JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Area Under Curves

Split the curve $y=x|x|+1$ by the sign of $x$: for $x\ge 0$ it is $y=x^2+1$, and for $x<0$ it is $y=-x^2+1$, both staying above the $x$-axis on $[-1,1]$. The area is ${\int }_{-1}^0(-x^2+1)dx+{\int }_0^1(x^2+1)dx$. The first integral is ${\left[-\frac{x^3}{3}+x\right]}_{-1}^0=\frac{2}{3}$ and the second is ${\left[\frac{x^3}{3}+x\right]}_0^1=\frac{4}{3}$, so the total area is $\frac{2}{3}+\frac{4}{3}=2$ square units.