The area bounded by the x-axis and the parabola \(y = 3x-x^2\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{9}{2}\) Sq.unit
The parabola meets the x-axis where
\(3x - x^2 = x(3 - x) = 0\), so \(x = 0\) or \(x = 3\)
Between these points of intersection the curve lies above the x-axis, so the area is
\( \displaystyle \int_{0}^{3} (3x - x^2)\,dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_{0}^{3} \)
\( \displaystyle = \frac{27}{2} - 9 = \frac{9}{2} \)
So the area is \(\frac{9}{2}\) square units.