The area bounded by the parabola \(y=x^2-3x\) with \(y \leq 0\) is

\(\frac{9}{2}\)

The correct answer is Option (4) → \(\frac{9}{2}\)

Points where $y=0$

$⇒x^2-3x=0$

$⇒x(x-3)=0$

$⇒x=0\,or\,x=3$

$A=\int\limits_0^3(-(x^2-3x))dx$ [Absolute value of y]

$=\int\limits_0^3(3x-x^2)dx$

$=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3$

$=\left[\frac{27}{2}-\frac{18}{2}\right]=\frac{9}{2}$ sq. units