The area bounded by the curve $y = x|x|$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ is

1. Here $y = -x^2$ for $x < 0$ and $y = x^2$ for $x > 0$, so the curve dips below the axis on $[-1, 0]$. 2. $\left|\int_{-1}^{0}(-x^2)\,dx\right| = \left|\left[-\frac{x^3}{3}\right]_{-1}^{0}\right| = \frac{1}{3}$. 3. $\int_{0}^{1}x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$. 4. Area $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$. 5. The integrand is odd, so $\int_{-1}^{1} = 0$; that cancellation gives the trap $0$, not the geometric area. _Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.6_