question_answer

The area enclosed between the curves \[y={{\log }_{e}}(x+e),x={{\log }_{e}}\left( \frac{1}{y} \right)\] and the x-axis is

A) 2 sq. units

B) 1 sq. units

C) 4 sq. units

D) None of these

Correct Answer: A

Solution :

[a] \[y={{\log }_{e}}(x+e),x={{\log }_{e}}\left( \frac{1}{y} \right)\] or \[y={{e}^{-x}}\]. For \[y={{\log }_{e}}(x+e),\] shift the graph of \[y={{\log }_{e}}x,e\] units to the hand side. Required area \[=\int\limits_{1-e}^{0}{{{\log }_{e}}(x+e)dx+\int\limits_{0}^{\infty }{{{e}^{-x}}dx}}\] \[=\left| x{{\log }_{e}}(x+e) \right|_{1-e}^{0}-\int\limits_{1-e}^{0}{\frac{x}{x+e}dx-\left| {{e}^{-x}} \right|_{0}^{\infty }}\]                     \[=\int\limits_{0}^{1-e}{\left( 1-\frac{e}{x+e} \right)dx-{{e}^{-\infty }}+{{e}^{0}}}\] \[=\left| x-e\log (x+e) \right|_{0}^{1-e}-0+1\] \[=1-e+e\log e+1=2\] sq. units.