Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is
- $e-1$
- $\int_1^e \ln (e+1-y) d y$
- $e-\int_0^1 e^x d x$
- $\int_1^e \ln y d y$
Solution

$ \begin{aligned} & =\int_e^q \ln t(-d t) \\ & \text { [put } e+1-y=t \Rightarrow-d y=d t \text { ] } \\ & =\int_1^e \ln t d t=\int_1^e \ln y d y=1 \end{aligned} $
Asked in: JEE Advanced 2009 (Paper 1)