The area (in sq. units) of the region bounded by the curve \(y = 9-x^{2}\) and the x-axis is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 36
The parabola meets the x-axis where \(9-x^{2} = 0\), that is at x = −3 and x = 3, and lies above the x-axis in between:
\( \displaystyle A = \int_{-3}^{3}(9-x^{2})\,dx \)
\( \displaystyle = \left[9x-\frac{x^{3}}{3}\right]_{-3}^{3} \)
\( = (27-9)-(-27+9) \)
\( = 36 \)
Shortcut: The area cut off from a parabola by a chord perpendicular to its axis is two-thirds of the enclosing rectangle. Here the rectangle is 6 wide (x from −3 to 3) and 9 high, so \(A = \frac{2}{3}\times54 = 36\).