∫_0^2(|x|+|x-2|) dx =

The correct answer is Option (2) → 4

On [0, 2], the modulus signs are removed as follows:

    \(x \ge 0\), so \(|x| = x\)
    \(x - 2 \le 0\), so \(|x - 2| = 2 - x\)

So the integrand is constant on [0, 2]:

    \(|x| + |x - 2| = x + 2 - x = 2\)

    \( \displaystyle \int_{0}^{2} 2\,dx = 2\Big[x\Big]_{0}^{2} = 4 \)