JEE Main · 2005 · 08 May 2005 · Offline · Area Under Curves

The curve $y=\ln (x+e)$ meets the $y$-axis at $(0,1)$, since $\ln e=1$, and the $x$-axis where $\ln (x+e)=0$, that is at $x=1-e$. The enclosed area is ${\int }_{1-e}^0\ln (x+e)dx$. An antiderivative, found by parts, is $(x+e)\ln (x+e)-x$; evaluating it at $0$ gives $e\ln e=e$, and at $1-e$ gives $1\cdot \ln 1-(1-e)=e-1$. The area is therefore $e-(e-1)=1$.