∫_-1^1 ≤ft(|x - 2| + |x|)dx =

The correct answer is Option (2) → 5

For −1 ≤ x ≤ 1, x − 2 is negative, so the modulus opens as |x − 2| = 2 − x:

    \( \displaystyle \int_{-1}^{1} |x - 2|\,dx = \int_{-1}^{1} (2 - x)\,dx \)

    \( \displaystyle = \left[2x - \frac{x^2}{2}\right]_{-1}^{1} \)

    \( \displaystyle = \left(2 - \frac{1}{2}\right) - \left(-2 - \frac{1}{2}\right) = 4 \)

|x| is an even function, and |x| = x for x ≥ 0:

    \( \displaystyle \int_{-1}^{1} |x|\,dx = 2\int_{0}^{1} x\,dx \)

    \( \displaystyle = 2\left[\frac{x^2}{2}\right]_{0}^{1} = 1 \)

Adding the two parts:

    \( \displaystyle \int_{-1}^{1}\left(|x - 2| + |x|\right)dx = 4 + 1 = 5 \)