A rod of length L = 1.5 m has a mass per unit length λ = λ₀(1 + 2x/L), where x is the distance from…

Step 1. Mass. $\int_0^L \left(1 + \dfrac{2x}{L}\right)dx = L + L = 2L$ (times $\lambda_0$).

Step 2. Moment. $\int_0^L x\left(1 + \dfrac{2x}{L}\right)dx = \dfrac{L^2}{2} + \dfrac{2L^2}{3} = \dfrac{7L^2}{6}$.

Step 3. Divide. $x_{cm} = \dfrac{7L^2/6}{2L} = \dfrac{7L}{12} = \dfrac{7 \times 1.5\ \text{m}}{12} = 0.875\ \text{m}$.