \cdot \left(\int_{1/2}^{1} y \cdot (2-(1/y))\;dy + \int_{1}^{2} y \cdot (2-y)\;dy\right)$