The integrand is $f(x) = 1 - |x|$. We observe that $f(-x) = 1 - |-x| = 1 - |x| = f(x)$. Since the function is even, the integral over the symmetric interval $[-1, 1]$ can be simplified as: $$ \int_{-1}^{1} (1 - |x|) dx = 2 \int_{0}^{1} (1 - |x|) dx $$
Alternatively, consider the symmetry of the function $f(x) = 1 - |x|$. The graph is symmetric about the y-axis. The area under the curve from $-1$ to $1$ is twice the area from $-1$ to $0$ or twice the area from $0$ to $1$. In the interval $[-1, 0]$, $|x| = -x$. Therefore, the integrand becomes $1 - (-x) = 1 + x$. Thus, the integral can be written as: $$ \int_{-1}^{1} (1 - |x|) dx = 2 \int_{-1}^{0} (1 + x) dx $$
Comparing our result $2 \int_{-1}^{0} (1 + x) dx$ with the given options, we find that it matches option (C).
Final Answer: C
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