The area bounded by the curve $y=3 x-x^{2}$ and the $x$-axis is\

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Professor PC Thomas Classes · Answer

(e) : We have, $y=3 x-x^{2}$ Put $3 x-x^{2}=0$\ $\Rightarrow x(x-3)=0 \Rightarrow x=0$ and $x=3$\ $	herefore \quad$ Required area $=\int_{0}^{3} y d x$\ \ $=\int_{0}^{3}\left(3 x-x^{2}ight) d x=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}ight]_{0}^{3}$\ $=\left[\frac{27}{2}-9ight]=\frac{9}{2}$ sq.units\