Consider a conductor of length l with a circular cross section, as...

Consider an elemental disc of thickness $dx$ at a distance $x$ from the end with radius $a$.

The radius of this disc is given by $r = a + \dfrac{b-a}{l}x$.

The resistance of this elemental disc is $dR = \dfrac{\rho dx}{\pi r^2}$.

The total resistance of the conductor is obtained by integrating $dR$ from $x = 0$ to $x = l$:

$R = \int_{0}^{l} \dfrac{\rho dx}{\pi \left(a + \dfrac{b-a}{l}x\right)^2}$

Let $y = a + \dfrac{b-a}{l}x$. Then $dy = \dfrac{b-a}{l}dx$.

When $x = 0$, $y = a$, and when $x = l$, $y = b$.

Substituting these into the integral:

$R = \dfrac{\rho l}{\pi (b-a)} \int_{a}^{b} y^{-2} dy$

$R = \dfrac{\rho l}{\pi (b-a)} \left[ -\dfrac{1}{y} \right]_{a}^{b}$

$R = \dfrac{\rho l}{\pi (b-a)} \left( \dfrac{1}{a} - \dfrac{1}{b} \right)$

$R = \dfrac{\rho l}{\pi (b-a)} \left( \dfrac{b-a}{ab} \right)$

$R = \dfrac{\rho l}{\pi a b}$