A) \[4-{{\log }_{e}}2\]
B) \[\frac{1}{4}+{{\log }_{e}}2\]
C) \[3-{{\log }_{e}}2\]
D) \[\frac{15}{4}-{{\log }_{e}}2\]
Correct Answer: D
Solution :
[d]: The point of intersection of the curves\[y={{x}^{3}}\]and\[y=1/x\]is (1,1) \[\therefore \]The required area \[=\int\limits_{1}^{2}{\left( {{x}^{3}}-\frac{1}{x} \right)dx}\] \[=\left[ \frac{{{x}^{4}}}{4}-\log x \right]_{1}^{2}=4-{{\log }_{e}}2-\frac{1}{4}\] \[=\left( \frac{15}{4}-{{\log }_{e}}2 \right)\text{sq}\text{.}\,\text{units}\]You need to login to perform this action.
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