The area (in sq. units) of the region bounded by the curve y = |x^2-1…

The correct answer is Option (1) → 4

Since \(x^{2}-1 \lt 0\) for \(|x| \lt 1\), the modulus changes the formula at x = ±1:

    \(|x| \le 1:\ \ y = 1-x^{2}\)
    \(1 \le |x| \le 2:\ \ y = x^{2}-1\)

The curve is symmetric about the y-axis, so the area is twice the part from 0 to 2:

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

    \( \displaystyle \int_{1}^{2}(x^{2}-1)\,dx = \left[\frac{x^{3}}{3}-x\right]_{1}^{2} = \frac{2}{3}-\left(-\frac{2}{3}\right) = \frac{4}{3} \)

    \( \displaystyle A = 2\left(\frac{2}{3}+\frac{4}{3}\right) = 4 \)

Why the other options are incorrect

• Option (2) \(\displaystyle\frac{4}{3}\) — Incorrect. This is \(\int_{-2}^{2}(x^{2}-1)\,dx\), which leaves out the modulus, so the middle part is subtracted instead of added.

• Option (3) \(\displaystyle\frac{8}{3}\) — Incorrect. This is only the two outer parts, \(2\times\frac{4}{3}\).

• Option (4) 2 — Incorrect. This is only the half of the region to the right of the y-axis.