JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 8)
Area of the region bounded by the curve $$y = {e^x}$$ and lines $$x=0$$ and $$y=e$$ is
$$e-1$$
$$\int\limits_1^e {\ln \left( {e + 1 - y} \right)dy} $$
$$e - \int\limits_0^1 {{e^x}dx} $$
$$\int\limits_1^e {\ln y\,dy} $$
Explanation
The required area is obtained as follows:
$$\int\limits_1^e {\ln y\,dy = (y\ln y - y)_1^e = (e - e) - \{ - 1\} = 1} $$.
Also,
$$\int\limits_1^e {\ln y\,dy = \int\limits_1^e {\ln (e + 1 - y)dy} } $$
Further, the area bounded by the region is
$$ = e \times 1 - \int\limits_0^e {{e^x}dx} $$
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