The area of region bounded by the curves $y=x^2+2$, y = -x, x = 0 and x = 1 is

$\frac{17}{6}$

The correct answer is Option (3) - $\frac{17}{6}$

$y=x^2+2, y = -x$,

$x = 0$ and $x = 1$

area = $\int\limits_0^1x^2+2dx+\frac{1}{2}×1×1$

$=\left[\frac{x^3}{3}+2x\right]_0^1+\frac{1}{2}$

$=\frac{1}{3}+2+\frac{1}{2}$

$=\frac{12+2+3}{6}$

$=\frac{17}{6}$ sq. units