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: |
$\frac{16}{3}$ sq.units 15 sq.units $\frac{21}{2}$ sq.units 12 sq.units |
15 sq.units |
The correct answer is Option (2) → 15 sq.units Given curves: $y = x^2 + 2$ and x-axis, between $x=0$ and $x=3$ Area = $\int_{0}^{3} (x^2 + 2) \, dx$ $\int (x^2 + 2) dx = \int x^2 dx + \int 2 dx = \frac{x^3}{3} + 2x$ Evaluate from 0 to 3: $\left[ \frac{3^3}{3} + 2*3 \right] - \left[ \frac{0^3}{3} + 2*0 \right] = (9 + 6) - 0 = 15$ Area of the region = 15 square units |