A) 1 sq. unit
B) 2 sq. unit
C) 4 sq. unit
D) none of these
Correct Answer: C
Solution :
clearly, the curve \[y=x{{(3-x)}^{2}}\] has maximum at \[x=1\] and minimum at \[x=3.\] |
\[\therefore \]Req. area \[=\int_{1}^{3}{x{{(3-x)}^{2}}dx}\] |
\[=\int_{1}^{3}{({{x}^{3}}-6{{x}^{2}}+9x)dx}\] |
\[=\left[ \frac{{{x}^{4}}}{4}-2{{x}^{3}}+\frac{9{{x}^{2}}}{2} \right]_{1}^{3}=4sq\,unit.\] |
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