∫_-2^1|x^2-1|dx equals:

The correct answer is Option (2) → 8/3

Inside the limits, \( x^2-1 \) changes sign at x = −1 (x = 1 is the upper limit), so the integral is split at x = −1:

    \( \displaystyle -2\le x\le -1:\ |x^2-1|=x^2-1 \)
    \( \displaystyle -1\le x\le 1:\ |x^2-1|=1-x^2 \)

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

    \( \displaystyle =\left(-\frac{1}{3}+1\right)-\left(-\frac{8}{3}+2\right) \)

    \( \displaystyle =\frac{2}{3}+\frac{2}{3} \)

    \( \displaystyle =\frac{4}{3} \)

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

    \( \displaystyle =\frac{2}{3}-\left(-\frac{2}{3}\right) \)

    \( \displaystyle =\frac{4}{3} \)

Adding the two parts:

    \( \displaystyle \int_{-2}^{1}|x^2-1|\,dx=\frac{4}{3}+\frac{4}{3}=\frac{8}{3} \)