The area enclosed between the curve y=loge(x+e) and the coordinate axes is:
Solutions · 1
Official solution
The graph of the curve \[y = {\log _e}\left( {x + e} \right)\] is as shown in the fig.
Required area
\[A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx\]
put \[x + e = t \Rightarrow dx = dt\]
also At x=1−e,t=1
At \[x = 0,\,\,t = e\]
\[\therefore\] \[A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]\]
e−e−0+1=1
Hence the required area is 1 square unit.