Question:

Given, <br> $f(x) = \begin{cases} -x-1, & x < -1 \\ 1+x, & -1 \le x < 0 \\ 1-x, & 0 \le x \le 1 \\ x-1, & x > 1 \end{cases}$ ; <br> $f(x-1) = \begin{cases} -x, & x-1 < -1 & \Rightarrow x < 0 \\ x, & -1 \le x-1 < 0 & \Rightarrow 0 \le x < 1 \\ 2-x, & 0 \le x-1 \le 1 & \Rightarrow 1 \le x \le 2 \\ x-2, & x-1 > 1 & \Rightarrow x > 2 \end{cases}$ <br> Similarly <br> $f(x+1) = \begin{cases} -x-2, & x+1 < -1 & \Rightarrow x < -2 \\ x+2, & -1 \le x+1 < 0 & \Rightarrow -2 \le x < -1 \\ -x, & 0 \le x+1 \le 1 & \Rightarrow -1 \le x \le 0 \\ x, & x+1 > 1 & \Rightarrow x > 0 \end{cases}$ <br> $\Rightarrow g(x) = f(x-1) + f(x+1) = \begin{cases} -2x-2 & x < -2 \\ 2, & -2 \le x < -1 \\ -2x, & -1 \le x \le 0 \\ 2x, & 0 < x < 1 \\ 2, & 1 < x \le 2 \\ 2x-2, & 2 < x \end{cases}$ <br> Clearly $g(x)$ is even, <br> Now $\int_{-3}^{5} g(x) dx = 2 \int_{0}^{3} g(x) dx + \int_{3}^{5} g(x) dx$ <br> $= 2 \left( \int_{0}^{1} 2x dx + \int_{1}^{2} 2 dx + \int_{2}^{3} (2x-2) dx \right) + \int_{3}^{5} (2x-2) dx = 24$