• # question_answer The area bounded by the curve $y=x{{(3-x)}^{2}},$ the x-axis and the ordinates of the maximum and minimum points of the curve, is given by A) 1 sq. unit B) 2 sq. unit C) 4 sq. unit           D) none of these

 clearly, the curve $y=x{{(3-x)}^{2}}$ has maximum at $x=1$ and minimum at $x=3.$ $\therefore$Req. area $=\int_{1}^{3}{x{{(3-x)}^{2}}dx}$ $=\int_{1}^{3}{({{x}^{3}}-6{{x}^{2}}+9x)dx}$ $=\left[ \frac{{{x}^{4}}}{4}-2{{x}^{3}}+\frac{9{{x}^{2}}}{2} \right]_{1}^{3}=4sq\,unit.$