The area of the region bounded by the curves \(y = x^2 + 2\) and x-axis, between \(x = 0\) and \(x = 3\) in the first quadrant is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15 sq.units
The curve \(y = x^2 + 2\) lies above the x-axis, so the area under the curve is
\( \displaystyle \text{Area} = \int_{0}^{3} (x^2 + 2)\,dx \)
\( \displaystyle = \left[\frac{x^3}{3} + 2x\right]_{0}^{3} \)
\(= (9 + 6) - 0 = 15\)
So the area is 15 square units.