Answer
(a)
\begin{align}
\int_{-2}^{0}f(x)dx = 2
\end{align}
(b)
\begin{align}
\int_{-2}^{2}f(x)dx = 4
\end{align}
(c)
\begin{align}
\int_{0}^{6}f(x)dx = 10
\end{align}
(d)
\begin{align}
\int_{-4}^{6}f(x)dx = 10
\end{align}
Work Step by Step
The given piecewise function is
\begin{align}
f(x) = \begin{cases}
\mid x-2\mid, &\quad x\geq0 \\
x+2, &\quad x\lt0 \\
\end{cases}
\end{align}
(a)
\begin{align}
\int_{-2}^{0}f(x)dx = \int_{-2}^{0} (x+2) dx = \big[\frac{x^{2}}{2} + 2x \big]_{-2}^{0} = 2
\end{align}
(b)
\begin{align}
\int_{-2}^{2}f(x)dx = \int_{-2}^{0} (x+2) dx + \int_{0}^{2} \mid x-2 \mid dx = 2 + \int_{0}^{2} (-x+2) dx = 2 + \big[\frac{-x^{2}}{2} + 2x \big]_{0}^{2} = 2 + 2 = 4
\end{align}
(c)
\begin{align}
\int_{0}^{6}f(x)dx = \int_{0}^{6} \mid x-2 \mid dx = \int_{0}^{2} (-x+2) dx + \int_{2}^{6} (x-2) dx = 2 + \big[\frac{x^{2}}{2} - 2x \big]_{2}^{6} = 2 + 8 = 10
\end{align}
(d)
\begin{align}
\int_{-4}^{6}f(x)dx = \int_{-4}^{0} (x+2) dx + \int_{0}^{6} \mid x-2 \mid dx = \big[\frac{x^{2}}{2} + 2x \big]_{-4}^{0} + 10 = 0 + 10 = 10
\end{align}
