A thin bar of length L has a mass per unit length $$\lambda$$, that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is $$\lambda_0$$, then the distance of the centre of mass from the lighter end is:

$$\lambda(x) = \lambda_0 + kx$$

$$M = \int_0^L \lambda(x) dx = \int_0^L (\lambda_0 + kx) dx = \left[ \lambda_0 x + \frac{kx^2}{2} \right]_0^L$$

$$M = \lambda_0 L + \frac{kL^2}{2} \implies kL^2 = 2(M - \lambda_0 L) \quad \dots (1)$$

$$x_{cm} = \frac{1}{M} \int_0^L x \lambda(x) dx$$

$$x_{cm} = \frac{1}{M} \int_0^L (x\lambda_0 + kx^2) dx = \frac{1}{M} \left[ \frac{\lambda_0 x^2}{2} + \frac{kx^3}{3} \right]_0^L$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{k L^3}{3} \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (kL^2) \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (2M - 2\lambda_0 L) \right]$$

$$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{2ML}{3} - \frac{2\lambda_0 L^2}{3} \right]$$

$$x_{cm} = \frac{2L}{3} + \frac{\lambda_0 L^2}{M} \left( \frac{1}{2} - \frac{2}{3} \right)$$

$$x_{cm} = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$$