Question: The area of the region
$A=[(x, y): 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1]$
in sq. units, is :
$\frac{2}{3}$
$\frac{1}{3}$
2
$\frac{4}{3}$
Correct Option: , 3
Solution:
The graph is a follows
$\int_{-1}^{0}\left(-x^{2}+1\right) d x+\int_{0}^{1}\left(x^{2}+1\right) d x=2$
