1. The problem is to evaluate the integral $$\int 30|x-1| \, dx$$. 2. Recall that the absolute value function $|x-1|$ can be expressed as a piecewise function: $$|x-1| = \begin{cases} x-1 & \text{if } x \geq 1 \\ -(x-1) & \text{if } x < 1 \end{cases}$$ 3. Therefore, the integral splits into two cases depending on the domain of $x$: - For $x \geq 1$, $$\int 30|x-1| \, dx = \int 30(x-1) \, dx$$ - For $x < 1$, $$\int 30|x-1| \, dx = \int 30(-(x-1)) \, dx = \int 30(1-x) \, dx$$ 4. Compute the integral for $x \geq 1$: $$\int 30(x-1) \, dx = 30 \int (x-1) \, dx = 30 \left( \frac{x^2}{2} - x \right) + C = 15x^2 - 30x + C$$ 5. Compute the integral for $x < 1$: $$\int 30(1-x) \, dx = 30 \int (1-x) \, dx = 30 \left( x - \frac{x^2}{2} \right) + C = 30x - 15x^2 + C$$ 6. Thus, the final answer is the piecewise function: $$\int 30|x-1| \, dx = \begin{cases} 15x^2 - 30x + C & \text{if } x \geq 1 \\ 30x - 15x^2 + C & \text{if } x < 1 \end{cases}$$ This represents the antiderivative of the given function.