A) 2 sq. unit
B) 4 sq. unit
C) 6 sq. unit
D) 8 sq. unit
Correct Answer: C
Solution :
[c] Equation of given curve is \[{{y}^{2}}=12x\] At \[y=6,36=12x\Rightarrow x=3\] \[\therefore \] Required area \[=\int_{0}^{3}{({{y}_{1}}-{{y}_{2}})}dx\] where \[{{y}_{1}}\] represents line and \[{{y}_{2}}\] represents the curve. \[=\int_{0}^{3}{\left( 6-\sqrt{12x} \right)dx=\left[ 6x \right]_{0}^{3}-\sqrt{12}\left[ \frac{2{{x}^{3/2}}}{3} \right]_{0}^{3}}\] \[=\left[ 6\times 3 \right]-\frac{\sqrt{12}\times 2\times \sqrt{27}}{3}=18-12=6\] sq. unit
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