The area enclosed between the curve $y=x^2+2$ and $x$-axis between $x=0$ and $x=3$ is :
Answer & explanation
Correct answer: option 2
$y = x^2 + 2$, 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