A) 1 sq. unit
B) 2 sq. unit
C) 4 sq. unit
D) None of these
Correct Answer: C
Solution :
[c] Clearly, the curve \[y=x{{(3-x)}^{2}}\] has maximum at \[x=1\] and minimum at \[x=3\]. \[\therefore \] Req. area \[=\int_{1}^{3}{x{{(3-x)}^{2}}dx}\] \[=\int_{1}^{3}{({{x}^{3}}-6{{x}^{2}}+9x)dx}\] \[=\left[ \frac{{{x}^{4}}}{4}-2{{x}^{3}}+\frac{9{{x}^{2}}}{2} \right]_{1}^{3}=4\] sq. unit.You need to login to perform this action.
You will be redirected in
3 sec