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
The correct answer is Option (2) → 15 sq. units
Since \(x^2 + 2 \gt 0\) for every x, the curve lies above the x-axis, so the area is the definite integral:
\( \displaystyle \text{Area} = \int_{0}^{3}\left(x^2+2\right)dx \)
\( \displaystyle = \left[\frac{x^3}{3}+2x\right]_{0}^{3} \)
\( = 9 + 6 \)
\( = 15 \text{ sq. units} \)