The area enclosed between the curve $y=x^2+2$ and $x$-axis between $x=0$ and $x=3$ is :

14 sq. units 15 sq. units 16 sq. units 18 sq. units

15 sq. units

$y = x^2 + 2$,  x axis    x = 0,  x = 3 So   $I =\int\limits_0^3 x^2+2 d x$ $=\left[\frac{x^3}{3}+2 x\right]_0^3$ $=\frac{3^3}{3}+2 \times 3-0-0$ = 9 + 6 = 15 sq. units