A) \[\frac{7}{6}sq.\,\,\text{units}\]
B) \[\frac{11}{3}sq.\,\,\text{units}\]
C) \[\frac{19}{3}sq.\,\,\text{units}\]
D) \[14\,\,sq.\,\,\text{units}\]
Correct Answer: A
Solution :
\[\text{Area}=\frac{1}{2}+\int\limits_{3}^{4}{\sqrt{4-x}\,dx}\]\[=\frac{1}{2}+\left( -\frac{{{(4-x)}^{3/2}}}{-\frac{3}{2}} \right)_{3}^{4}=\frac{1}{2}-\frac{2}{3}\,(0-1)=\frac{7}{6}\] |
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