The value of ∫_-1^1(x+|x|)^2 dx is

The correct answer is Option (1) → \(\displaystyle\frac{4}{3}\)

Expanding, and using \(|x|^{2} = x^{2}\),

    \((x+|x|)^{2} = x^{2} + 2x|x| + |x|^{2} = 2x^{2} + 2x|x|\)

Here \(2x^{2}\) is an even function and \(2x|x|\) is an odd function, so the odd part contributes 0 on \([-1, 1]\):

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

    \( \displaystyle = 2\cdot 2\int_{0}^{1}x^{2}\,dx = 4\cdot\frac{1}{3} = \frac{4}{3} \)

The same value follows directly: \(x+|x| = 0\) for \(x \lt 0\) and \(= 2x\) for \(x \ge 0\), so the integral is \(\int_{0}^{1}4x^{2}\,dx = \frac{4}{3}\).

Why the other options are incorrect

• Option (2) \(\displaystyle\frac{2}{3}\) — Incorrect. It integrates \(2x^{2}\) over \([0, 1]\) only, forgetting the doubling for an even function.

• Option (3) 0 — Incorrect. The integrand is a square, never negative, and it is positive for \(x \gt 0\), so the integral cannot be 0.

• Option (4) \(\displaystyle\frac{8}{3}\) — Incorrect. It treats \(2x|x|\) as \(2x^{2}\), which is wrong for negative x.