question_answer

The area bounded by the curve \[y=1+{{\log }_{e}}\,\,x,\] the x-axis and the straight line x = e is equal to (in square units)

A)  \[3e-2\]         

B)  \[e\]

C)  \[e-\frac{1}{e}\]             

D)  \[e+\frac{1}{e}\]

Correct Answer: B

Solution :

Given equation of curve, \[y=1+{{\log }_{e}}x\] ?.(i) and straight line \[x=e\] ?..(ii) On solving Eqs. (i) and (ii), we get \[y=1+{{\log }_{e}}e=1+1=2\] So, the intersection point of both curve is \[(e,2).\] Area of \[O'AB=\int_{1}^{e}{y\,\,dx}=\int_{1}^{e}{(1+{{\log }_{e}}x)dx}\] \[=\{x+x{{\log }_{e}}\,x-x\}_{1}^{e}\] \[=[x\,\,{{\log }_{e}}x]_{1}^{e}=e{{\log }_{e}}e-0\] \[=e\]