A) 2/3 sq. unit
B) 4/3 sq. unit
C) 5/3 sq. unit
D) 4/5 sq. unit
Correct Answer: B
Solution :
[b] Given curve is \[y=4x-{{x}^{2}}-3\] Since, area bounded by x-axis \[\therefore y=0\] \[\Rightarrow 4x-{{x}^{2}}-3=0\Rightarrow {{x}^{2}}-4x+3=0\] \[\Rightarrow {{x}^{2}}-3x-x+3=0\Rightarrow (x-3)(x-1)=0\] \[\Rightarrow x=1,3\] \[\therefore \] Required area \[=\int_{1}^{3}{(4x-{{x}^{2}}-3)dx}\] \[\left. =\frac{4{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-3x \right|_{1}^{3}=\left( \frac{36}{2}-\frac{27}{3}-9 \right)\] \[-\left( \frac{4}{2}-\frac{1}{3}-3 \right)\] \[=(18-9-9)-\left( 2-\frac{10}{3} \right)=0-\left( \frac{-4}{3} \right)=\frac{4}{3}\] sq. unit.
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